Question

a 58.0 g sample of unknown composition was carefully analyzed and found to consist of 24.3 g of carbon, 3.072 grams of hydrogen, 5.406 g of oxygen, and the remainder of nitrogen. determine the empirical formula of the unknown compound.

Explanation:

Step1: Calculate mass of nitrogen

Mass of N = $58.0 - 24.3 - 3.072 - 5.406 = 25.222$ g

Step2: Convert mass to moles

Moles of C: $\frac{24.3}{12.01} \approx 2.023$ mol
Moles of H: $\frac{3.072}{1.008} = 3.048$ mol
Moles of O: $\frac{5.406}{16.00} \approx 0.3379$ mol
Moles of N: $\frac{25.222}{14.01} \approx 1.800$ mol

Step3: Divide by smallest mole value

Divide all by 0.3379:
C: $\frac{2.023}{0.3379} \approx 6$
H: $\frac{3.048}{0.3379} \approx 9$
O: $\frac{0.3379}{0.3379} = 1$
N: $\frac{1.800}{0.3379} \approx 5$

Answer:

$\text{C}_6\text{H}_9\text{O}\text{N}_5$