experimental molar mass meaning

3.1 Formula Mass and the Mole Concept

Learning objectives.

• Calculate formula masses for covalent and ionic compounds
• Define the amount unit mole and the related quantity Avogadro’s number Explain the relation between mass, moles, and numbers of atoms or molecules, and perform calculations deriving these quantities from one another

We can argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton’s atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances.

Formula Mass

In an earlier chapter, we described the development of the atomic mass unit, the concept of average atomic masses, and the use of chemical formulas to represent the elemental makeup of substances. These ideas can be extended to calculate the formula mass of a substance by summing the average atomic masses of all the atoms represented in the substance’s formula.

Formula Mass for Covalent Substances

For covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform (CHCl 3 ), a covalent compound once used as a surgical anesthetic and now primarily used in the production of tetrafluoroethylene, the building block for the "anti-stick" polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms. Figure 3.2 outlines the calculations used to derive the molecular mass of chloroform, which is 119.37 amu.

Likewise, the molecular mass of an aspirin molecule, C 9 H 8 O 4 , is the sum of the atomic masses of nine carbon atoms, eight hydrogen atoms, and four oxygen atoms, which amounts to 180.15 amu ( Figure 3.3 ).

Example 3.1

Computing molecular mass for a covalent compound, check your learning, formula mass for ionic compounds.

Ionic compounds are composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter. The formula mass for an ionic compound is calculated in the same way as the formula mass for covalent compounds: by summing the average atomic masses of all the atoms in the compound’s formula. Keep in mind, however, that the formula for an ionic compound does not represent the composition of a discrete molecule, so it may not correctly be referred to as the “molecular mass.”

As an example, consider sodium chloride, NaCl, the chemical name for common table salt. Sodium chloride is an ionic compound composed of sodium cations, Na + , and chloride anions, Cl − , combined in a 1:1 ratio. The formula mass for this compound is computed as 58.44 amu (see Figure 3.4 ).

Note that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing or additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses.

Example 3.2

Computing formula mass for an ionic compound.

The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, H 2 O, and hydrogen peroxide, H 2 O 2 , are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole , which remains indispensable in modern chemical science.

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (such as atoms, molecules, and ions) as the number of atoms in a sample of pure 12 C weighing exactly 12 g. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth.

The number of entities composing a mole has been experimentally determined to be 6.02214179 × × 10 23 , a fundamental constant named Avogadro’s number ( N A ) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being 6.022 × × 10 23 /mol.

Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (see Figure 3.5 ).

Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, 12 C, the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single 12 C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of 12 C contains 1 mole of 12 C atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, 12 C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu ( Figure 3.6 ).

While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water weighing about 0.03 g (see Figure 3.7 ). Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules.

Link to Learning

The mole is used in chemistry to represent 6.022 × × 10 23 of something, but it can be difficult to conceptualize such a large number. Watch this video and then complete the “Think” questions that follow. Explore more about the mole by reviewing the information under “Dig Deeper.”

The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.

Example 3.3

Deriving moles from grams for an element.

The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):

The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”

The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.

Example 3.4

Deriving grams from moles for an element.

In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):

The result is in agreement with our expectations, around 0.04 g Ar.

Example 3.5

Deriving number of atoms from mass for an element.

Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth N A , or approximately 10 22 Cu atoms. Carrying out the two-step computation yields:

The factor-label method yields the desired cancellation of units, and the computed result is on the order of 10 22 as expected.

4.586 × × 10 22 Au atoms

Example 3.6

Deriving moles from grams for a compound.

The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, C 2 H 5 O 2 N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:

The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass yields:

This result is consistent with our rough estimate.

Example 3.7

Deriving grams from moles for a compound.

The molar mass for this compound is computed to be 176.124 g/mol. The given number of moles is a very small fraction of a mole (~10 −4 or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:

This is consistent with the anticipated result.

Example 3.8

Deriving the number of atoms and molecules from the mass of a compound.

Given that saccharin has a molar mass of 183.18 g/mol, how many saccharin molecules are in a 40.0-mg (0.0400-g) sample of saccharin? How many carbon atoms are in the same sample?

Using the provided mass and molar mass for saccharin yields:

The compound’s formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided sample is:

9.545 × × 10 22 molecules C 4 H 10 ; 9.545 × × 10 23 atoms H

How Sciences Interconnect

Counting neurotransmitter molecules in the brain.

The brain is the control center of the central nervous system ( Figure 3.9 ). It sends and receives signals to and from muscles and other internal organs to monitor and control their functions; it processes stimuli detected by sensory organs to guide interactions with the external world; and it houses the complex physiological processes that give rise to our intellect and emotions. The broad field of neuroscience spans all aspects of the structure and function of the central nervous system, including research on the anatomy and physiology of the brain. Great progress has been made in brain research over the past few decades, and the BRAIN Initiative, a federal initiative announced in 2013, aims to accelerate and capitalize on these advances through the concerted efforts of various industrial, academic, and government agencies (more details available at www.whitehouse.gov/share/brain-initiative).

Specialized cells called neurons transmit information between different parts of the central nervous system by way of electrical and chemical signals. Chemical signaling occurs at the interface between different neurons when one of the cells releases molecules (called neurotransmitters) that diffuse across the small gap between the cells (called the synapse) and bind to the surface of the other cell. These neurotransmitter molecules are stored in small intracellular structures called vesicles that fuse to the cell wall and then break open to release their contents when the neuron is appropriately stimulated. This process is called exocytosis (see Figure 3.10 ). One neurotransmitter that has been very extensively studied is dopamine, C 8 H 11 NO 2 . Dopamine is involved in various neurological processes that impact a wide variety of human behaviors. Dysfunctions in the dopamine systems of the brain underlie serious neurological diseases such as Parkinson’s and schizophrenia.

One important aspect of the complex processes related to dopamine signaling is the number of neurotransmitter molecules released during exocytosis. Since this number is a central factor in determining neurological response (and subsequent human thought and action), it is important to know how this number changes with certain controlled stimulations, such as the administration of drugs. It is also important to understand the mechanism responsible for any changes in the number of neurotransmitter molecules released—for example, some dysfunction in exocytosis, a change in the number of vesicles in the neuron, or a change in the number of neurotransmitter molecules in each vesicle.

Significant progress has been made recently in directly measuring the number of dopamine molecules stored in individual vesicles and the amount actually released when the vesicle undergoes exocytosis. Using miniaturized probes that can selectively detect dopamine molecules in very small amounts, scientists have determined that the vesicles of a certain type of mouse brain neuron contain an average of 30,000 dopamine molecules per vesicle (about 5 × 10 −20 5 × 10 −20 mol or 50 zmol). Analysis of these neurons from mice subjected to various drug therapies shows significant changes in the average number of dopamine molecules contained in individual vesicles, increasing or decreasing by up to three-fold, depending on the specific drug used. These studies also indicate that not all of the dopamine in a given vesicle is released during exocytosis, suggesting that it may be possible to regulate the fraction released using pharmaceutical therapies. 1

• 1 Omiatek, Donna M., Amanda J. Bressler, Ann-Sofie Cans, Anne M. Andrews, Michael L. Heien, and Andrew G. Ewing. “The Real Catecholamine Content of Secretory Vesicles in the CNS Revealed by Electrochemical Cytometry.” Scientific Report 3 (2013): 1447, accessed January 14, 2015, doi:10.1038/srep01447.

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Molar Mass: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: May 10, 2023

As a fundamental concept in chemistry, the molar mass is essential for understanding the relationships between mass, moles, and Avogadro’s number. In the following post, we’ll cover the basics, including how to find molar mass, its relationship to Avogadro’s number and moles, and how to convert mass into moles.

What We Review

Avogadro’s Number and the Mole

Avogadro’s number is a fundamental constant that represents the number of particles (atoms, molecules, ions) in one mole of a substance. Specifically, the number is defined as 6.022 \times 10^{23} particles per mole, and it is named after the Italian scientist Amedeo Avogadro.

The Mole and its Relationship to Avogadro’s Number

A mole is a unit of measurement that is used to express the amount of a substance in chemistry. One mole of a substance contains Avogadro’s number of particles. For instance, one mole of carbon dioxide ( CO_2 ) contains 6.022 \times 10^{23} molecules of CO_2 .

