# Quiz - Empirical Formula | Fundamentals of Chemical Reactions

1

Analysis of a compound known to contain only Mg, P, and O gives this analysis: 21.8% Mg, 27.7% P, 50.3% O. What is its empirical formula?

1. Convert mass percentages to moles:

• Assume 100 g of the compound.
• Mass of Mg = 21.8 g, Mass of P = 27.7 g, Mass of O = 50.3 g

2. Calculate the number of moles of each element:

• Moles of Mg:  ​≈ 0.897 mol
• Moles of P:  ≈ 0.895 mol
• Moles of O:  ​≈ 3.144 mol

3. Determine the simplest whole-number ratio:

• Divide each by the smallest number of moles calculated (0.895).
• Mg: $\frac{0.897}{0.895}$ ​≈ 1
• P: $\frac{0.895}{0.895}$ ≈ 1
• O: $\frac{3.144}{0.895}$ ≈ 3.51

4. Simplify the ratio:

• The ratio of Mg:P:O is approximately 1:1:3.51.
• To get whole numbers, multiply each by 2 to approximate 2:2:7.

5. Empirical formula:

• The simplest whole-number ratio is approximately Mg2​P2​O7​.
2

A chloride salt of rhenium contains 63.6% Re by mass. What is its empirical formula?

1. Assume you have 100 g of the compound.

• Mass of Re = 63.6 g
• Mass of Cl = 100 g - 63.6 g = 36.4 g

2. Convert these masses to moles using the atomic masses of Re and Cl.

• Moles of Re:  ​≈ 0.341 mol
• Moles of Cl:  ≈ 1.027 mol

3. Determine the simplest whole-number ratio of the elements.

• Ratio of Re to Cl: $\frac{0.341}{0.341}$ ≈ 1
• Ratio of Cl to Re: $\frac{1.027}{0.341}$ ≈ 3

The simplest whole-number ratio of Re to Cl is approximately 1:3, indicating the empirical formula is ReCl3

3

An oxide of manganese contains 2.29 g of manganese per gram of oxygen. What is the empirical formula of this compound?

1. Given Data:

• Mass of manganese per gram of oxygen: 2.29 g Mn / 1 g O

2. Convert masses to moles:

• Moles of Mn:  ≈ 0.341 mol
• Moles of Cl:  ≈ 1.027 mol

3. Determine the simplest whole-number ratio of the elements.

• Ratio of Re to Cl: $\frac{0.341}{0.341}$ ≈ 1
• Ratio of Cl to Re: $\frac{1.027}{0.341}$ ≈ 3

The simplest whole-number ratio of Re to Cl is approximately 1:3, indicating the empirical formula is ReCl3

4

The percentages by mass of C, H, and Cl in a compound are C 52.2%, H 3.7%, and Cl 44.1%. How many carbon atoms are in the simplest formula of the compound?

1. Assume 100 g of the compound:

• Mass of C = 52.2 g
• Mass of H = 3.7 g
• Mass of Cl = 44.1 g

2. Convert these masses to moles:

• Moles of C:  ​≈ 4.35 mol
• Moles of H:  ≈ 3.67 mol
• Moles of Cl:  ​≈ 1.24 mol

3. Determine the simplest whole-number ratio:

• Divide each by the smallest number of moles calculated (1.24).
• C: $\frac{4.35}{1.24}$ ​≈ 3.51
• H: $\frac{3.67}{1.24}$ ≈ 2.96
• Cl: $\frac{1.24}{1.24}$ = 1

4. Simplify the ratio:

• The ratio of C : H : Cl is approximately 3.5 : 3 : 1.
• To get whole numbers, multiply each by 2 to approximate 7 : 6 : 2.

5. Empirical formula:

• The simplest whole-number ratio is approximately C7H6Cl2​.
5

The mass percentages in a compound are carbon 57.48%, hydrogen 4.22% and oxygen 38.29%. What is its empirical formula?

1. Assume 100 g of the compound:

• Mass of C = 57.48 g
• Mass of H = 4.22 g
• Mass of O = 38.29 g

2. Convert these masses to moles:

• Moles of C:  ​≈ 4.79 mol
• Moles of H:  ≈ 4.19 mol
• Moles of O:  ​≈ 2.39 mol

3. Determine the simplest whole-number ratio:

• Divide each by the smallest number of moles calculated (2.39).
• C: $\frac{4.79}{2.39}$ ​≈ 2
• H: $\frac{4.19}{2.39}$ ≈ 1.75
• O: $\frac{2.39}{2.39}$ = 1

4. Simplify the ratio:

• The ratio of C : H : O is approximately 2 : 1.75 : 1.
• To get whole numbers, multiply each by 4 to approximate 8 : 7 : 4.

5. Empirical formula:

• The simplest whole-number ratio is approximately C8H7O4​.