# Empirical formulas molecular formulas problems and solutions

1. What is the empirical formula of a compound that has 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen by mass? Solution: Based on mass percentage, assume 100 g of the compound. This will give us 40.0 g of Carbon, 6.7 g of Hydrogen, and 53.3 g of Oxygen. Converting these to moles, we have about 3.33 moles of Carbon, 6.67 moles of Hydrogen, and 3.33 moles of Oxygen. Dividing each by the smallest number of moles (3.33), we get a ratio of 1:2:1. Hence, the empirical formula is CH₂O.

2. A compound has a molar mass of 180 g/mol and an empirical formula of CH₂O. What is its molecular formula? Solution: The empirical formula mass of CH₂O is 30 g/mol. The ratio of the molar mass to the empirical formula mass is 180/30 = 6. Therefore, the molecular formula is C₆H₁₂O₆.
3. What is the empirical formula of a compound that has 49.48% Carbon, 5.19% Hydrogen, and 45.33% Oxygen by mass? Solution: Assume 100 g of the compound to convert percentages to grams. Then convert grams to moles to get a ratio of 1:2:1. Hence, the empirical formula is C₄H₈O₄.
4. A compound has a molar mass of 78 g/mol and an empirical formula of CH. What is its molecular formula? Solution: The empirical formula mass of CH is 13 g/mol. The ratio of the molar mass to the empirical formula mass is 78/13 = 6. Therefore, the molecular formula is C₆H₆.
5. What is the empirical formula of a compound that has 87.5% Nitrogen and 12.5% Hydrogen by mass? Solution: Assume 100 g of the compound. This gives us 87.5 g of Nitrogen and 12.5 g of Hydrogen. Converting these to moles, we get a ratio of 6.25:12.5, which simplifies to 1:2. Hence, the empirical formula is NH₂.
6. A compound has a molar mass of 92 g/mol and an empirical formula of NO₂. What is its molecular formula? Solution: The empirical formula mass of NO₂ is 46 g/mol. The ratio of the molar mass to the empirical formula mass is 92/46 = 2. Therefore, the molecular formula is N₂O₄.
7. What is the empirical formula of a compound that has 72.4% Iron and 27.6% Oxygen by mass? Solution: Assume 100 g of the compound. This gives us 72.4 g of Iron and 27.6 g of Oxygen. Converting these to moles, we get a ratio of approximately 1:1.5. Multiplying the ratio by 2 to get whole numbers, the empirical formula is Fe₂O₃.
8. A compound has a molar mass of 342.3 g/mol and an empirical formula of C₆H₁₀O₅. What is its molecular formula? Solution: The empirical formula mass of C₆H₁₀O₅ is 162.15 g/mol. The ratio of the molar mass to the empirical formula mass is 342.3/162.15 = 2. Therefore, the molecular formula is C₁₂H₂₀O₁₀.
9. What is the empirical formula of a compound that has 36.84% Nitrogen and 63.16% Oxygen by mass? Solution: Assume 100 g of the compound. This gives us 36.84 g of Nitrogen and 63.16 g of Oxygen. Converting these to moles, we get a ratio of 2.63:3.94, which simplifies to 2:3. Hence, the empirical formula is N₂O₃.
10. A compound has a molar mass of 180.2 g/mol and an empirical formula of CH₂O. What is its molecular formula? Solution: The empirical formula mass of CH₂O is 30.03 g/mol. The ratio of the molar mass to the empirical formula mass is 180.2/30.03 = 6. Therefore, the molecular formula is C₆H₁₂O₆.
11. What is the empirical formula of a compound that has 60% Carbon, 13.33% Hydrogen, and 26.67% Oxygen by mass? Solution: Assume 100 g of the compound. This gives us 60 g of Carbon, 13.33 g of Hydrogen, and 26.67 g of Oxygen. Converting these to moles, we get a ratio of 5:12:1.67. This simplifies to 3:7:1 when divided by the smallest number of moles (1.67). Hence, the empirical formula is C₃H₇O.
12. A compound has a molar mass of 46 g/mol and an empirical formula of CH₃O. What is its molecular formula? Solution: The empirical formula mass of CH₃O is 31 g/mol. The ratio of the molar mass to the empirical formula mass is 46/31 = 1.5. Because this isn’t a whole number, the empirical formula is the same as the molecular formula, which is CH₃O.
13. What is the empirical formula of a compound that has 52.2% Carbon, 13.1% Hydrogen, and 34.7% Oxygen by mass? Solution: Assume 100 g of the compound. This gives us 52.2 g of Carbon, 13.1 g of Hydrogen, and 34.7 g of Oxygen. Converting these to moles, we get a ratio of 4.34:13.0:2.17, which simplifies to 2:6:1. Hence, the empirical formula is C₂H₆O.
14. A compound has a molar mass of 98 g/mol and an empirical formula of CH₂. What is its molecular formula? Solution: The empirical formula mass of CH₂ is 14 g/mol. The ratio of the molar mass to the empirical formula mass is 98/14 = 7. Therefore, the molecular formula is C₇H₁₄.
15. What is the empirical formula of a compound that has 41.39% Carbon, 3.47% Hydrogen, 8.29% Nitrogen, and 46.85% Oxygen by mass? Solution: Assume 100 g of the compound. This gives us 41.39 g of Carbon, 3.47 g of Hydrogen, 8.29 g of Nitrogen, and 46.85 g of Oxygen. Converting these to moles, we get a ratio of 3.44:3.44:0.59:2.92. This simplifies to 6:6:1:5 when divided by the smallest number of moles (0.59). Hence, the empirical formula is C₆H₆NO₅.
16. A compound has a molar mass of 132 g/mol and an empirical formula of C₂H₅O. What is its molecular formula? Solution: The empirical formula mass of C₂H₅O is 45 g/mol. The ratio of the molar mass to the empirical formula mass is 132/45 = 2.93. Because this isn’t a whole number, the empirical formula is the same as the molecular formula, which is C₂H₅O.
17. What is the empirical formula of a compound that has 71.43% Chlorine and 28.57% Carbon by mass? Solution: Assume 100 g of the compound. This gives us 71.43 g of Chlorine and 28.57 g of Carbon. Converting these to moles, we get a ratio of 2.02:2.38, which simplifies to 1:1. Hence, the empirical formula is CCl.
18. A compound has a molar mass of 180.2 g/mol and an empirical formula of C₂H₄O₂. What is its molecular formula? Solution: The empirical formula mass of C₂H₄O₂ is 60.05 g/mol. The ratio of the molar mass to the empirical formula mass is 180.2/60.05 = 3. Therefore, the molecular formula is C₆H₁₂O₆.
19. What is the empirical formula of a compound that has 30.4% Nitrogen and 69.6% Oxygen by mass? Solution: Assume 100 g of the compound. This gives us 30.4 g of Nitrogen and 69.6 g of Oxygen. Converting these to moles, we get a ratio of 2.17:4.35, which simplifies to 1:2. Hence, the empirical formula is NO₂.
20. A compound has a molar mass of 176 g/mol and an empirical formula of CH₂O. What is its molecular formula? Solution: The empirical formula mass of CH₂O is 30 g/mol. The ratio of the molar mass to the empirical formula mass is 176/30 = 5.87. Because this isn’t a whole number, the empirical formula is the same as the molecular formula, which is CH₂O.