Using Avogadro’s Number

You can see that by using Avogadro’s number, we can calculate the number of particles in a given number of moles of a substance. For example, let’s say we have 2 moles of carbon. We can use Avogadro’s number to determine the number of atoms in 2 moles of carbon:

Similarly, we can use Avogadro’s number to determine the number of molecules in a given number of moles of a substance. For example, let’s say we have 3 moles of water ( H_2O ). We can use Avogadro’s number to determine the number of water molecules in 3 moles of water:

In summary, by using Avogadro’s number, we can convert between the number of particles in a substance and the amount of the substance in moles. This allows us to make calculations that relate the mass, volume, and number of particles of a substance, which are important concepts in chemistry.

For a more in-depth view of Avogadro’s number and the mole, check out the following TED-Ed video:

Molar Mas s

Molar mass is the mass of one mole of a substance and is expressed in grams per mole ( \text{g/mol} ). Chemists denote molar mass with the symbol \text{M} . It is a useful quantity in chemistry because it allows us to relate the mass of a substance to the number of particles present in it.

How to Find Molar Mass Using the Periodic Table

The molar mass of an element is equal to its atomic mass in atomic mass units ( \text{amu} ) converted to grams per mole. For example, the molar mass of carbon is 12.01\text{ g/mol} , which is the atomic mass of carbon ( 12.01\text{ amu} ) converted to grams per mole.

How to Find Molar Mass of a Compound

To calculate the molar mass of a compound, we need to add up the molar masses of all the elements in the compound. For example, the molar mass of water ( H_2O ) can be calculated by adding up the molar masses of two hydrogen atoms and one oxygen atom. The atomic mass of hydrogen is 1.01\text{ amu} , and the atomic mass of oxygen is 16.00\text{ amu} . Therefore:

Practice Finding Molar Mass

Example 1: simple compound.

Let’s walk step-by-step for finding the molar mass of calcium chloride ( CaCl_2 ):

• First, write out the chemical formula for the compound. In this case, the formula for calcium chloride is CaCl_2 .
• Then, find the atomic mass of each element in the compound. You can find the atomic mass of each element on the periodic table. For calcium ( Ca ), the atomic mass is 40.08\text{ g/mol} . For chlorine ( Cl ), the atomic mass is 35.45\text{ g/mol} .
• Next, multiply the atomic mass of each element by the number of atoms of that element in the compound. In this case, there is one calcium atom ( Ca ) and two chlorine atoms ( Cl ) in calcium chloride. So, we multiply the atomic mass of calcium ( 40.08\text{ g/mol} ) by 1 , and the atomic mass of chlorine ( 35.45\text{ g/mol} ) by 2 :
• Calcium ( Ca ): 40.08\text{ g/mol} \times 1 = 40.08\text{ g/mol}
• Chlorine ( Cl ): 35.45\text{ g/mol} \times 2 = 70.90 \text{g/mol}
• Lastly, add the atomic masses of each element in the compound to get the molar mass of the compound. In this case, we add the atomic masses of calcium and chlorine to get the molar mass of calcium chloride:

So the molar mass of calcium chloride ( CaCl_2 ) is 110.98\text{ g/mol} .

Example 2: Complex Compound

For our second example, let’s find the molar mass of glucose, which has the chemical formula C_6H_{12}O_6 .

• First, write down the chemical formula of the compound.
• Then, look up the atomic masses of each element in the periodic table.
• Carbon (C) = 12.01\text{ g/mol}
• Hydrogen (H) = 1.01\text{ g/mol}
• Oxygen (O) = 16.00\text{ g/mol}
• Next. multiply the atomic mass of each element by the number of atoms of that element in the compound.
• 6\text{ carbon atoms} \times 12.01\text{ g/mol} = 72.06\text{ g/mol}
• 12\text{ hydrogen atoms} \times 1.01\text{ g/mol} = 12.12\text{ g/mol}
• 6\text{ oxygen atoms} \times 16.00\text{ g/mol} = 96.00\text{ g/mol}
• Lastly, add up the molar masses of each element in the compound:

Therefore, the molar mass of glucose ( C_6H_{12}O_6 ) is 180.18\text{ g/mol} .

Converting Between Mass and Moles

How to convert mass into moles.

Once we have determined the molar mass of a substance, we can use it to convert a given mass of the substance into the number of moles. This can be useful for determining the amount of a substance needed for a chemical reaction, or for calculating the mass of a product formed in a reaction.

To convert mass into moles, we use the following formula:

On the other hand, we can also find the mass for a given amount of moles using:

Let’s say we have 25\text{ grams} of sodium chloride ( NaCl ), and we want to convert it to moles. First, we’ll find the molar mass as follows:

• 1\text{ sodium atom} \times 22.99\text{ g/mol} = 22.99\text{ g/mol}
• 1\text{ chlorine atom} \times 35.45\text{ g/mol} = 35.45\text{ g/mol}

Adding these up gives us 58.44\text{ g/mol} .

Then we can use the formula to calculate the number of moles of sodium chloride:

Therefore, we have 0.428\text{ moles} of sodium chloride in 25\text{ grams} of sodium chloride.

In our next example, we will find the mass of 2.5\text{ moles} of calcium carbonate ( CaCO_3 ).

First, identify the molar mass of the compound.

• Calcium has a molar mass of 40.08\text{ g/mol}
• Carbon has a molar mass of 12.01\text{ g/mol}
• Oxygen has a molar mass of 16.00\text{ g/mol}

Then, use the formula for finding mass using moles:

Next, plug in the given values and solve for mass:

Therefore, the mass of 2.5\text{ moles} of calcium carbonate is 250.23\text{ g} .

Example 3: Finding the Mass of a Number of Molecules

For our last example, let’s pull all of these concepts together to determine the mass of a certain number of a given molecule.

Let’s say we have 2.5 \times 10^{24} molecules of glucose ( C_6H_{12}O_6) . We want to find the mass of this amount of glucose.

First, find the molar mass of glucose by adding up the atomic masses of all the atoms in the molecule:

Therefore the molar mass of glucose is 180.18\text{ g/mol} .

Then, use Avogadro’s number to find the number of moles of glucose:

Next, use the formula \text{mass} = \text{moles} \times\text{molar mass} to find the mass of 4.15\text{ moles} of glucose:

Therefore, the mass of 2.5 \times 10^{24}\text{ molecules} of glucose is 748.4\text{ g} .

In conclusion, understanding molar mass is an essential concept in chemistry as it allows us to relate the amount of a substance to the number of particles present in it. We learned about Avogadro’s number and how it relates to the mole, which is a unit used to express the amount of a substance. We also learned how to calculate the molar mass of a compound using the periodic table and how to convert mass into moles. The lessons learned in this post will continue through the study of chemical reactions and equations.

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7.4 Determining Empirical and Molecular Formulas

Learning objectives.

By the end of this section, you will be able to:

• Determine the empirical formula of a compound
• Determine the molecular formula of a compound

Determination of Empirical Formulas

As previously mentioned, the most common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, we must keep in mind that chemical formulas represent the relative numbers , not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. To accomplish this, we can use molar masses to convert the mass of each element to a number of moles. We then consider the moles of each element relative to each other, converting these numbers into a whole-number ratio that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:

[latex]1.17 \;\text{g C} \times \frac{1 \;\text{mol C}}{12.01 \;\text{g C}} = 0.142 \;\text{mol C}[/latex]

[latex]0.287 \;\text{g H} \times \frac{1 \;\text{mol H}}{1.008 \;\text{g H}} = 0.284 \;\text{mol H}[/latex]

Thus, we can accurately represent this compound with the formula C 0.142 H 0.248 . Of course, per accepted convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:

(Recall that subscripts of “1” are not written but rather assumed if no other number is present.)

The empirical formula for this compound is thus CH 2 . This may or not be the compound’s molecular formula as well; however, we would need additional information to make that determination (as discussed later in this section).

A molecular formula is the true formula for a compound.  It lists how many atoms of each element are in the compound.  The empirical formula is the simplest or most reduced ratio of elements in a compound. If a compound’s chemical formula cannot be reduced any further, then the empirical formula is the same as the molecular formula.  

Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:

[latex]\text{Cl}_{0.150}\text{O}_{0.525} \; = \; \text{Cl}_{\frac{0.150}{0.150}} \; \text{O}_{\frac{0.525}{0.150}} = \text{ClO}_{3.5}[/latex]

In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, we must multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl 2 O 7 as the final empirical formula.

