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# 10.3: Chemical Formulas

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Calculate the percent composition of a compound either from mass data or from the chemical formula. Use percent composition to calculate the mass of an element in a certain sample of a compound.
• Calculate the percent of water in a hydrate.
• Determine the empirical formula of a compound from percent composition data.
• Determine the molecular formula of a compound from the empirical formula and the molar mass.

## Lesson Vocabulary

• hydrate
• percent composition

## Check Your Understanding

### Recalling Prior Knowledge

• How are the grams of an element or of a compound converted to moles?
• What is an empirical formula? What is a molecular formula and how does it relate to an empirical formula?

In previous chapters, you have learned about chemical nomenclature – naming compounds and writing correct chemical formulas. In this lesson, you will learn how the subscripts in a chemical formula represent the mole ratio between the elements in a compound.

## Percent Composition

Packaged foods that you eat typically have nutritional information provided on the label. The label of a popular brand of peanut butter (Figure below) reveals that one serving size is considered to be 32 g. The label also gives the masses of various types of compounds that are present in each serving. One serving contains 8 g of protein, 16 g of fat, and 1 g of sugar. This information can be used to determine the composition of the peanut butter on a percent by mass basis. For example, to calculate the percent of protein in the peanut butter, we could do the following:

\begin{align*}\dfrac{8 \ \text{g protein}}{32 \ \text{g}} \times 100\% = 25\% \ \text{protein}\end{align*}

Foods like peanut butter provide nutritional information on the label in the form of masses of different types of compounds present per serving.

In a similar way, chemists often need to know what elements are present in a compound and in what percentage. The percent composition is the percent by mass of each element in a compound. It is calculated in a similar way that we just saw for the peanut butter.

\begin{align*}\% \ \text{by mass} = \frac{\text{mass of element}}{\text{mass of compound}} \times 100\%\end{align*}

### Percent Composition from Mass Data

Sample Problem 10.11 shows the calculation of the percent composition of a compound based on mass data.

Sample Problem 10.11: Percent Composition from Mass

A certain newly synthesized compound is known to contain the elements zinc and oxygen. When a 20.00 g sample of the sample is decomposed, 16.07 g of zinc remains. Determine the percent composition of the compound.

Step 1: List the known quantities and plan the problem.

Known

• mass of compound = 20.00 g
• mass of Zn = 16.07 g

Unknown

• percent Zn = ? %
• percent O = ? %

Subtract to find the mass of oxygen in the compound. Divide each element’s mass by the mass of the compound to find the percent by mass.

Step 2: Calculate

Mass of oxygen = 20.00 g – 16.07 g = 3.93 g O

\begin{align*} & \% \ \text{Zn} = \frac{16.07 \ \text{g Zn}}{20.00 \ \text{g}} \times 100\% = 80.35\% \ \text{Zn} \\ & \% \ \text{O} = \frac{3.93 \ \text{g O}}{20.00 \ \text{g}} \times 100\% = 19.65\% \ \text{O} \end{align*}

The calculations make sense because the sum of the two percentages adds up to 100%. By mass, the compound is mostly zinc.

Practice Problems
1. What is the percent composition of a compound that consists of 13.18 g of carbon and 3.32 g of hydrogen?
2. 5.00 g of aluminum is reacted with 7.00 g of fluorine to form a compound. When the compound is isolated, its mass is found to be 10.31 g, with 1.69 g of aluminum left unreacted. Determine the percent composition of the compound.

### Percent Composition from a Chemical Formula

The percent composition of a compound can also be determined from the formula of the compound. The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound. That is divided by the molar mass of the compound and multiplied by 100%.

\begin{align*}\% \ \text{by mass} = \frac{\text{mass of element in 1 mol}}{\text{molar mass of compound}} \times 100\%\end{align*}

The percent composition of a given compound is always the same as long as the compound is pure.

Sample Problem 10.12: Percent Composition from Chemical Formula

Dichlorine heptoxide (Cl2O7) is a highly reactive compound used in some organic synthesis reactions. Calculate the percent composition of dichlorine heptoxide.

Step 1: List the known quantities and plan the problem.

Known

• mass of Cl in 1 mol Cl2O7 = 70.90 g
• mass of O in 1 mol Cl2O7 = 112.00 g
• molar mass of Cl2O7 = 182.90 g/mol

Unknown

• percent Cl = ? %
• percent O = ? %

Calculate the percent by mass of each element by dividing the mass of that element in 1 mole of the compound by the molar mass of the compound and multiplying by 100%.

