Question

1. a compound is discovered that is 64.9% carbon, 13.5% hydrogen, and 21.6% oxygen. if its molar mass is 74 g/mol, what are its empirical and molecular formulas?

Explanation:

Step1: Assume 100g of compound

Mass of C: $64.9\ \text{g}$, H: $13.5\ \text{g}$, O: $21.6\ \text{g}$

Step2: Calculate moles of each element

Moles of C: $\frac{64.9\ \text{g}}{12.01\ \text{g/mol}} \approx 5.40\ \text{mol}$
Moles of H: $\frac{13.5\ \text{g}}{1.008\ \text{g/mol}} \approx 13.4\ \text{mol}$
Moles of O: $\frac{21.6\ \text{g}}{16.00\ \text{g/mol}} = 1.35\ \text{mol}$

Step3: Find mole ratio (divide by smallest)

Ratio C: $\frac{5.40}{1.35} = 4$
Ratio H: $\frac{13.4}{1.35} \approx 10$
Ratio O: $\frac{1.35}{1.35} = 1$

Step4: Determine empirical formula

Empirical formula: $\text{C}_4\text{H}_{10}\text{O}$

Step5: Calculate empirical formula mass

Empirical mass: $(4 \times 12.01) + (10 \times 1.008) + 16.00 = 74.12\ \text{g/mol}$

Step6: Find multiplier for molecular formula

Multiplier: $\frac{74\ \text{g/mol}}{74.12\ \text{g/mol}} \approx 1$

Answer:

Empirical formula: $\text{C}_4\text{H}_{10}\text{O}$
Molecular formula: $\text{C}_4\text{H}_{10}\text{O}$