# 2.3.2: Formula Mass, Percent Composition, and the Mole


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
• Compute the percent composition of a compound

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 (CHCl3), 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 $$\PageIndex{1}$$ outlines the calculations used to derive the molecular mass of chloroform, which is 119.37 amu.

Figure $$\PageIndex{1}$$: The average mass of a chloroform molecule, CHCl3, is 119.37 amu, which is the sum of the average atomic masses of each of its constituent atoms. The model shows the molecular structure of chloroform.

Likewise, the molecular mass of an aspirin molecule, C9H8O4, 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 $$\PageIndex{2}$$).

Figure $$\PageIndex{2}$$: The average mass of an aspirin molecule is 180.15 amu. The model shows the molecular structure of aspirin, C9H8O4.

Example $$\PageIndex{1}$$: Computing Molecular Mass for a Covalent Compound

Ibuprofen, C13H18O2, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular mass (amu) for this compound?

Solution

Molecules of this compound are comprised of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. Following the approach described above, the average molecular mass for this compound is therefore:

Exercise $$\PageIndex{1}$$

Acetaminophen, C8H9NO2, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass (amu) for this compound?

151.16 amu

### 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 (Figure $$\PageIndex{3}$$).

Figure $$\PageIndex{3}$$: Table salt, NaCl, contains an array of sodium and chloride ions combined in a 1:1 ratio. Its formula mass is 58.44 amu.

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 $$\PageIndex{2}$$: Computing Formula Mass for an Ionic Compound

Aluminum sulfate, Al2(SO4)3, is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass (amu) of this compound?

Solution

The formula for this compound indicates it contains Al3+ and SO42− ions combined in a 2:3 ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, Al2S3O12. Following the approach outlined above, the formula mass for this compound is calculated as follows:

Exercise $$\PageIndex{2}$$

Calcium phosphate, Ca3(PO4)2, is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate?

310.18 amu

## The Mole

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, H2O, and hydrogen peroxide, H2O2, 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 12C 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 $$6.02214179 \times 10^{23}$$, a fundamental constant named Avogadro’s number (NA) 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 $$6.022 \times 10^{23}/\ce{mol}$$.

Video $$\PageIndex{1}$$: What is Avogadro's Number?

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) (Figure $$\PageIndex{4}$$).

Figure $$\PageIndex{4}$$: Each sample contains $$6.022 \times 10^{23}$$ atoms —1.00 mol of atoms. From left to right (top row): 65.4 g zinc, 12.0 g carbon, 24.3 g magnesium, and 63.5 g copper. From left to right (bottom row): 32.1 g sulfur, 28.1 g silicon, 207 g lead, and 118.7 g 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, 12C, the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single 12C atom weighs 12 amu (its atomic mass is 12 amu). The former definition of the mole was that a mole was 12 g of 12C contains 1 mole of 12C 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, 12C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure $$\PageIndex{4}$$). On May 20, 2019 the definition was permanently changed to Avogadro's number: a mole is $$6.02214179 \times 10^{23}$$ of any object, from atoms to apples.2

Figure $$\PageIndex{5}$$: Each sample contains $$6.022 \times 10^{23}$$ molecules or formula units—1.00 mol of the compound or element. Clock-wise from the upper left: 130.2 g of C8H17OH (1-octanol, formula mass 130.2 amu), 454.4 g of HgI2 (mercury(II) iodide, formula mass 454.4 amu), 32.0 g of CH3OH (methanol, formula mass 32.0 amu) and 256.5 g of S8 (sulfur, formula mass 256.5 amu). (credit: Sahar Atwa).

Table $$\PageIndex{1}$$: Mass of one mole of elements
Element Average Atomic Mass (amu) Molar Mass (g/mol) Atoms/Mole
C 12.01 12.01 $$6.022 \times 10^{23}$$
H 1.008 1.008 $$6.022 \times 10^{23}$$
O 16.00 16.00 $$6.022 \times 10^{23}$$
Na 22.99 22.99 $$6.022 \times 10^{23}$$
Cl 33.45 35.45 $$6.022 \times 10^{23}$$

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 after a rainfall. 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.

Video $$\PageIndex{2}$$: The mole is used in chemistry to represent $$6.022 \times 10^{23}$$ 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 $$\PageIndex{3}$$: 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?

Solution

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 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:”

$\mathrm{4.7\; \cancel{g} K \left ( \dfrac{mol\; K}{39.10\;\cancel{g}}\right)=0.12\;mol\; K} \nonumber$

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

Exercise $$\PageIndex{3}$$: Beryllium

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?

