Question

1. what is the empirical formula for a compound that is 83.7 % carbon and 16.3 % hydrogen?

Explanation:

Step1: Assume 100g of compound

Mass of C = 83.7 g, Mass of H = 16.3 g

Step2: Convert mass to moles

Moles of C: $\frac{83.7\ \text{g}}{12.01\ \text{g/mol}} \approx 6.97\ \text{mol}$
Moles of H: $\frac{16.3\ \text{g}}{1.008\ \text{g/mol}} \approx 16.17\ \text{mol}$

Step3: Divide by smallest mole value

Ratio of C: $\frac{6.97}{6.97} = 1$
Ratio of H: $\frac{16.17}{6.97} \approx 2.32 \approx \frac{7}{3}$

Step4: Scale to whole numbers

Multiply both ratios by 3: C = $1 \times 3 = 3$, H = $\frac{7}{3} \times 3 = 7$

Answer:

$\text{C}_3\text{H}_7$