In summary, empirical formulas are derived from experimentally measured element masses by:

• Deriving the number of moles of each element from its mass
• Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
• Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained

Figure 7.4a outlines this procedure in flow chart fashion for a substance containing elements A and X.

Example 7.4a

Determining a compound’s empirical formula from the masses of its elements.

A sample of the black mineral hematite (Figure 7.3b), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?

For this problem, we are given the mass in grams of each element. Begin by finding the moles of each:

Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:

The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe 1 O 1.5 ). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:

The empirical formula is Fe 2 O 3 .

Exercise 7.4a

What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?

Check Your Answer [1]

Watch Calculating Percent Composition and Empirical Formulas

Video source: Mr. Causey. (2013, March 22). Calculating percent composition and empirical formulas [Video]. YouTube.

Deriving Empirical Formulas from Percent Composition

Finally, with regard to deriving empirical formulas, consider instances in which a compound’s percent composition is available rather than the absolute masses of the compound’s constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.

Example 7.4b

Determining an empirical formula from percent composition.

The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O (Figure 7.3c). What is the empirical formula for this gas?

Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. The calculation is “most convenient” because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element’s mass percentage. This numerical equivalence results from the definition of the “percentage” unit, whose name is derived from the Latin phrase per centum meaning “by the hundred.” Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:

The molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:

Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO 2 .

Exercise 7.4b

What is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O?

Check Your Answer [2]

Derivation of molecular formulas.

Recall that empirical formulas are symbols representing the relative numbers of a compound’s elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.

Molecular formulas are derived by comparing the compound’s molecular or molar mass to its empirical formula mass . As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If we know the molecular (or molar) mass of the substance, we can divide this by the empirical formula mass in order to identify the number of empirical formula units per molecule, which we designate as n :

The molecular formula is then obtained by multiplying each subscript in the empirical formula by n , as shown by the generic empirical formula A x B y :

For example, consider a covalent compound whose empirical formula is determined to be CH 2 O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:

Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:

Note that this same approach may be used when the molar mass (g/mol) instead of the molecular mass (amu) is used. In this case, we are merely considering one mole of empirical formula units and molecules, as opposed to single units and molecules.

Example 7.4c

Determination of the molecular formula for nicotine.

Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?

Determining the molecular formula from the provided data will require comparison of the compound’s empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound’s empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:

Next, we calculate the molar ratios of these elements relative to the least abundant element, N.

The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C 5 H 7 N. The empirical formula mass for this compound is therefore 81.13 amu/formula unit, or 81.13 g/mol formula unit.

We calculate the molar mass for nicotine from the given mass and molar amount of compound:

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:

Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:

Exercise 7.4c

What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 amu?

Check Your Answer [3]

Glucose, an example of empirical vs molecular formulas.

Consider glucose, the sugar that circulates in our blood to provide fuel for the body and brain. Results from combustion analysis of glucose report that glucose contains 39.68% carbon and 6.58% hydrogen. Because combustion occurs in the presence of oxygen, it is impossible to directly determine the percentage of oxygen in a compound by using combustion analysis; other more complex methods are necessary. Assuming that the remaining percentage is due to oxygen, then glucose would contain 53.79% oxygen. A 100.0 g sample of glucose would therefore contain 39.68 g of carbon, 6.58 g of hydrogen, and 53.79 g of oxygen. To calculate the number of moles of each element in the 100.0 g sample, divide the mass of each element by its molar mass:

[latex]moles \, C = 39.68 \, g \, C \times {1 \, mol \, C \over 12.011 \, g \, C } = 3.304 \, mol \, C \label{3.3.4a}[/latex]

[latex]moles \, H = 6.58 \, g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H } = 6.53 \, mol \, H \label{3.3.4b}[/latex]

[latex]moles \, O = 53.79 \, g \, O \times {1 \, mol \, O \over 15.9994 \, g \, O } = 3.362 \, mol \, O \label{3.3.4c}[/latex]

Once again, the subscripts of the elements in the empirical formula are found by dividing the number of moles of each element by the number of moles of the element present in the smallest amount:

[latex]C: {3.304 \over 3.304} = 1.000 \, \, \, \, H: {6.53 \over 3.304} = 1.98 \, \, \, \, O: {3.362 \over 3.304} = 1.018[/latex]

The oxygen:carbon ratio is 1.018, or approximately 1, and the hydrogen:carbon ratio is approximately 2. The empirical formula of glucose is therefore CH 2 O, but what is its molecular formula?

Many known compounds have the empirical formula CH 2 O, including formaldehyde, which is used to preserve biological specimens and has properties that are very different from the sugar circulating in the blood. At this point, it cannot be known whether glucose is CH 2 O, C 2 H 4 O 2 , or any other (CH 2 O) n . However, the experimentally determined molar mass of glucose (180 g/mol) can be used to resolve this dilemma.

First, calculate the formula mass, the molar mass of the formula unit, which is the sum of the atomic masses of the elements in the empirical formula multiplied by their respective subscripts. For glucose,

[latex]\text {formula mass of}  CH_2O = \left [ 1 \, mol C \left ( {12.011 \, g \over 1 \, mol \, C} \right ) \right ] + \left [ 2 \, mol \, H \left ({1.0079 \, g \over 1 \, mol \, H }\right )\right ] + \left [ 1 \, mole \, O \left ( {15.5994 \, mol \, O \over 1 \, mol \, O} \right ) \right ] = 30.026 g \label{3.3.5}[/latex]

This is much smaller than the observed molar mass of 180 g/mol.

Second, determine the number of formula units per mole. For glucose, calculate the number of (CH 2 O) units—that is, the n in (CH 2 O) n —by dividing the molar mass of glucose by the formula mass of CH 2 O:

[latex]n={180 \, g \over 30.026\, g/CH_2O} = 5.99 \approx 6 CH_2O \, \text {formula units} \label{3.3.6}[/latex]

Each glucose contains six CH 2 O formula units, which gives a molecular formula for glucose of (CH 2 O) 6 , which is more commonly written as C 6 H 12 O 6 . The molecular structures of formaldehyde and glucose, both of which have the empirical formula CH 2 O, are shown in Figure 7.4d:

Scientists in Action: Ellen Henrietta Swallow Richards, PhD.

Ellen Henrietta Swallow Richards spent much of her career studying the quantity of substances within compounds whether it was in water, metals or food. She was able to determine the amount of nickel in various ores and the relative quantities of substances in food and water. Her work led to educating the population on nutrition from foods, clean water standards and how to chemically analyze ores.

Read more about Ellen’s history and contributions to chemistry provided by the American Society of Civil Engineers [New Tab] .

Key Equations

• [latex]\%\text{X} = \frac{\text{mass X}}{\text{mass commpound}} \times 100\%[/latex]
• [latex]\frac{\text{molecular or molar mass ( amu or} \;\frac{\text{g}}{\text{mol}})}{\text{empirical formula mass ( amu or} \;\frac{\text{g}}{\text{mol}})} = n \;\text{formula units/molecule}[/latex]
• (A x B y ) n = A nx B ny

Attribution & References

Except where otherwise noted, this page is adapted by Adrienne Richards and Samantha Sullivan Sauer from “ 6.2 Determining Empirical and Molecular Formulas ” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0 . ​Access for free at Chemistry (OpenStax) ​

• The Section: “Glucose, an example of empirical vs molecular formulas” is adapted from “ Empirical and Molecular Formulas ” In Chemistry 101A (LibreTexts) ,   licensed under CC BY-NC-SA 4.0.
• N 2 O 5 ↵
• CH 2 O ↵
• C 8 H 10 N 4 O 2 ↵

simplest or most reduced ratio of elements in a compound

true formula for a compound; lists how many atoms of each element are in the compound

sum of average atomic masses for all atoms represented in an empirical formula

7.4 Determining Empirical and Molecular Formulas Copyright © 2023 by Gregory Anderson; Caryn Fahey; Jackie MacDonald; Adrienne Richards; Samantha Sullivan Sauer; J.R. van Haarlem; and David Wegman is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Chapter 7. The Mole Concept

7.1 the mole concept, learning objectives.

By the end of this section, you will be able to:

• Define the amount unit mole and the related quantity Avogadro’s number
• Explain the relation between mass, moles, and numbers of atoms or molecules, and perform calculations deriving these quantities from one another

The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, H 2 O, and hydrogen peroxide, H 2 O 2 , are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole , which remains indispensable in modern chemical science.

