1. 10.3 Percent Composition and Chemical Formulas
Read the lesson title aloud to students.

2. Percent Composition The % composition of a compound describes the relative amounts of the elements in the compound.
Review with students examples of percentages and how one calculates the percent of a number.
Tell students: The percent composition of a compound describes the relative amounts of different elements in the compound.
Ask: What must be true about the sum of the masses of all the elements in a compound?
Answer: It must equal the mass of the compound.
Ask: Does it matter how big a sample you use for the percent composition calculation?
Answer: No.
Click to reveal the pie charts comparing potassium chromate and potassium dichromate.
Point out that these compounds contain the same three elements, but the percent compositions of the compounds are different.
Ask: Why is the ratio of the percent by mass of potassium and chromium in potassium dichromate not 1:1?
Answer: The percent by mass is based on mass, and potassium and chromium have different atomic masses.

3. Calculating Percent Composition
When a gram sample of a compound containing only magnesium and oxygen is decomposed, 5.40 grams of oxygen is obtained. What is the percent composition of this compound?
Have students identify the knowns and unknowns for the problem.
Click to reveal the knowns and unknowns.
Ask: Is any additional information needed to determine the percent by mass of oxygen? If so, what information is needed?
Answer: No additional information is needed.
Ask: Is any additional information needed to determine the percent by mass of magnesium? If so, what information is needed?
Answer: Yes; the mass of magnesium in the compound is needed.
Ask: How could this additional information be determined?
Answer: Because magnesium is the only other element in the compound, its mass can be determined by subtracting the mass of oxygen in the compound from the total mass.

4. Calculating Percent Composition, cont.
When a gram sample of a compound containing only magnesium and oxygen is decomposed, 5.40 grams of oxygen is obtained. What is the percent composition of this compound? =
Ask: What equation can be used to solve this problem?
Answer: the equation that expresses the percent by mass of an element in terms of the mass of the element and the mass of the compound
Click to reveal the equation.
Walk students through substituting the known values into the equation.
Click to reveal the equation for calculation of percent by mass of oxygen.
Ask a volunteer to write the percent by mass of oxygen.
Click to reveal the correct answer.
Walk students through how to calculate the mass of magnesium through subtraction.
Click to reveal the equation for calculating percent by mass of magnesium.
Ask a volunteer to write the percent by mass of magnesium.
Ask: In what unit of measure is the answer expressed?
Answer: The answer is a dimensionless quantity.
Explain to students that a percent sign is not a symbol that represents a unit. The symbol simply means “per hundred.”
39.7% O =
60.3% Mg

5. Percent Composition From a Formula
Calculate the percent composition of propane (C3H8).
Tell students: It is also possible to calculate percent composition from a chemical formula.
Tell students: In the last problem, we were given information about the mass of a specific sample of a compound. In this example, we are given only the chemical formula.
Point out that since percent composition is dimensionless, we can assume any mass of a compound we want. Therefore, we can calculate percent by mass based on the mass of one mole of the compound.
Click to reveal the equation for calculating percent by mass based on molar mass.
Ask: What information has to be derived from the molecular formula of propane to use this equation to solve the problem?
Answer: the mass of carbon and the mass of hydrogen in one mole of the compound
Ask: What information is needed from the periodic table?
Answer: the molar masses of carbon and hydrogen
Have students identify the knowns and unknowns for the problem.
Click to reveal the knowns and unknowns.

6. Calculating Percent Composition: Propane
Calculate the percent composition of propane (C3H8). =
81.8% C
Walk students through substituting the known values into the equation.
Click to reveal the equation for calculation of percent by mass of carbon.
Ask a volunteer to write the percent by mass of carbon.
Click to reveal the correct answer.
Click to reveal the equation for calculating percent by mass of hydrogen.
Ask a volunteer to write the percent by mass of hydrogen. =
18% H

7. Mass From Percent Composition
Calculate the mass of carbon and the mass of hydrogen in grams of propane (C3H8).
Point out that propane is the same compound that was used in the previous example.
Ask: Why is the percent composition of hydrogen not necessary to solve this problem?
Answer: The problem asks only for the mass of carbon in propane. Only the total mass of the propane sample and the percent composition of carbon in propane are necessary for this calculation.
Have students identify the knowns and unknowns for the problem.
Click to reveal the knowns and unknowns.

