Question

(true formulas)
name _______________
solve the problems below

1. the empirical formula of a compound is $ce{no_{2}}$. its molecular mass is 92 g/mol. what is its molecular formula?
2. the empirical formula of a compound is $ce{ch_{2}}$. its molecular mass is 70 g/mol. what is its molecular formula?
3. a compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. its molecular mass is 60. g/mol. what is its molecular formula?
4. a compound is 64.9% carbon, 13.5% hydrogen and 21.6% oxygen. its molecular mass is 74 g/mol. what is its molecular formula?
5. a compound is 54.5% carbon, 9.1% hydrogen and 36.4% oxygen. its molecular mass is 88 g/mol. what is its molecular formula?

Explanation:

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Problem 1

Step1: Calculate empirical formula mass

Empirical formula: $\text{NO}_2$
Mass = $14 + 2\times16 = 46$ g/mol

Step2: Find scaling factor $n$

$n = \frac{\text{Molecular Mass}}{\text{Empirical Mass}} = \frac{92}{46} = 2$

Step3: Multiply formula by $n$

Molecular formula = $2\times\text{NO}_2 = \text{N}_2\text{O}_4$

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Problem 2

Step1: Calculate empirical formula mass

Empirical formula: $\text{CH}_2$
Mass = $12 + 2\times1 = 14$ g/mol

Step2: Find scaling factor $n$

$n = \frac{\text{Molecular Mass}}{\text{Empirical Mass}} = \frac{70}{14} = 5$

Step3: Multiply formula by $n$

Molecular formula = $5\times\text{CH}_2 = \text{C}_5\text{H}_{10}$

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Problem 3

Step1: Convert % to moles (100g sample)

Moles of C: $\frac{40.0}{12} \approx 3.33$ mol
Moles of H: $\frac{6.7}{1} = 6.7$ mol
Moles of O: $\frac{53.3}{16} \approx 3.33$ mol

Step2: Find mole ratio (divide by smallest)

Ratio C:H:O = $\frac{3.33}{3.33}:\frac{6.7}{3.33}:\frac{3.33}{3.33} = 1:2:1$
Empirical formula: $\text{CH}_2\text{O}$

Step3: Calculate empirical formula mass

Mass = $12 + 2\times1 + 16 = 30$ g/mol

Step4: Find scaling factor $n$

$n = \frac{60}{30} = 2$

Step5: Multiply formula by $n$

Molecular formula = $2\times\text{CH}_2\text{O} = \text{C}_2\text{H}_4\text{O}_2$

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Problem 4

Step1: Convert % to moles (100g sample)

Moles of C: $\frac{64.9}{12} \approx 5.41$ mol
Moles of H: $\frac{13.5}{1} = 13.5$ mol
Moles of O: $\frac{21.6}{16} = 1.35$ mol

Step2: Find mole ratio (divide by smallest)

Ratio C:H:O = $\frac{5.41}{1.35}:\frac{13.5}{1.35}:\frac{1.35}{1.35} \approx 4:10:1$
Empirical formula: $\text{C}_4\text{H}_{10}\text{O}$

Step3: Calculate empirical formula mass

Mass = $4\times12 + 10\times1 + 16 = 74$ g/mol

Step4: Find scaling factor $n$

$n = \frac{74}{74} = 1$

Step5: Multiply formula by $n$

Molecular formula = $1\times\text{C}_4\text{H}_{10}\text{O} = \text{C}_4\text{H}_{10}\text{O}$

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Problem 5

Step1: Convert % to moles (100g sample)

Moles of C: $\frac{54.5}{12} \approx 4.54$ mol
Moles of H: $\frac{9.1}{1} = 9.1$ mol
Moles of O: $\frac{36.4}{16} \approx 2.28$ mol

Step2: Find mole ratio (divide by smallest)

Ratio C:H:O = $\frac{4.54}{2.28}:\frac{9.1}{2.28}:\frac{2.28}{2.28} \approx 2:4:1$
Empirical formula: $\text{C}_2\text{H}_4\text{O}$

Step3: Calculate empirical formula mass

Mass = $2\times12 + 4\times1 + 16 = 44$ g/mol

Step4: Find scaling factor $n$

$n = \frac{88}{44} = 2$

Step5: Multiply formula by $n$

Molecular formula = $2\times\text{C}_2\text{H}_4\text{O} = \text{C}_4\text{H}_8\text{O}_2$

Answer:

1. $\text{N}_2\text{O}_4$
2. $\text{C}_5\text{H}_{10}$
3. $\text{C}_2\text{H}_4\text{O}_2$
4. $\text{C}_4\text{H}_{10}\text{O}$
5. $\text{C}_4\text{H}_8\text{O}_2$