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (atoms, molecules, ions, etc.) as the number of atoms in a sample of pure 12 C weighing exactly 12 g. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth.

The number of entities composing a mole has been experimentally determined to be [latex]6.02214179\times {10}^{23}[/latex], a fundamental constant named Avogadro’s number ( N A ) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being [latex]6.022\times {10}^{23}\text{/mol}[/latex].

Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (see Figure 1).

Figure 1. Each sample contains 6.022 × 10 23 atoms—1.00 mol of atoms. From left to right (top row): 65.4g zinc, 12.0g carbon, 24.3g magnesium, and 63.5g copper. From left to right (bottom row): 32.1g sulfur, 28.1g silicon, 207g lead, and 118.7g tin. (credit: modification of work by Mark Ott)

Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, 12 C, the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu . Per the amu definition, a single 12 C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of 12 C contains 1 mole of 12 C atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, 12 C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure 2).

Figure 2. Each sample contains 6.02 × 10 23 molecules or formula units—1.00 mol of the compound or element. Clock-wise from the upper left: 130.2g of C 8 H 17 OH (1-octanol, formula mass 130.2 amu), 454.9g of HgI 2 (mercury(II) iodide, formula mass 459.9 amu), 32.0g of CH 3 OH (methanol, formula mass 32.0 amu) and 256.5g of S 8 (sulfur, formula mass 256.6 amu). (credit: Sahar Atwa)

Figure 3. A single drop of water.

While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water weighing about 0.03 g (see Figure 3). The number of molecules in a single droplet of water is roughly 100 billion times greater than the number of people on earth.

Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules.

The mole is used in chemistry to represent [latex]6.022\times {10}^{23}[/latex] of something, but it can be difficult to conceptualize such a large number. Watch this video to learn more.

The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.

Example 1: Deriving Moles from Grams for an Element

According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?

The mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu, and so its molar mass is 39.10 g/mol. The given mass of K (4.7 g) is a bit more than one-tenth the molar mass (39.10 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol.

The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”

[latex]4.7\cancel{\text{ g K}}\left(\frac{\text{1 mol K}}{\text{39.10 }\cancel{\text{g K}}}\right)=0.12\text{ mol K}[/latex]

The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.

Check Your Learning

Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g?

Example 2: Deriving Grams from Moles for an Element

A liter of air contains [latex]9.2\times {10}^{-4}[/latex] mol argon. What is the mass of Ar in a liter of air?

In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):

[latex]9.2\times {10}^{-4}\cancel{\text{mol Ar}}\left(\frac{\text{39.95 }\text{g Ar}}{\text{1 }\cancel{\text{mol Ar}}}\right)=\text{0.037 }\text{g Ar}[/latex]

The result is in agreement with our expectations as noted above, around 0.04 g Ar.

What is the mass of 2.561 mol of gold?

Example 3: Deriving Number of Atoms from Mass for an Element

Copper is commonly used to fabricate electrical wire (Figure 7). How many copper atoms are in 5.00 g of copper wire?

Figure 7. Copper wire is composed of many, many atoms of Cu. (credit: Emilian Robert Vicol)

Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth N A , or approximately 10 22 Cu atoms. Carrying out the two-step computation yields:

[latex]\text{5.00 }\cancel{\text{g Cu}}\left(\frac{\text{1 }\cancel{\text{mol Cu}}}{\text{63.55 }\cancel{\text{g Cu}}}\right)\left(\frac{\text{6.022}\times{10}^{23}\text{ }\text{atoms Cu}}{\text{1 }\cancel{\text{mol Cu}}}\right)=4.74\times {10}^{22}\text{ atoms Cu}[/latex]

The factor-label method yields the desired cancellation of units, and the computed result is on the order of 10 22 as expected.

A prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?

Example 4: Deriving Moles from Grams for a Compound

Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula C 2 H 5 O 2 N. How many moles of glycine molecules are contained in 28.35 g of glycine?

The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass yields:

[latex]\text{28.35 }\cancel{\text{g }\text{C}_{2}\text{H}_{5}\text{O}_{2}\text{N}}\left(\frac{\text{1 }\text{mol }\text{C}_{2}\text{H}_{5}\text{O}_{2}\text{N}}{\text{75.07 }\cancel{\text{g }\text{C}_{2}\text{H}_{5}\text{O}_{2}\text{N}}}\right)=0.378\text{ mol }\text{C}_{2}\text{H}_{5}\text{O}_{2}\text{N}[/latex]

This result is consistent with our rough estimate.

How many moles of sucrose, C 12 H 22 O 11 , are in a 25-g sample of sucrose?

Example 5: Deriving Grams from Moles for a Compound

Vitamin C is a covalent compound with the molecular formula C 6 H 8 O 6 . The recommended daily dietary allowance of vitamin C for children aged 4–8 years is [latex]1.42\times {10}^{-4}\text{mol.}[/latex] What is the mass of this allowance in grams?

The molar mass for this compound is computed to be 176.124 g/mol. The given number of moles is a very small fraction of a mole (~10 -4 or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:

[latex]1.42\times {10}^{-4}\cancel{\text{mol }\text{C}_{6}\text{H}_{8}\text{O}_{6}}\left(\frac{\text{176.124 }\text{g }\text{C}_{6}\text{H}_{8}\text{O}_{6}}{\text{1 }\cancel{\text{mol }\text{C}_{6}\text{H}_{8}\text{O}_{6}}}\right)=\text{0.0250 }\text{g }\text{C}_{6}\text{H}_{8}\text{O}_{6}[/latex]

This is consistent with the anticipated result.

What is the mass of 0.443 mol of hydrazine, N 2 H 4 ?

Example 6: Deriving the Number of Atoms and Molecules from the Mass of a Compound

A packet of an artificial sweetener contains 0.0400 g of saccharin (C 7 H 5 NO 3 S). (a) Given that saccharin has a molar mass of 183.18 g/mol, how many saccharin molecules are in a 0.0400-g sample of saccharin?

Using the provided mass and molar mass for saccharin yields:

[latex]\text{0.0400 }\cancel{\text{g }{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}}\left(\frac{\text{1 }\cancel{\text{mol }}{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}}{\text{183.18 }\cancel{\text{g }{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}}}\right)\left(\frac{6.022\times {10}^{23}\text{ molecules}{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S }}{\text{1 }\cancel{\text{mol }{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}}}\right)=1.31\times {10}^{20}\text{molecules }{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}[/latex]

How many C 4 H 10 molecules are contained in 9.213 g of this compound?

[latex]9.545\times {10}^{22}\text{molecules }{\text{C}}_{4}{\text{H}}_{10}[/latex]

Chemical Formulas and the Mole

Suppose we want to know the number of hydrogen atoms found in a dozen CH 4 molecule. For each CH 4 molecule there are four hydrogen atoms, we determine this from the chemical formula, there is one carbon atom per one CH 4 molecule. As shown below, if we have one dozen molecules of CH 4 we can determine that we will have a total of 48 hydrogen’s using dimensional analysis.

[latex]12\text{ molecules CH}_{4}\left(\frac{\text{4 }\text{atoms H}}{\text{1 molecule CH}_{4}}\right)=48\text{ }\text{atoms H}[/latex]

Since is unlikely a chemist will be working with a single molecule, and much more likely to be working molecules on the scale of Avogadro’s number, it makes more sense for us to relate the number of moles of H to the moles of CH 4 .   If we had 12 mole sample of CH 4 , how many moles of H atoms will be present in the sample?  We can derive a mole to mole relationship between H and CH 4 using the chemical formula.

[latex]\left(\frac{\text{4 }\text{mol H}}{\text{1 mol CH}_{4}}\right)[/latex]

Therefore, if we have a 12 mol sample of CH 4 , we can calculate the moles of H present in the sample:

[latex]\text{12 }\cancel{\text{mol CH}_{4}}\left(\frac{\text{4 }\text{mol H}}{\text{1 }\cancel{\text{mol CH}_{4}}}\right)=48 \text{ mol H} [/latex]

All we need to derive a mol to mol relationship between a compound and an element within that compound is the chemical formula.

Example 7: Converting between moles of a compound and moles of an element within the compound

An organic compound, commonly known as strawberry aldehyde, is used in the flavor industry in artificial fruit flavors. Given the formula C 12 H 14 O 3 , determine the moles of each element within 2.50 moles of strawberry aldehyde?