0.360 mol

Example $$\PageIndex{4}$$: Deriving Grams from Moles for an Element

A liter of air contains $$9.2 \times 10^{−4}$$ mol argon. What is the mass of Ar in a liter of air?

Solution

The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10−3) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):

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

$\mathrm{9.2 \times10^{-4}\; \cancel{mol} \; Ar \left( \dfrac{39.95\;g}{\cancel{mol}\;Ar} \right)=0.037\;g\; Ar} \nonumber$

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

Exercise $$\PageIndex{4}$$

What is the mass of 2.561 mol of gold?

504.4 g

Example $$\PageIndex{6}$$: Deriving Number of Atoms from Mass for an Element

Copper is commonly used to fabricate electrical wire (Figure $$\PageIndex{6}$$). How many copper atoms are in 5.00 g of copper wire?

Figure $$\PageIndex{6}$$: Copper wire is composed of many, many atoms of Cu. (credit: Emilian Robert Vicol)

Solution

The number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu, and then using Avogadro’s number (NA) to convert this molar amount to number of Cu atoms:

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 NA, or approximately 1022 Cu atoms. Carrying out the two-step computation yields:

$\mathrm{5.00\:\cancel{g}\:Cu\left(\dfrac{\cancel{mol}\:Cu}{63.55\:\cancel{g}}\right)\left(\dfrac{6.022\times10^{23}\:atoms}{\cancel{mol}}\right)=4.74\times10^{22}\:atoms\: of\: copper}$

The factor-label method yields the desired cancellation of units, and the computed result is on the order of 1022 as expected.

Exercise $$\PageIndex{6}$$

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?

$$4.586 \times 10^{22}\; Au$$ atoms

﻿

Example $$\PageIndex{7}$$: 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 C2H5O2N. How many moles of glycine molecules are contained in 28.35 g of glycine?

Solution

We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example $$\PageIndex{6}$$:

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, C2H5O2N, 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:

$\mathrm{28.35\:\cancel{g}\:glycine\left(\dfrac{mol\: glycine}{75.07\:\cancel{g}}\right)=0.378\:mol\: glycine} \nonumber$

This result is consistent with our rough estimate.

Exercise $$\PageIndex{7}$$

How many moles of sucrose, $$C_{12}H_{22}O_{11}$$, are in a 25-g sample of sucrose?

0.073 mol

Example $$\PageIndex{8}$$: Deriving Grams from Moles for a Compound

Vitamin C is a covalent compound with the molecular formula C6H8O6. The recommended daily dietary allowance of vitamin C for children aged 4–8 years is 1.42 × 10−4 mol. What is the mass of this allowance in grams?

Solution

As for elements, the mass of a compound can be derived from its molar amount as shown:

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:

$\mathrm{1.42\times10^{-4}\:\cancel{mol}\:vitamin\: C\left(\dfrac{176.124\:g}{\cancel{mol}\:vitamin\: C}\right)=0.0250\:g\: vitamin\: C} \nonumber$

This is consistent with the anticipated result.

Exercise $$\PageIndex{8}$$

What is the mass of 0.443 mol of hydrazine, $$N_2H_4$$?

14.2 g

Example $$\PageIndex{9}$$: Deriving the Number of Molecules from the Compound Mass

A packet of an artificial sweetener contains 40.0 mg of saccharin (C7H5NO3S), which has the structural formula:

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?

Solution

The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Example $$\PageIndex{8}$$, and then multiplying by Avogadro’s number:

Using the provided mass and molar mass for saccharin yields:

$\mathrm{0.0400\:\cancel{g}\:\ce{C7H5NO3S}\left(\dfrac{\cancel{mol}\:\ce{C7H5NO3S}}{183.18\:\cancel{g}\:\ce{C7H5NO3S}}\right)\left(\dfrac{6.022\times10^{23}\:\ce{C7H5NO3S}\:molecules}{1\:\cancel{mol}\:\ce{C7H5NO3S}}\right)}\\ =\mathrm{1.31\times10^{20}\:\ce{C7H5NO3S}\:molecules}$

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

$\mathrm{1.31\times10^{20}\:\ce{C7H5NO3S}\: molecules\left(\dfrac{7\:C\: atoms}{1\:\ce{C7H5NO3S}\: molecule}\right)=9.20\times10^{21}\:C\: atoms} \nonumber$

Exercise $$\PageIndex{9}$$

How many $$C_4H_{10}$$ molecules are contained in 9.213 g of this compound? How many hydrogen atoms?

﻿
• $$9.545 \times 10^{22}\; \text{molecules}\; C_4H_{10}$$
• $$9.545 \times 10^{23 }\;\text{atoms}\; H$$

Video $$\PageIndex{3}$$: A preview of some of the uses we will have for moles in upcoming units

## Percent Composition

In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.

