5.3 Formula Mass

Learning Objectives

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

Calculate formula masses for covalent and ionic compounds

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 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 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 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 2).

Figure 2. The average mass of an aspirin molecule is 180.15 amu. The model shows the molecular structure of aspirin, C₉H₈O₄.

Example 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?

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Check Your Learning

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?

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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).

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 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 non-significant 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 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?

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Check Your Learning

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?

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Key Concepts and 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 covalent compound is also called the molecular mass.

Exercises

What is the total mass (amu) of carbon in each of the following molecules?
1. CH₄
2. CHCl₃
3. C₁₂H₁₀O₆
4. CH₃CH₂CH₂CH₂CH₃
What is the total mass of hydrogen in each of the molecules?
1. CH₄
2. CHCl₃
3. C₁₂H₁₀O₆
4. CH₃CH₂CH₂CH₂CH₃
Calculate the molecular or formula mass of each of the following:
1. P₄
2. H₂O
3. Ca(NO₃)₂
4. CH₃CO₂H (acetic acid)
5. C₁₂H₂₂O₁₁ (sucrose, cane sugar).
Determine the molecular mass of the following compounds:
Determine the molecular mass of the following compounds:
Which molecule has a molecular mass of 28.05 amu?
Show Selected Answers
1. Each molecule has the following mass (amu) of carbon.
1. [latex]1\times 12.01\text{ amu}=12.01\text{ amu}[/latex]
2. [latex]1\times 12.01\text{ amu}=12.01\text{ amu}[/latex]
3. [latex]12\times 12.01\text{ amu}=144.12\text{ amu}[/latex]
4. [latex]5\times 12.01\text{ amu}=60.05\text{ amu}[/latex]
3. The molecular or formula masses are as follows:
1. [latex]4\times 30.974\text{ amu}=123.896\text{ amu}[/latex]
2. [latex]2\times 1.008\text{ amu}+15.999\text{ amu}=18.015\text{ amu}[/latex]
3. [latex]40\times 0.078\text{ amu}+2\times 14.007\text{ amu}+6\times 15.999\text{ amu}=164.086\text{ amu}[/latex]
4. [latex]2\times 12.011\text{ amu}+4\times 1.008\text{ amu}+2\times 15.999\text{ amu}=60.052\text{ amu}[/latex]
5. [latex]12\times 12.011\text{ amu}+22\times 1.008\text{ amu}\times 11\times 15.999\text{ amu}=342.297\text{ amu}[/latex]
5. The molecular mass of each compound is as follows:
1. C₄H₈[latex]\begin{array}{ll}4\text{C}\times 12.011\hfill & =48.044\text{amu}\hfill \\ 8\text{H}\times 1.0079\hfill & =\underline{8.06352\text{amu}}\hfill \\ \hfill & =56.107\text{amu}\hfill \end{array}[/latex]
2. C₄H₆[latex]\begin{array}{ll}4\text{C}\times 12.011\hfill & =48.044\text{amu}\hfill \\ 6\text{H}\times 1.0079\hfill & =\underline{6.0474\text{amu}}\hfill \\ \hfill & =54.091\text{amu}\hfill \end{array}[/latex]
3. H₂Si₂Cl₄[latex]\begin{array}{ll}\hfill 2\text{H}\times 1.0079& =2.01558\text{amu}\hfill \\ 2\text{Si}\times 28.0855\hfill & =56.1710\text{amu}\hfill \\ 4\text{Cl}\times 35.4527\hfill & =\underline{141.8108\text{amu}}\hfill \\ \hfill & =199.9976\text{amu}\hfill \end{array}[/latex]
4. H₃PO₄[latex]\begin{array}{ll}\hfill 3\text{H}\times 1.0079& =3.0237\text{amu}\hfill \\ 1\text{P}\times 30.973762\hfill & =30.973762\text{amu}\hfill \\ \hfill 4\text{O}\times 15.9994& =\underline{63.9976\text{amu}}\hfill \\ \hfill & =97.9950\text{amu}\hfill \end{array}[/latex]