Since we are being asked to determine the number of moles of each element (C, H, and O) within a mole sample of C 12 H 14 O 3 , we can use a mole to mole ratio from the chemical formula.

Moles of carbon: For one mole of C 12 H 14 O 3 , there will be twelve moles of C.

[latex]\text{2.50 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}\left(\frac{\text{12 }\text{mol C}}{\text{1 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}}\right)=30 \text{ mol C} [/latex]

Moles of hydrogen: For one mole of C 12 H 14 O 3 , there will be fourteen moles of H.

[latex]\text{2.50 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}\left(\frac{\text{14 }\text{mol H}}{\text{1 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}}\right)=35 \text{ mol H} [/latex]

Moles of oxygen: For one mole of C 12 H 14 O 3 , there will be three moles of O.

[latex]\text{2.50 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}\left(\frac{\text{3 }\text{mol O}}{\text{1 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}}\right)=7.5 \text{ mol O} [/latex]

How many moles of nitrogen are in 10.0 moles of N 2 O 4 ?

[latex]\text{10.0 }\cancel{\text{mol N}_{2}\text{O}_{4}}\left(\frac{\text{2 }\text{mol N}}{\text{1 }\cancel{\text{mol N}_{2}\text{O}_{4}}}\right)=20.0 \text{ mol N} [/latex]

We can use mol to mol ratios in combination with other mole conversions, such as Avogadro’s number and molar mass.

Example 8: Converting between grams of a compound and grams of an element within the compound

Isoamyl acetate, C 7 H 14 O 2 , is an organic ester commonly referred to as banana oil due to it strong banana odor.  How many grams of carbon will be in a 0.500 g sample of isoamyl acetate?

Since we are given grams of isoamyl acetate, we need the molar mass of isoamyl acetate.  Using the periodic table we can calculate the molar mass of isoamyl acetate to be 130.18 g/mol.  Since we are looking for grams of carbon, we also need the molar mass of carbon (12.01 g/mol).  We also need to utilize a mole to mole ratio from a chemical formula since we are looking for “units” of carbon within “units” of the larger isoamyl acetate sample.  Using the provided formula, we can derive the mole to mole:  [latex]\left(\frac{\text{7 }\text{mol C}}{\text{1 }\text{mol C}_{7}\text{H}_{14}\text{O}_{2}}\right) [/latex]

[latex]\text{0.500 }\cancel{\text{g C}_{7}\text{H}_{14}\text{O}_{2}}\left(\frac{\text{1 }\cancel{\text{mol C}_{7}\text{H}_{14}\text{O}_{2}}}{\text{130.18 }\cancel{\text{g C}_{7}\text{H}_{14}\text{O}_{2}}}\right)\left(\frac{\text{7 }\cancel{\text{mol C}}}{\text{1 }\cancel{\text{mol C}_{7}\text{H}_{14}\text{O}_{2}}}\right)\left(\frac{\text{12.01 }\text{g C}}{\text{1 }\cancel{\text{mol C}}}\right)=0.323 \text{ g C} [/latex]

How many grams of oxygen are in 5.0 g of N 2 O 4 ?

[latex]\text{5.0 }\cancel{\text{g N}_{2}\text{O}_{4}}\left(\frac{\text{1 }\cancel{\text{mol N}_{2}\text{O}_{4}}}{\text{92.02 }\cancel{\text{g N}_{2}\text{O}_{4}}}\right)\left(\frac{\text{4 }\cancel{\text{mol O}}}{\text{1 }\cancel{\text{mol N}_{2}\text{O}_{4}}}\right)\left(\frac{\text{16.00 }\cancel{\text{g O}}}{\text{1 }\cancel{\text{mol O}}}\right)=3.5 \text{ g O} [/latex]

Example 9: Converting between grams of a compound and atoms of an element within the compound

Cow’s and other livestock are responsible for nearly 40% of global methane (CH 4 ) emissions.  A cow produces on average, 260.0 g of methane per day.  How many atoms of hydrogen is present in a 260.0 g sample of methane?

Since we have grams of methane given, we will need the molar mass of methane (16.04 g/mol).  We are looking for atoms of H, meaning we will need Avogadro’s number, and since we are going from “units” of methane to “units” of carbon, we will need a mole to mole ratio [latex]\left(\frac{\text{4 }\text{mol H}}{\text{1 }\text{mol C}\text{H}_{4}}\right) [/latex]

[latex]\text{260.0 }\cancel{\text{g C}\text{H}_{4}}\left(\frac{\text{1 }\cancel{\text{mol C}\text{H}_{4}}}{\text{16.04 }\cancel{\text{g C}\text{H}_{4}}}\right)\left(\frac{\text{4 }\cancel{\text{mol H}}}{\text{1 }\cancel{\text{mol C}\text{H}_{4}}}\right)\left(\frac{6.022\times {10}^{23}\text{atoms H}}{\text{1 }\cancel{\text{mol H}}}\right)=3.905\times {10}^{25} \text{ atoms H} [/latex]

How many atoms of carbon is present in a 260.0 g sample of methane?

[latex]\text{260.0 }\cancel{\text{g C}\text{H}_{4}}\left(\frac{\text{1 }\cancel{\text{mol C}\text{H}_{4}}}{\text{16.04 }\cancel{\text{g C}\text{H}_{4}}}\right)\left(\frac{\text{1 }\cancel{\text{mol C}}}{\text{1 }\cancel{\text{mol C}\text{H}_{4}}}\right)\left(\frac{6.022\times {10}^{23}\text{atoms C}}{\text{1 }\cancel{\text{mol C}}}\right)=9.761\times {10}^{24} \text{ atoms C} [/latex]

Key Concepts and Summary

A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be [latex]6.022\times {10}^{23}[/latex], a quantity called Avogadro’s number. The mass in grams of 1 mole of substance is its molar mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula mass (amu) and molar mass (g/mol) for any substance are numerically equivalent (for example, one H 2 O molecule weighs approximately18 amu and 1 mole of H 2 O molecules weighs approximately 18 g).

• Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula.
• Which has the largest number of molecules? Explain why.
• Which has the greatest mass? Explain why.
• number of atoms of F in 1.5 moles of F
• number of atoms of Al in 5.2 mol Al
• number of molecules of CO 2 in 0.67 mol of CO 2
• number of molecules of C 2 H 5 OH in 0.0250 mol of C 2 H 5 OH
• number of formula units of NaCl in 0.050 mol of NaCl
• number of formula units of Ca 3 (PO 4 ) 2 in 3.40 mol of Ca 3 (PO 4 ) 2
• Sc 2 (SO 4 ) 3
• CH 3 COCH 3 (acetone)
• C 6 H 12 O 6 (glucose)
• number of moles of Cu in 5.50 g Cu
• number of moles of S in 30.2 g S
• number of moles of CCl 4 in 0.250 g CCl 4
• number of moles of C 12 H 22 O 11 in 100.0 g C 12 H 22 O 11
• number of moles of Na 2 S in 12.0 g Na 2 S
• number of moles of Ca 3 (PO 4 ) 2 in 20.0 g Ca 3 (PO 4 ) 2
• Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C 2 H 5 OH), 0.60 mol of formic acid (HCO 2 H), or 1.0 mol of water (H 2 O)? Explain why.
• Which contains the greatest number of moles of oxygen atoms: 1 mol of ethanol (C 2 H 5 OH), 1 mol of formic acid (HCO 2 H), or 1 mol of water (H 2 O)? Explain why.
• 25.0 g of propylene, C 3 H 6
• [latex]3.06\times {10}^{-3}\text{g}[/latex] of the amino acid glycine, C 2 H 5 NO 2
• 25 lb of the herbicide Treflan, C 13 H 16 N 2 O 4 F (1 lb = 454 g)
• 0.125 kg of the insecticide Paris Green, Cu 4 (AsO 3 ) 2 (CH 3 CO 2 ) 2
• 325 mg of aspirin, C 6 H 4 (CO 2 H)(CO 2 CH 3 )
• 2.12 g of potassium bromide, KBr
• 0.1488 g of phosphoric acid, H 3 PO 4
• 23 kg of calcium carbonate, CaCO 3
• 78.452 g of aluminum sulfate, Al 2 (SO 4 ) 3
• 0.1250 mg of caffeine, C 8 H 10 N 4 O 2
• The approximate minimum daily dietary requirement of the amino acid leucine, C 6 H 13 NO 2 , is 1.1 g. What is this requirement in moles?
• 0.600 mol of oxygen atoms
• 0.600 mol of oxygen molecules, O 2
• 0.600 mol of ozone molecules, O 3
• A 55-kg woman has [latex]7.5\times {10}^{-3}\text{mol}[/latex] of hemoglobin (molar mass = 64,456 g/mol) in her blood. How many hemoglobin molecules is this? What is this quantity in grams?
• Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, ZrSiO 4 , a semiprecious stone.
• Determine which of the following contains the greatest mass of hydrogen: 1 mol of CH 4 , 0.6 mol of C 6 H 6 , or 0.4 mol of C 3 H 8 .
• Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO 4 , 266 g of Al 2 Cl 6 , or 225 g of Al 2 S 3 .
• Diamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats (1 carat = 200 mg). How many atoms are present in the diamond?
• The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone
• One 55-gram serving of a particular cereal supplies 270 mg of sodium, 11% of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance?
• A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C 12 H 22 O 11 ) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar?
• What mass of fluorine atoms in mg was present?
• How many fluorine atoms were present?
• 20.0 g of H 2 O (18.02 g/mol)
• 77.0 g of CH 4 (16.06 g/mol)
• 68.0 g of CaH 2 (42.09 g/mol)
• 100.0 g of N 2 O (44.02 g/mol)
• 84.0 g of HF (20.01 g/mol)

1. Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.

3.The number of “particles” of each substance is as follows:

• atoms of F: [latex]\text{1.5 }\cancel{\text{mol F}}\left(\frac{\text{6.022}\times{10}^{23}\text{ }\text{atoms F}}{\text{1 }\cancel{\text{mol F}}}\right)=9.0\times {10}^{23}\text{ atoms F}[/latex]
• atoms of Al: [latex]\text{5.2 }\cancel{\text{mol Al}}\left(\frac{\text{6.022}\times{10}^{23}\text{ }\text{atoms Al}}{\text{1 }\cancel{\text{mol Al}}}\right)=3.1\times {10}^{24}\text{ atoms Al}[/latex]
• molecules of CO 2 : [latex]\text{0.67 }\cancel{\text{mol CO}_{2}}\left(\frac{\text{6.022}\times{10}^{23}\text{ }\text{ molecules CO}_{2}}{\text{1 }\cancel{\text{mol CO}_{2}}}\right)=4.0\times {10}^{23}\text{ molecules CO}_{2}[/latex]
• molecules of C 2 H 5 OH: [latex]\text{0.0250 }\cancel{\text{mol C}_{2}\text{H}_{5}\text{OH}}\left(\frac{\text{6.022}\times{10}^{23}\text{ }\text{molecules C}_{2}\text{H}_{5}\text{OH}}{\text{1 }\cancel{\text{mol C}_{2}\text{H}_{5}\text{OH}}}\right)=1.51\times {10}^{22}\text{ molecules C}_{2}\text{H}_{5}\text{OH}[/latex]
• formula units of NaCl: [latex]\text{0.050 }\cancel{\text{mol NaCl}}\left(\frac{\text{6.022}\times{10}^{23}\text{ }\text{ formula units NaCl}}{\text{1 }\cancel{\text{mol NaCl}}}\right)=3.0\times {10}^{22}\text{ formula units NaCl}[/latex]
• formula units of Ca 3 (PO 4 ) 2 : [latex]\text{3.40 }\cancel{\text{mol Ca}_{3}\left(\text{PO}_{4}\right )\text{}_{2}}\left(\frac{\text{6.022}\times{10}^{23}\text{ }\text{formula units Ca}_{3}\left(\text{PO}_{4}\right )\text{}_{2}}{\text{1 }\cancel{\text{mol Ca}_{3}\left(\text{PO}_{4}\right )\text{}_{2}}}\right)=2.05\times {10}^{24}\text{ formula units Ca}_{3}\left(\text{PO}_{4}\right )\text{}_{2}[/latex]

4. The molar mass of each substance is as follows:

• HF molar mass = 20.01 g/mol
• NH 3  molar mass = 17.03 g/mol
• HNO 3  molar mass = 63.02 g/mol
•  Ag 2 SO 4  molar mass = 311.87 g/mol
• B(OH) 3  molar mass = 61.83 g/mol
• S 8  molar mass = 256.56 g/mol
• C 5 H 12  molar mass = 72.15 g/mol
• Sc 2 (SO 4 ) 3   molar mass = 378.13 g/mol
• CH 3 COCH 3 (acetone)  molar mass = 58.08 g/mol
• C 6 H 12 O 6 (glucose)  molar mass = 180.16 g/mol

5. The moles of each substance is as follows:

• moles of Cu: [latex]\text{5.50 }\cancel{\text{g Cu}}\left(\frac{\text{1 mol Cu}}{\text{63.55 }\cancel{\text{g Cu}}}\right)=0.0865\text{ mol Cu}[/latex]
• moles of S: [latex]\text{30.2 }\cancel{\text{g S}}\left(\frac{\text{1 mol S}}{\text{32.07 }\cancel{\text{g S}}}\right)=0.942\text{ mol S}[/latex]
• moles of CCl 4 : [latex]\text{0.250 }\cancel{\text{g CCl}_{4}}\left(\frac{\text{1 mol CCl}_{4}}{\text{153.81 }\cancel{\text{g CCl}_{4}}}\right)=0.00163\text{ mol CCl}_{4}[/latex]
• moles of C 12 H 22 O 11 : [latex]\text{100.0 }\cancel{\text{g C}_{12}\text{H}_{22}\text{O}_{11}}\left(\frac{\text{1 mol C}_{12}\text{H}_{22}\text{O}_{11}}{\text{342.30 }\cancel{\text{g C}_{12}\text{H}_{22}\text{O}_{11}}}\right)=0.2921\text{ mol C}_{12}\text{H}_{22}\text{O}_{11}[/latex]
• moles of Na 2 S: [latex]\text{12.0 }\cancel{\text{g Na}_{2}\text{S}}\left(\frac{\text{1 mol Na}_{2}\text{S}}{\text{78.05 }\cancel{\text{g Na}_{2}\text{S}}}\right)=0.154\text{ mol Na}_{2}\text{S}[/latex]
• moles of Ca 3 (PO 4 ) 2 : [latex]\text{20.0 }\cancel{\text{g Ca}_{3}\left(\text{PO}_{4}\right )\text{}_{2}}\left(\frac{\text{1}\text{ }\text{mol Ca}_{3}\left(\text{PO}_{4}\right )\text{}_{2}}{\text{310.18 }\cancel{\text{g Ca}_{3}\left(\text{PO}_{4}\right )\text{}_{2}}}\right)=0.0645 \text{ mol Ca}_{3}\left(\text{PO}_{4}\right )\text{}_{2}[/latex]

13. The mass of each compound is as follows:

• [latex]0.600\cancel{\text{mol}}\times 15.9994\text{g/}\cancel{\text{mol}}=9.60\text{g}[/latex]
• [latex]0.600\cancel{\text{mol}}\times 2\times 15.994\text{g/}\cancel{\text{mol}}=19.2\text{g}[/latex]
• [latex]0.600\cancel{\text{mol}}\times 3\times 15.994\text{g/}\cancel{\text{mol}}=28.8\text{g}[/latex]

15. Determine the number of moles of each component. From the moles, calculate the number of atoms and the mass of the elements involved. Zirconium: [latex]0.3384\cancel{\text{mol}}\times 6.022\times {10}^{23}{\cancel{\text{mol}}}^{\cancel{-1}}=2.038\times 1023\text{atoms;}0.3384\cancel{\text{mol}}\times 91.224\text{g/}\cancel{\text{mol}}=30.87\text{g;}[/latex] Silicon: [latex]0.3384\cancel{\text{mol}}\times 6.022\times {10}^{23}{\cancel{\text{mol}}}^{\cancel{-1}}=2.038\times {10}^{23}\text{atoms;}0.3384\cancel{\text{mol}}\times 28.0855\text{g/}\cancel{\text{mol}}=9.504\text{g;}[/latex] Oxygen: [latex]4\times 0.3384\cancel{\text{mol}}\times 6.022\times {10}^{23}{\cancel{\text{mol}}}^{\cancel{-1}}=8.151\times {10}^{23}\text{atoms;}4\times 0.3384\cancel{\text{mol}}\times 15.9994\text{g/}\cancel{\text{mol}}=21.66\text{g}[/latex]

17. Determine the molar mass and, from the grams present, the moles of each substance. The compound with the greatest number of moles of Al has the greatest mass of Al.