The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

$\mathrm{\%H=\dfrac{mass\: H}{mass\: compound}\times100\%}$

$\mathrm{\%C=\dfrac{mass\: C}{mass\: compound}\times100\%}$

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

$\mathrm{\%H=\dfrac{2.5\:g\: H}{10.0\:g\: compound}\times100\%=25\%}$

$\mathrm{\%C=\dfrac{7.5\:g\: C}{10.0\:g\: compound}\times100\%=75\%}$

Example $$\PageIndex{10}$$: Calculation of Percent Composition

Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

Solution

To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

$\mathrm{\%C=\dfrac{7.34\:g\: C}{12.04\:g\: compound}\times100\%=61.0\%} \nonumber$

$\mathrm{\%H=\dfrac{1.85\:g\: H}{12.04\:g\: compound}\times100\%=15.4\%} \nonumber$

$\mathrm{\%N=\dfrac{2.85\:g\: N}{12.04\:g\: compound}\times100\%=23.7\%} \nonumber$

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

Exercise $$\PageIndex{10}$$

A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?

12.1% C, 16.1% O, 71.8% Cl

## Determining Percent Composition from Formula Mass

Video $$\PageIndex{4}$$: A video overview of how to calculate percent composition of a compound based on its chemical formula.

Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:

$\mathrm{\%N=\dfrac{14.01\:amu\: N}{17.03\:amu\:NH_3}\times100\%=82.27\%}$

$\mathrm{\%H=\dfrac{3.024\:amu\: N}{17.03\:amu\:NH_3}\times100\%=17.76\%}$

This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example $$\PageIndex{2}$$. As long as we know the chemical formula of the substance in question, we can easily derive percent composition from the formula mass or molar mass.

Example $$\PageIndex{11}$$: Determining Percent Composition from a Molecular Formula

Aspirin is a compound with the molecular formula C9H8O4. What is its percent composition?

Solution

To calculate the percent composition, we need to know the masses of C, H, and O in a known mass of C9H8O4. It is convenient to consider 1 mol of C9H8O4 and use its molar mass (180.159 g/mole, determined from the chemical formula) to calculate the percentages of each of its elements:

\begin{align*} \%\ce C&=\mathrm{\dfrac{9\:mol\: C\times molar\: mass\: C}{molar\: mass\:\ce{C9H18O4}}\times100=\dfrac{9\times12.01\:g/mol} \nonumber{180.159\:g/mol}\times100=\dfrac{108.09\:g/mol}{180.159\:g/mol}\times100} \nonumber\\ \%\ce C&=\mathrm{60.00\,\%\,C} \nonumber \end{align*}

\begin{align*} \%\ce H&=\mathrm{\dfrac{8\:mol\: H\times molar\: mass\: H}{molar\: mass\:\ce{C9H18O4}}\times 100=\dfrac{8\times 1.008\:g/mol} \nonumber {180.159\:g/mol}\times 100=\dfrac{8.064\:g/mol}{180.159\:g/mol}\times 100} \nonumber\\ \%\ce H&=4.476\,\%\,\ce H\nonumber \end{align*}

\begin{align*} \%\ce O&=\mathrm{\dfrac{4\:mol\: O\times molar\: mass\: O}{molar\: mass\: \ce{C9H18O4}}\times 100=\dfrac{4\times 16.00\:g/mol} \nonumber{180.159\:g/mol}\times 100=\dfrac{64.00\:g/mol}{180.159\:g/mol}\times 100} \nonumber \\ \%\ce O&=35.52\% \nonumber \end{align*}

Note that these percentages sum to equal 100.00% when appropriately rounded.

Exercise $$\PageIndex{11}$$

To three significant digits, what is the mass percentage of iron in the compound $$Fe_2O_3$$?

﻿

69.9% Fe

## Summary

Video $$\PageIndex{5}$$: A summary of the ideas from this unit, including a look at Avogardro's number, the mole, and calculating formula masses.

The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. 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 6.022 × 1023, 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 H2O molecule weighs approximately 18 amu and 1 mole of H2O molecules weighs approximately 18 g).

## Footnotes

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.
2. Read more about the redefinition of SI units including the kilogram here (Laura Howe, CE&N, Nov. 16, 2018).

## Glossary

experimentally determined value of the number of entities comprising 1 mole of substance, equal to 6.022 × 1023 mol−1
formula mass
sum of the average masses for all atoms represented in a chemical formula; for covalent compounds, this is also the molecular mass
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 12C
molar mass
mass in grams of 1 mole of a substance
percent composition
percentage by mass of the various elements in a compound

• Adelaide Clark, Oregon Institute of Technology
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