a sample of a compound contains 32.0 g c and 8.0 g h. its molar mass is 30.0 g/mol. what is the compounds…

Answer:

C. $C_2H_6$

Explanation:

Step1: Calculate moles of C and H

Moles of C: $n_{C}=\frac{32.0\ g}{12.0\ g/mol}= 2.67\ mol$ Moles of H: $n_{H}=\frac{8.0\ g}{1.0\ g/mol}=8.0\ mol$

Step2: Find mole - ratio

Divide by the smaller number of moles (2.67 mol). For C: $\frac{2.67\ mol}{2.67\ mol}=1$ For H: $\frac{8.0\ mol}{2.67\ mol}\approx3$ Empirical formula is $CH_3$.

Step3: Determine molecular formula

Empirical - formula mass of $CH_3$ is $12 + 3\times1=15\ g/mol$. Let $n=\frac{\text{Molar mass}}{\text{Empirical - formula mass}}=\frac{30.0\ g/mol}{15\ g/mol}=2$. Molecular formula is $(CH_3)_2 = C_2H_6$.