2.6: Formula Mass and the Mole Concept

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Page ID: 451252

Lisa Sharpe Elles
University of Kansas

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Learning Objectives

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

Calculate formula masses for molecular 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.

Many 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

An earlier chapter of this text 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.

A molecular 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 molecular 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.

A table and diagram are shown. The table is made up of six columns and five rows. The header row reads: “Element,” “Quantity,” a blank space, “Average atomic mass (a m u),” a blank space, and “Subtotal (a m u).” The first column contains the symbols “C,” “H,” “C l” and a blank, merged cell that runs the width of the first five columns. The second column contains the numbers “1,” “1,” and “3” as well as the merged cell. The third column contains the multiplication symbol in each cell except for the last, merged cell. The fourth column contains the numbers “12.01,” “1.008,” and “35.45” as well as the merged cell. The fifth column contains the symbol “=” in each cell except for the last, merged cell. The sixth column contains the values “12.01,” “1.008,” “106.35,” and “119.37.” There is a thick black line below the number 106.35. The merged cell under the first five columns reads “Molecular mass.” To the left of the table is a diagram of a molecule. Three green spheres are attached to a slightly smaller black sphere, which is also attached to a smaller white sphere. The green spheres lie beneath and to the sides of the black sphere while the white sphere is located straight up from the black sphere. — 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 $\PageIndex{2}$).

Example $\PageIndex{1}$

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

Solution

Molecules of this compound are composed 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 molecule and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass (amu) for this compound?

Answer: 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 (see Figure $\PageIndex{3}$).

A table and diagram are shown. The table is made up of six columns and four rows. The header row reads: “Element,” “Quantity,” a blank space, “Average atomic mass (a m u),” a blank space and “Subtotal (a m u).” The first column contains the symbols “N a”, “C l,” and a merged cell. The merged cell runs the length of the first five columns. The second column contains the numbers “1” and “1” as well as the merged cell. The third column contains the multiplication symbol in each cell except for the last, merged cell. The fourth column contains the numbers “22.99” and “35.45” as well as the merged cell. The fifth column contains the symbol “=” in each cell except for the last, merged cell. The sixth column contains the values “22.99,” “35.45,” and “58.44.” There is a thick black line below the number “35.45.” The merged cell under the first five columns reads “Formula mass.” To the left of the table is a diagram of a chemical structure. The diagram shows green and purple spheres placed in an alternating pattern, making up the corners of eight stacked cubes to form one larger cube. The green spheres are slightly smaller than the purple spheres. — 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 3.2: 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 $\PageIndex{1}$

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?

Answer: 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, sophisticated instruments 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 sample of matter. 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. A mole of substance is that amount in which there are $6.02214076 \times 10^{23}$.

This figure contains eight different substances displayed on white circles. The amount of each substance is visibly different. — Figure $\PageIndex{4}$ Each sample contains 6.022 $\times$ 10²³ 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)

Molar Mass

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). A mole of ¹²C weighs 12 g (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 $\PageIndex{5}$).

This photo shows two vials filled with a colorless liquid. It also shows two bowls: one filled with an off-white powder and one filled with a bright red powder. — Figure $\PageIndex{5}$ Each sample contains 6.02 $\times$ 10²³ 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)

Element	Average Atomic Mass (amu)	Molar Mass (g/mol)	Atoms/Mole
C	12.01	12.01	6.022 $\times$ 10²³
H	1.008	1.008	6.022 $\times$ 10²³
O	16.00	16.00	6.022 $\times$ 10²³
Na	22.99	22.99	6.022 $\times$ 10²³
Cl	35.45	35.45	6.022 $\times$ 10²³

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 $\PageIndex{6}$). 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.

A close-up photo of a water droplet on a leaf is shown. The water droplet is not perfectly spherical. — Figure $\PageIndex{6}$ The number of molecules in a single droplet of water is roughly 100 billion times greater than the number of people on earth. (credit: “tanakawho”/Wikimedia commons)

How Big is a Mole?

Video $\PageIndex{1}$: 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.

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, as demonstrated in the next several example problems.

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):

A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, “Mass of K atoms ( g )” while the one on the right contains the phrase, “Moles of K atoms ( mol ).” There is a phrase under the arrow that says, “Divide by 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 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?

Answer: 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):

A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, “Moles of A r atoms ( mol )” while the one on the right contains the phrase, “Mass of A r atoms ( g ).” There is a phrase under the arrow that says “Multiply by molar mass ( g / mol ).”

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 $\PageIndex{1}$

What is the mass of 2.561 mol of gold?

Answer: 504.4 g

Example $\PageIndex{5}$: Deriving Number of Atoms from Mass for an Element

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

A close-up photo of a spool of copper wire is shown. — Figure $\PageIndex{7}$: 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 (N_A) 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 N_A, or approximately 10²² Cu atoms. Carrying out the two-step computation yields:

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

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

Exercise $\PageIndex{5}$

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?

Answer: 4.586 × × 10²² Au atoms

Example $\PageIndex{6}$: 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

Derive the number of moles of a compound from its mass following the same procedure used for an element in Example $\PageIndex{3}$:

A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, “Mass of C subscript 2 H subscript 5 O subscript 2 N ( g )” while the box on the right contains the phrase, “Moles of C subscript 2 H subscript 5 O subscript 2 N ( mol ).” There is a phrase under the arrow that says “Divide by molar mass (g / mol).”

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:

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

Exercise $\PageIndex{6}$

How many moles of sucrose, C₁₂H₂₂O₁₁, are in a 25-g sample of sucrose?

Answer: 0.073 mol

Example $\PageIndex{7}$: 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 $\times$ 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:

A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, “Moles of vitamin C ( mol )” while the one the right contains the phrase, “Mass of vitamin C ( g )”. There is a phrase under the arrow that says “Multiply by molar mass (g / mol).”

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 $\PageIndex{7}$

What is the mass of 0.443 mol of hydrazine, N₂H₄?

Answer: 14.2 g

Example $\PageIndex{8}$: Deriving the Number of Atoms and Molecules from the Mass of a Compound

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 $\PageIndex{6}$, and then multiplying by Avogadro’s number:

Using the provided mass and molar mass for saccharin yields:

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

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 $\PageIndex{8}$

How many C₄H₁₀ molecules are contained in 9.213 g of this compound? How many hydrogen atoms?

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

Summary

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 molecular 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 individual H₂O molecule weighs approximately 18 amu and one mole of H₂O molecules weighs approximately 18 g).

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Example \(\PageIndex{1}\)

Solution

Exercise \(\PageIndex{1}\)

Example 3.2: Computing Formula Mass for an Ionic Compound

Solution

Exercise \(\PageIndex{1}\)

How Big is a Mole?

Example \(\PageIndex{3}\): Deriving Moles from Grams for an Element

Solution

Exercise \(\PageIndex{3}\)

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

Solution

Exercise \(\PageIndex{1}\)

Example \(\PageIndex{5}\): Deriving Number of Atoms from Mass for an Element

Solution

Exercise \(\PageIndex{5}\)

Example \(\PageIndex{6}\): Deriving Moles from Grams for a Compound

Solution

Exercise \(\PageIndex{6}\)

Example \(\PageIndex{7}\): Deriving Grams from Moles for a Compound

Solution

Exercise \(\PageIndex{7}\)

Example \(\PageIndex{8}\): Deriving the Number of Atoms and Molecules from the Mass of a Compound

Solution

Exercise \(\PageIndex{8}\)