Step 2: Calculate

\begin{align*} &\% \ \text{Cl} = \frac{70.90 \ \text{g Cl}}{182.90 \ \text{g}} \times 100\% = 38.76\% \ \text{Cl} \\ &\% \ \text{O} = \frac{112.00 \ \text{g O}}{182.90 \ \text{g}} \times 100\% = 61.24\% \ \text{O} \end{align*}

The percentages add up to 100%.

Practice Problem
1. Calculate the percent composition of the following compounds:
1. magnesium fluoride, MgF2
2. silver nitrate, AgNO3

Percent composition can also be used to determine the mass of a certain element that is contained in any mass of a compound. In the previous sample problem, it was found that the percent composition of dichlorine heptoxide is 38.76% Cl and 61.24% O. Suppose that you needed to know the masses of chlorine and oxygen present in a 12.50 g sample of dichlorine heptoxide. You can set up a conversion factor based on the percent by mass of each element.

\begin{align*} & 12.50 \ \text{g Cl}_2\text{O}_7 \times \frac{38.76 \ \text{g Cl}}{100 \ \text{g Cl}_2\text{O}_7} = 4.845 \ \text{g Cl} \\ & 12.50 \ \text{g Cl}_2\text{O}_7 \times \frac{61.24 \ \text{g O}}{100 \ \text{g Cl}_2\text{O}_7} = 7.655 \ \text{g O} \end{align*}

The sum of the two masses is 12.50 g, the mass of the sample size.

### Percent of Water in a Hydrate

Many ionic compounds naturally contain water as part of the crystal lattice structure. A hydrate is a compound that has one or more water molecules bound to each formula unit. Ionic compounds that contain a transition metal are often highly colored. Interestingly, it is common for the hydrated form of a compound to be of a different color than the anhydrous form, which has no water in its structure. A hydrate can usually be converted to the anhydrous compound by heating. Figure below show the anhydrous compound cobalt(II) chloride to be blue while the hydrate is a distinctive magenta color.

On the left is anhydrous cobalt(II) chloride, CoCl2. On the right is the hydrated form of the compound called cobalt(II) chloride hexahydrate, CoCl2•6H2O.

The hydrated form of cobalt(II) chloride contains six water molecules in each formula unit. The name of the compound is cobalt(II) chloride hexahydrate and its formula is CoCl2•6H2O. The formula for water is set apart at the end of the formula with a dot, followed by a coefficient that represents the number of water molecules per formula unit.

It is useful to know the percent of water contained within a hydrate. Sample problem 10.13 demonstrates the procedure.

Sample Problem 10.13: Percent of Water in a Hydrate

Find the percent water in cobalt(II) chloride hexahydrate, CoCl2•6H2O.

Step 1: List the known quantities and plan the problem.

The mass of water in the hydrate is the coefficient (6) multiplied by the molar mass of H2O. The molar mass of the hydrate is the molar mass of the CoCl2 plus the mass of water.

Known

• mass of H2O in 1 mol hydrate = 108.12 g
• molar mass of hydrate = 237.95 g/mol

Unknown

• percent H2O = ? %

Calculate the percent by mass of water by dividing the mass of H2O in 1 mole of the hydrate by the molar mass of the hydrate and multiplying by 100%.

Step 2: Calculate

\begin{align*}\% \ \text{H}_2\text{O} = \frac{108.12 \ \text{g H}_2\text{O}}{237.95 \ \text{g}} \times 100\% = 45.44\% \ \text{H}_2\text{O}\end{align*}

Nearly half of the mass of the hydrate is composed of water molecules within the crystal.

Practice Problem
1. Gypsum is a soft mineral used as in plaster and is composed of calcium sulfate dihydrate. Calculate the percent of water in calcium sulfate dihydrate, CaSO4•2H2O.

## Empirical Formulas

Recall that an empirical formula is one that shows the lowest whole-number ratio of the elements in a compound. Because the structure of ionic compounds is an extended three-dimensional network of positive and negative ions, all formulas of ionic compounds are empirical. However, we can also consider the empirical formula of a molecular compound. Ethene is a small hydrocarbon compound with the formula C2H4 (Figure below). While C2H4 is its molecular formula and represents its true molecular structure, it has an empirical formula of CH2. The simplest ratio of carbon to hydrogen in ethene is 1:2. There are two ways to view that ratio. Considering one molecule of ethene, the ratio is 1 carbon atom for every 2 atoms of hydrogen. Considering one mole of ethene, the ratio is 1 mole of carbon for every 2 moles of hydrogen. So the subscripts in a formula represent the mole ratio of the elements in that formula.