8. Mass From Percent Composition, cont.
Calculate the mass of carbon and the mass of hydrogen in grams of propane (C3H8).
Remind students that in the previous problem, they calculated that the percent by mass of carbon in propane is 81.8 percent.
Ask: If the percent by mass of carbon is 81.8 percent, how many grams of carbon are there in 100 grams of propane?
Answer: 81.8 grams
Point out to students that since there are 81.8 grams of carbon in 100 grams of propane, the ratio 81.8 g C/100 g propane is equal to 1 and can be used as a conversion factor.
Click to reveal the conversion factor.
Review with students that the same reasoning can be used to identify the conversion factor for the mass of hydrogen.
Review with students how to set up the calculation for the mass of carbon in the sample.
Click to reveal the calculation.
Ask a volunteer to describe how to set up the calculation for hydrogen.
Click to reveal the equation.
Ask a volunteer to write the mass of hydrogen.
Click to reveal the correct answer.
15 g H

9. Empirical Formulas
An empirical formula gives the lowest whole-number ratio of the atoms of the elements in a compound. This is not usually the real formula, it is a mathematical formula.
Review the definition of empirical formula with students.
Ask: What is the ratio of carbon to hydrogen in ethyne?
Answer: 1:1
Ask: What is the ratio of carbon to hydrogen in styrene?
Point out that the ratio of carbon to hydrogen is the same in both ethyne and styrene, so these two compounds have the same empirical formula even though they are different compounds.
Ask: What is the empirical formula for ethyne and styrene?
Answer: CH
Ethyne (C2H2)
Styrene (C8H8)

10. Calculating Empirical Formula
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound?
Have students identify the knowns and unknowns for the problem.
Click to reveal the knowns and unknowns.
Ask: What are some possible empirical formulas for the compound analyzed in this problem?
Answer: NO, NO2, NO3, N2O, N4O, N2O3, N2O5

11. Calculating Empirical Formula, cont.
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound?
Ask: How is the mass of each element converted into moles?
Answer: The mass of each compound is divided by the molar mass of the compound.
Click to reveal the calculation for the number of moles of each element.
Explain that the number of moles found in this first step represents the ratio of atoms in the compound.
Ask: Why is 1.85:4.63 not an empirical formula for the compound?
Answer: The empirical formula is the lowest whole-number ratio of elements. This ratio must be converted into a ratio of two whole numbers.
Click to reveal how to convert the initial molar amounts into a ratio of a number to one.
Ask: How can you convert 2.5 into a whole number?
Answer: multiply by 2
Click to reveal the calculation for the mole ratio and the empirical formula.

12. Molecular Formulas
The molecular formula for a compound shows the actual numbers of atoms in a molecule of the compound.
Use the figure to explain how different substances can have the same empirical formula. Point out that the molecular formula of a compound is not always the lowest whole-number ratio of its constituent elements. For example, ethanoic acid has the molecular formula C2H4O2, but the subscripts must be divided by two to yield the empirical formula, CH2O.
Point out that the three compounds shown all have the same empirical formula, but they are distinct compounds.

13. Calculating Molecular Formula
Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N.
Have students identify the knowns and unknowns for the problem.
Click to reveal the knowns and unknowns.
Ask: What are some possible molecular formulas for the compound analyzed in this problem?
Answer: CH4N, C2H8N2, C3H12N3

14. Calculating Molecular Formula, cont.
Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N.
Review the concept of empirical formula mass (efm) to students.
Click to reveal how to calculate the efm for the compound.
Ask: Is the efm less than or equal to the molar mass of the compound?
Answer: less than the molar mass
Ask: Could the empirical formula mass ever be greater than the molar mass of a compound?
Answer: No.
Ask: By what factor do you have to multiply the efm by to get the molar mass of the compound?
Answer: 2
Click to reveal the ratio of the molar mass to the efm.
Point out that if the molar mass is twice the efm, then the molecular formula must contain twice as many atoms as the empirical formula.
Click to reveal the molecular formula calculation.