• Molar mass AlPO 4 : 26.981539 + 30.973762 + 4(15.9994) = 121.9529 g/mol
• Molar mass Al 2 Cl 6 : 2(26.981539) + 6(35.4527) = 266.6793 g/mol
• Molar mass Al 2 S 3 : 2(26.981539) + 3(32.066) = 150.161 g/mol

AlPO 4 : [latex]\frac{122\cancel{\text{g}}}{121.9529\cancel{\text{g}}{\text{mol}}^{-1}}=1.000\text{mol.}[/latex]

[latex]\text{mol Al}=1\times 1.000\text{mol}=1.000\text{mol}[/latex]

Al 2 Cl 6 : [latex]\frac{266\text{g}}{266.6793\text{g}{\text{mol}}^{-1}}=0.997\text{mol}[/latex]

[latex]\text{mol Al}=2\times 0.997\text{mol}=1.994\text{mol}[/latex]

Al 2 S 3 : [latex]\frac{225\cancel{\text{g}}}{150.161\cancel{\text{g}}{\text{mol}}^{-1}}=1.50\text{mol}[/latex]

[latex]\text{mol Al}=2\times 1.50\text{mol}=3.00\text{mol}[/latex]

19. Determine the number of grams present in the diamond and from that the number of moles. Find the number of carbon atoms by multiplying Avogadro’s number by the number of moles:

[latex]\frac{3104\cancel{\text{carats}}\times \frac{200\cancel{\text{mg}}}{1\cancel{\text{carat}}}\times \frac{1\cancel{\text{g}}}{1000\cancel{\text{mg}}}}{12.011\cancel{\text{g}}\cancel{{\text{mol}}^{-1}}\left(6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\right)}=3.113\times {10}^{25}\text{C atoms}[/latex]

21. Determine the molar mass of sugar. 12(12.011) + 22(1.00794) + 11(15.9994) = 342.300 g/mol; Then [latex]0.0278\text{mol}\times 342.300\text{g/mol}=9.52\text{g sugar.}[/latex] This 9.52 g of sugar represents [latex]\frac{11.0}{60.0}[/latex] of one serving or

[latex]\frac{60.0\text{g serving}}{11.0\cancel{\text{g sugar}}}\times 9.52\cancel{\text{g sugar}}=51.9\text{g cereal.}[/latex]

This amount is [latex]\frac{51.9\text{g cereal}}{60.0\text{g serving}}=0.865[/latex] servings, or about 1 serving.

Avogadro’s number ( N A ): experimentally determined value of the number of entities comprising 1 mole of substance, equal to [latex]6.022\times {10}^{23}{\text{mol}}^{-1}[/latex]

molar mass: mass in grams of 1 mole of a substance

mole: amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of 12 C

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Empirical Formula: Definition and Examples

How to read the element ratio in an empirical formula

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The empirical formula of a compound is defined as the formula that shows the ratio of elements present in the compound, but not the actual numbers of atoms found in the molecule. The ratios are denoted by subscripts next to the element symbols.

Also Known As: The empirical formula is also known as the  simplest formula  because the subscripts are the smallest whole numbers that indicate the ratio of elements.

Key Takeaways: Empirical Formula

• The empirical formula is also known as the simplest formula in chemistry.
• It gives the smallest whole number ratio of elements in a compound using subscripts following element symbols.
• In some cases, the empirical formula is the same as the molecular formula, which gives the actual number of atoms in a compound (e.g., H 2 O).
• Otherwise, the molecular formula is a multiple of the empirical formula (e.g., CH 2 O is the empirical formula for glucose, C 6 H 12 O 6 ).

Empirical Formula Examples

Glucose has a molecular formula of C 6 H 12 O 6 . It contains 2 moles of hydrogen for every mole of carbon and oxygen. The empirical formula for glucose is CH 2 O.

The molecular formula of ribose is C 5 H 10 O 5 , which can be reduced to the empirical formula CH 2 O.

How to Determine Empirical Formula

• Begin with the number of grams of each element, which you usually find in an experiment or have given in a problem.
• To make the calculation easier, assume the total mass of a sample is 100 grams, so you can work with simple percentages. In other words, set the mass of each element equal to the percent. The total should be 100 percent.
• Use the molar mass you get by adding up the atomic weight of the elements from the periodic table to convert the mass of each element into moles.
• Divide each mole value by the small number of moles you obtained from your calculation.
• Round each number you get to the nearest whole number. The whole numbers are the mole ratio of elements in the compound, which are the subscript numbers that follow the element symbol in the chemical formula.

Sometimes determining the whole number ratio is tricky and you'll need to use trial and error to get the correct value. For values close to x.5, you'll multiply each value by the same factor to obtain the smallest whole number multiple. For example, if you get 1.5 for a solution, multiply each number in the problem by 2 to make the 1.5 into 3. If you get a value of 1.25, multiply each value by 4 to turn the 1.25 into 5.

Using Empirical Formula to Find Molecular Formula

You can use the empirical formula to find the molecular formula if you know the molar mass of the compound. To do this, calculate the empirical formula mass and then divide the compound molar mass by the empirical formula mass. This gives you the ratio between the molecular and empirical formulas. Multiply all of the subscripts in the empirical formula by this ratio to get the subscripts for the molecular formula.

Empirical Formula Example Calculation

A compound is analyzed and calculated to consist of 13.5 g Ca, 10.8 g O, and 0.675 g H. Find the empirical formula of the compound.

Start by converting the mass of each element into moles by looking up the atomic numbers from the periodic table. The atomic masses of the elements are 40.1 g/mol for Ca, 16.0 g/mol for O, and 1.01 g/mol for H.

13.5 g Ca x (1 mol Ca / 40.1 g Ca) = 0.337 mol Ca

10.8 g O x (1 mol O / 16.0 g O) = 0.675 mol O

0.675 g H x (1 mol H / 1.01 g H) = 0.668 mol H

Next, divide each mole amount by the smallest number or moles (which is 0.337 for calcium) and round to the nearest whole number:

0.337 mol Ca / 0.337 = 1.00 mol Ca

0.675 mol O / 0.337 = 2.00 mol O

0.668 mol H / 0.337 = 1.98 mol H which rounds up to 2.00

Now you have the subscripts for the atoms in the empirical formula:

Finally, apply the rules of writing formulas to present the formula correctly. The cation of the compound is written first, followed by the anion. The empirical formula is properly written as Ca(OH) 2

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Molar Mass Calculator

What is molar mass, molar mass vs. molecular weight, how to find the molar mass formula for any compound, examples of how to calculate the molar masses of nacl, naoh and h₂o, how to use this molar mass calculator.

Our molar mass calculator comes to the rescue if you need to quickly check the weight of 1 mole of any element or chemical compound and you are unable to use the periodic table . Simply select one by one the elements from the list and give the number of atoms in their molecular formula to get the molar mass in a flash.

Many similar tools are case-sensitive, so you must carefully enter the formula of the compound by hand to get the result. In our tool, there is no exception whether you want to count the molar mass of CO 2 or maybe the molar mass of NaOH – you will get the correct result the first time!

Here we will tell you more about what the molar mass is and how to find the molar mass of any compound.

All substances are made up of atoms or molecules . In chemistry, it is crucial to accurately measure their quantities . To determine the number of reactants and products of chemical reactions, we use the SI base unit of the mole , the symbol mol .

One mole is the amount of a substance containing as many molecules (atoms, molecules, ions) as there are atoms in 12 g of the carbon isotope 12 C . This fixed number called the Avogadro's constant or Avogadro number (N A ), is equal exactly to 6.02214076×10 23 elementary objects, which can be atoms, molecules, ions, or electrons. Here read more about how to calculate moles .

Molar mass μ is a physical quantity that tells us what the mass of one mole of a substance is. We can calculate it in a simple way by dividing the mass of the substance m by its amount in moles n : μ = m/n

The SI unit of molar mass is kg/mol , but the g/mol unit is more commonly used.