Ball-and-stick model of ethene, C2H4.

In a procedure called elemental analysis, an unknown compound can be analyzed in the laboratory in order to determine the percentages of each element contained within it. These percentages can be transformed into the mole ratio of the elements, which leads to the empirical formula. The steps to be taken are outlined below.

1. Assume a 100 g sample of the compound so that the given percentages can be directly converted into grams.
2. Use each element’s molar mass to convert the grams of each element to moles.
3. In order to find a whole-number ratio, divide the moles of each element by whichever of the moles from step 2 is the smallest.
4. If all the moles at this point are whole numbers (or very close), the empirical formula can be written with the moles as the subscript of each element.
5. In some cases, one or more of the moles calculated in step 3 will not be whole numbers. Multiply each of the moles by the smallest whole number that will convert each into a whole number. Write the empirical formula.

Sample Problem 10.14: Determining the Empirical Formula of a Compound

A compound of iron and oxygen is analyzed and found to contain 69.94% iron and 30.06% oxygen. Find the empirical formula of the compound.

Step 1: List the known quantities and plan the problem.

Known

• % of Fe = 69.94%
• % of O = 30.06%

Unknown

• Empirical formula = Fe?O?

Steps to follow are outlined in the text.

Step 2: Calculate

1. Assume a 100 g sample.

→ 69.94 g Fe

→ 30.06 g O

2. Convert to moles.

\begin{align*} & 69.94 \ \text{g Fe} \times \frac{1 \ \text{mol Fe}}{55.85 \ \text{g Fe}} = 1.252 \ \text{mol Fe} \\ & 30.06 \ \text{g O} \times \frac{1 \ \text{mol O}}{16.00 \ \text{g O}} = 1.879 \ \text{mol O} \end{align*}

3. Divide both moles by the smallest of the results.

\begin{align*} & \frac{1.252 \ \text{mol Fe}}{1.252} = 1 \ \text{mol Fe} \\ & \frac{1.879 \ \text{mol O}}{1.252} = 1.501 \ \text{mol O} \end{align*}

4. Since the moles of O is still not a whole number, both moles can be multiplied by 2, while rounding to a whole number.

\begin{align*} & 1 \ \text{mol Fe} \times 2 = 2 \ \text{mol Fe} \\ & 1.501 \ \text{mol O} \times 2 = 3 \ \text{mol} O \end{align*}

The empirical formula of the compound is Fe2O3.

The subscripts are whole numbers and represent the mole ratio of the elements in the compound. The compound is the ionic compound iron(III) oxide.

Practice Problem
1. Calculate the empirical formula of each compound from the percentages listed.
1. 63.65% N, 36.35% O
2. 81.68% C, 18.32% H

## Molecular Formulas

Molecular formulas give the kind and number of atoms of each element present in a molecular compound. In many cases, the molecular formula is the same as the empirical formula. The molecular formula of methane is CH4 and because it contains only one carbon atom, that is also its empirical formula. Sometimes, however, the molecular formula is a simple whole-number multiple of the empirical formula. Acetic acid is an organic acid that is the main component of vinegar. Its molecular formula is C2H4O2. Glucose is a simple sugar that cells use as a primary source of energy. Its molecular formula is C6H12O6. The structures of both molecules are shown in Figure below. They are very different compounds, yet both have the same empirical formula of CH2O.

Acetic acid (left) has a molecular formula of C2H4O2, while glucose (right) has a molecular formula of C6H12O6. Both have the empirical formula CH2O.

Empirical formulas can be determined from the percent composition of a compound. In order to determine its molecular formula, it is necessary to know the molar mass of the compound. Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds. In order to go from the empirical formula to the molecular formula, follow these steps:

1. Calculate the empirical formula mass (EFM), which is simply the molar mass represented by the empirical formula.
2. Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number.
3. Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.

Sample Problem 10.15: Determining the Molecular Formula of a Compound

The empirical formula of a compound of boron and hydrogen is BH3. Its molar mass is 27.7 g/mol. Determine the molecular formula of the compound.

