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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 percentage of a hydrate's mass that is due to water.
• 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

### Recalling Prior Knowledge

• How is the mass of an element or 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 molar 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 7 g of protein, 15 g of fat, and 3 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 perform the following calculation:

\begin{align*}\dfrac{7 \ \text{g protein}}{32 \ \text{g}} \times 100\% = 22\% \ \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 percentages. The percent composition is the percent by mass of each element in a compound. It is calculated in a way that is similar to what 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 how the percent composition of a compound can be calculated 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 compound is decomposed, 16.07 g of pure zinc remains. Determine the percent composition of the compound.

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

Known

• total mass of sample = 20.00 g
• mass of Zn in the sample = 16.07 g

Unknown

• percent Zn = ? %
• percent O = ? %

First, subtract the mass of the zinc from the total mass to find the mass of oxygen in the sample. Then, divide each element’s mass by the total mass 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. A sample of a given compound contains 13.18 g of carbon and 3.32 g of hydrogen. What is the percent composition of this compound?
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 its chemical formula. The subscripts in the formula are first used to calculate the mass of each element found in one mole of the compound. That value is then 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 a Chemical Formula

Dichlorine heptoxide (Cl2O7) is a highly reactive compound used in some 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 a sample whose total mass is known. 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 total sample.

### 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 its anhydrous form by heating. Figure below shows that the anhydrous compound cobalt(II) chloride is blue, while its 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, cobalt(II) chloride hexahydrate (CoCl2•6 H2O).

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•6 H2O. The formula for water is set apart at the end of the formula with a dot, preceded by a coefficient that represents the number of water molecules per formula unit.

It is sometimes useful to know what percent of a hydrate's mass is water. Sample problem 10.13 demonstrates the procedure for finding this value.

Sample Problem 10.13: Percent of Water in a Hydrate

What percent of cobalt(II) chloride hexahydrate (CoCl2•6 H2O) is water?

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

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

Known

• mass of H2O in 1 mole of 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 from water molecules within the crystal.

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

## 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, only empirical formulas are used to describe ionic compounds. 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. In each molecule of ethene, there is 1 carbon atom for every 2 atoms of hydrogen. Similarly, we can also say that in one mole of ethene, there is 1 mole of carbon for every 2 moles of hydrogen. The subscripts in a formula represent the molar ratio of the elements in that compound.

Ball-and-stick model of ethene, C2H4.

In a procedure called elemental analysis, an unknown compound can be analyzed in the laboratory to determine the percentages of each element contained within it. These values can be used to find the molar ratios of the elements, which gives us 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 the smallest value obtained in step 2.
4. If all the values at this point are whole numbers (or very close), each number is equal to the subscript of the corresponding element in the empirical formula.
5. In some cases, one or more of the values calculated in step 3 will not be whole numbers. Multiply each of them by the smallest number that will convert all values into whole numbers. Note that all values must be multiplied by the same number so that the relative ratios are not changed. These values can then be used to 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?

Follow the steps 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 values 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 numbers can be multiplied by 2. The results are now close enough to be rounded to the nearest 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 molar ratio of the elements in the compound. The unknown compound is 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 tell us how many atoms of each element are present in one molecule of a molecular compound. In many cases, the molecular formula is the same as the empirical formula. For example, the molecular formula of methane is CH4, and because 1:4 is the smallest whole-number ratio that can be written for this compound, 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 gives vinegar its distinctive taste and smell. Its molecular formula is C2H4O2. Glucose is a simple sugar that cells use as their 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, 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 also 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 of 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 that contains 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 its 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 determining 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 contains one or more water molecules in the crystal lattice for each formula unit. The percentage of a hydrate's mass that is composed of water can be calculated from its formula.
• Percent composition data can be used to determine a compound’s empirical formula, which is the molar ratio between the elements in the compound.
• The empirical formula and the molar mass of a substance can be used to determine its molecular formula, which is the number of each kind of atom in a single molecule of the compound.

## Lesson Review Questions

### Reviewing Concepts

1. How many moles of aluminum, sulfur, and oxygen atoms 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 molecular compounds.
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. A sample of a certain binary compound contains 6.93 g of silicon and 7.89 g of oxygen. What is the percent composition of the compound?
2. Calculate the percent composition for each 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 problem 6, calculate the mass of the indicated element in each of the following samples.
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 by mass?
1. KNO3
2. NO2
3. NH4Cl
4. Li3N
5. Find the percentage of the total mass that is due to water for each of the following hydrates.
1. ZnSO4•7 H2O
2. Mn(NO3)2•4 H2O
6. Find the empirical formulas of compounds with the following percent compositions.
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, 1.390 g O, and no other elements. Find its molecular formula.
9. 12.50 g of a hydrated form of copper(II) sulfate, CuSO4x H2O (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 value of 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.

• What are some ways to classify different types of chemical reactions?
• How are calculations with moles, grams, and volume involved in the analysis of chemical reactions?

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