It may seem as though the two quantities mean the same thing, but this is not true. Molecular weight and molar mass are different quantities, although numerically equal:

Molecular weight (or molecular mass ) is the mass of a molecule given in the dalton unit ( Da ) or the unified atomic mass unit ( u ). This one roughly corresponds to the mass of a single proton or neutron. For example, the molecular weight of CO 2 is 44.01 Da or 44.01 u.

Molar mass is the mass of one mole of a substance expressed in g/mol . Thus, the molar mass of CO 2 will be 44.01 g/mol.

Now we see that the values of molar mass and molecular mass are equal, but they have different units and define the mass of the substance slightly differently. Our molecular weight calculator will help you to find the molecular weight efficiently.

The general molar mass formula of a compound A x B y C z involves multiplying the number of atoms by the molar mass of that element and adding the mass of all the elements in the molecule to get the molar mass :

How do we know the molar masses of individual atoms? They are assigned to each element in this molar mass calculator or use the periodic table to read them.

Let's calculate the molar mass of glucose . When we want to calculate the molar mass of a selected substance, four steps are enough:

Write down the chemical formula of the glucose: C 6 H 12 O 6 .

Find the molar masses of carbon (C), hydrogen (H), and oxygen (O).

Count the number of atoms of each element in the compound.

Find the molar mass of glucose by multiplying the atomic masses of the atoms and their number, then find the sum:

μ = 6 × 12.01 g/mol + 12 × 1.0079 g/mol + 6 × 16 g/mol = 180.1548 g/mol

If you know how to calculate molar mass , learn about other ways to express the amount of substances , especially in solutions by visiting our molality calculator and molarity calculator .

🙋 The molar mass of an ion is the same as the sum of the molar masses of the elements belonging to the ion. For example, the molar mass of Fe³⁺ equals Fe or 55.85 g/mol.

We calculate the molar mass of one molecule of a substance by adding up the molar masses of the individual atoms that form the molecule. Equipped with this knowledge, let us calculate the molar mass of NaCl :

Similarly, we can calculate the molar mass of compounds containing more atoms. For example, the molar mass of NaOH will be:

If there is more than 1 atom of a given type in a molecule, we need to multiply the molar mass of that element by the number of atoms . For example, we calculate the molar mass of water (H 2 O) as:

Finding the molar mass is fairly quick for simple compounds, but if you are using more complex situations, our calculator will be an invaluable help. Let's see how it works by calculating the molar mass of water :

Write down the chemical formula of water: H 2 O.

Select "Hydrogen(H)" as the first element from the list, and set the number of atoms as 2 .

Repeat this. Select "Oxygen (O)" from the list and set the number of atoms as 1 .

An additional field will appear in the molar mass calculator if you want to add more elements to the formula.

You have the result, the molar mass of water is 18.0158 g/mol 👏

In the table below the calculation , you will find information about the molar mass of each element and their percentage in the molecule .

What are the units for molar mass?

The units of molar mass are grams per mole or g/mol . Molar mass is defined as the mass of one mole of a substance in grams. Sometimes the unit kg/mol is also used.

How can I find the molar mass of a compound?

To find the molar mass of a chosen compound , e.g., molar mass of HCl :

• Find the atomic mass of each element in the periodic table.
• Count the number of atoms of each element.
• Find the molar masses by multiplying the molar mass of the atoms by their number in the HCI , then add them together.

Therefore, the molar mass of HCl is 1 × 1.0079 g/mol + 1 × 35.45 g/mol = 36.4579 g/mol .

What is the molar mass of CO₂?

The molar mass of CO₂ is 44.01 g/mol . To calculate this:

Find molar masses of carbon ( 12.01 g/mol ) and oxygen ( 16 g/mol ).

Multiply the molar mass of each element by the number of atoms in the compound's formula, then add them together:

( 1 × 12.01 g/mol ) + ( 2 × 16 g/mol ) = 44.01 g/mol .

What is the molar mass of H₂O?

The molar mass of H₂O is 18.0158 g/mol. To find this, sum up the molar masses of 2 hydrogen atoms and 1 oxygen atom:

2 × 1.0079 g/mol + 12 × 16 g/mol = 18.0158 g/mol .

What is the molar mass of NaCl?

The molar mass of NaCl is 58.44 g/mol. To get this, sum up the molar masses of 1 sodium atom and 1 chlorine atom:

1 × 22.99 g/mol + 1 × 35.45 g/mol = 58.44 g/mol.

Hydrogen ion concentration

Ideal egg boiling, rate constant, steps to calories.

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• Chemistry Articles
• Vant Hoff Factor Equation Abnormal Molar Mass

Van't Hoff Factor and Abnormal Molar Mass

What is the van’t hoff factor.

The Van’t Hoff factor offers insight on the effect of solutes on the colligative properties of solutions. It is denoted by the symbol ‘i’. The Van’t Hoff factor can be defined as

the ratio of the concentration of particles formed when a substance is dissolved to the concentration of the substance by mass.

The extent to which a  substance associates or dissociates in a solution is described by the Van’t Hoff factor. For example, when a non-electrolytic substance is dissolved in water, the value of i is generally 1. However, when an ionic compound forms a solution in water, the value of i is equal to the total number of ions present in one formula unit of the substance.

For example, the Van’t Hoff factor of CaCl 2 is ideally 3, since it dissociates into one Ca 2+ ion and two Cl – ions. However, some of these ions associate with each other in the solution, leading to a decrease in the total number of particles in the solution.

This factor is named after the Dutch physical chemist Jacobus Henricus Van’t Hoff, who won the first Nobel Prize in chemistry. It is important to note that the measured value of the Van’t Hoff factor for electrolytic solutions is generally lower than the predicted value (due to the pairing of ions). The greater the charge on the ions, the greater the deviation.

Table of Contents

Effects of association/dissociation, abnormal molar masses.

• Frequently Asked Questions–FAQs
• Association is the joining of two or more particles to form one entity.
• An example of the association of two particles is the dimerization of carboxylic acids when dissolved in benzene.
• Dissociation refers to the splitting of a molecule into multiple ionic entities.
• For example, sodium chloride (NaCl) dissociates into Na + and Cl – ions when dissolved in water.

The effects of the association or dissociation of a solute on the solution, its colligative properties , and the Van’t Hoff factor are tabulated below.

The theoretical values of molecular mass , when calculated from the colligative properties of solutions, are sometimes found to differ from the experimentally obtained values. These values are often referred to as abnormal molar masses.

Van’t Hoff explained that when solutes are dissolved in a solvent they dissociate into ions. Since colligative properties depend only on the number of solute particles, the dissociation of solute molecules into ions results in an increase in the number of particles and hence affects the colligative properties.

When 1 mole of NaCl is dissolved in 1 Kg of water, if all the molecules of NaCl dissociate in water, there will be 1 mole of Cl – ions and 1 mole of Na + ions in the resulting solution (a total of 2 moles of ions in the solution). But while calculating the molar mass using the colligative properties, we consider only 1 mol of NaCl to be present in the solution.

Some of the substances tend to associate in aqueous state and for such molecules, the number of ions/molecules present in the solution is less than the actual number of molecules. So, for those substances that dissociate in solution, the observed molar mass will always be less than the real mass and for those substances that associate in solutions, the real mass will always be less than the observed molar mass.

The abnormality in the molecular mass can be explained as follows:

• The dissociation of solute molecules into multiple ions results in an increase in the number of particles. This, in turn, increases the colligative properties of the solution.
• Since the molar mass is inversely proportional to the colligative properties, its value tends to be lower than expected.
• When solute particles associate with each other, the total number of particles in the solution decreases, leading to a decrease in the colligative properties.
• In this case, the molar mass values obtained are higher than expected.

Recommended Video

Colligative properties & van ‘t hoff factor.

Frequently Asked Questions – FAQs

Under what condition van’t hoff factor ‘i’ is equal to unity, under what condition van’t hoff factor ‘i’ is less than one, what is the van’t hoff factor for a compound which undergoes tetramerisation in an organic solvent, arrange the following in increasing order of acidity: phenol, ethanol, and water., how much molecular mass of nacl is obtained experimentally using colligative properties, what is the van’t hoff factor for a compound which undergoes dimerisation in an organic solvent.

Thus, the Van’t Hoff factor is briefly described in this article. The abnormal molar mass values obtained from the colligative properties of solutions are also explained with the help of this factor. To learn more about this topic and other related topics, such as Raoult’s Law , register with BYJU’S and download the mobile application on your smartphone.

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