Step 1: List the known quantities and plan the problem.

Known

• empirical formula = BH3
• molar mass = 27.7 g/mol

Unknown

• molecular formula = ?

Step 2: Calculate

1. The empirical formula mass (EFM) = 13.84 g/mol
2. \begin{align*}\dfrac{\text{molar mass}}{\text{EFM}} = \dfrac{27.7}{13.84} = 2\end{align*}
3. BH3 × 2 = B2H6

The molecular formula of the compound is B2H6.

The molar mass of the molecular formula matches the molar mass of the compound.

Practice Problems
1. A compound with the empirical formula CH has a molar mass of 78 g/mol. Determine the molecular formula.
2. A compound is found to consist of 43.64% phosphorus and 56.36% oxygen. The molar mass of the compound is 284 g/mol. Find the molecular formula of the compound.

You can watch a video lecture about molecular and empirical formulas at http://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiometry/v/molecular-and-empirical-formulas.

You can watch a video lecture about molecular and empirical formulas from percent composition at http://www.khanacademy.org/science/physics/thermodynamics/v/molecular-and-empirical-forumlas-from-percent-composition.

## Lesson Summary

• The percent composition of a compound is the percent by mass of each of the elements in the compound. It can be calculated from mass data or from the chemical formula.
• A hydrate is an ionic compound that has one or more water molecules bound to each formula unit and part of the crystal lattice. The percent of water in a hydrate can be calculated from its formula.
• Percentage composition data can be used to determine a compound’s empirical formula, the mole ratio between the elements in the compound.
• The empirical formula and the molar mass can be used to determine the molecular formula, the actual number of each kind of atom in the molecule.

## Lesson Review Questions

### Reviewing Concepts

1. How many moles of atoms of each element are there in one mole of Al2(SO4)3?
2. How can a hydrate be converted into its anhydrous form?
3. Give the empirical formula for each of the following.
1. C8H18
2. H2O2
3. N2O4
4. C5H12
4. What is the relationship between a compound’s empirical formula and its molecular formula?

### Problems

1. What is the percent composition of a compound that contains 6.93 g of silicon and 7.89 g of oxygen?
2. Calculate the percent composition of the following compounds.
1. potassium bromide, KBr
2. ammonium chloride, NH4Cl
3. acetone, C3H6O
4. barium phosphate, Ba3(PO4)2
3. Using the answers from number 6, calculate the mass in each of the following.
1. potassium in 4.23 g of KBr
2. chlorine in 126 g of NH4Cl
3. carbon in 41.0 g of C3H6O
4. phosphorus in 39.6 g of Ba3(PO4)2
4. Which compound has the highest nitrogen content?
1. KNO3
2. NO2
3. NH4Cl
4. Li3N
5. Find the percentages of water in the following hydrates.
1. ZnSO4•7H2O
2. Mn(NO3)2•4H2O
6. Find the empirical formulas of compounds with the given percentages.
1. 38.35% Cl and 61.65% F
2. 59.35% Sr, 8.135% C, and 32.51% O
7. Find the molecular formula of each compound, given its empirical formula and molar mass.
1. CH2O, 120 g/mol
2. C2HCl, 181.5 g/mol
8. The molar mass of a compound is 92 g/mol. Analysis of a sample of the compound indicates that it contains 0.606 g N and 1.390 g O. Find its molecular formula.
9. 12.50 g of a hydrated form of copper(II) sulfate, CuSO4•xH2O (where x is unknown) is gently heated. When all the water has been driven off, the mass of the anhydrous copper(II) sulfate is found to be 7.99 g.
1. What is the mass of the water that was lost as a result of the heating?
2. Convert this mass of water to moles of water.
3. Convert the mass of the anhydrous CuSO4 to moles.
4. Divide the answer to b by the answer to c. This is the x in the formula of the hydrate. Write the formula of hydrated copper(II) sulfate.

You can watch video lectures about formulas from mass composition at http://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiometry/v/formula-from-mass-composition and another mass composition problem at http://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiometry/v/another-mass-composition-problem.

## Points to Consider

Chemical reactions are the essence of chemistry. We will describe chemical reactions both qualitatively and quantitatively.

• Can chemical reactions be classified according to different types of reactions?
• How will calculations with moles, grams, and volume be involved in the analysis of chemical reactions?

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