a sample of a compound contains 60.0 g c and 5.05 g h. its molar mass is 78.12 g/mol. what is the compounds…

Explanation:

Step1: Calculate moles of C and H

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{60.0\ g}{12.01\ g/mol}\approx5.0\ mol$ $n_{H}=\frac{m_{H}}{M_{H}}=\frac{5.05\ g}{1.01\ g/mol}\approx5.0\ mol$

Step2: Determine the empirical - formula ratio

The ratio of C to H is $\frac{n_{C}}{n_{H}}=\frac{5.0\ mol}{5.0\ mol}=1:1$, so the empirical formula is CH.

Step3: Calculate the empirical - formula mass

$M_{empirical}=12.01\ g/mol + 1.01\ g/mol=13.02\ g/mol$

Step4: Find the multiple (n)

$n=\frac{M_{molecular}}{M_{empirical}}=\frac{78.12\ g/mol}{13.02\ g/mol}=6$

Step5: Determine the molecular formula

Multiply the empirical - formula by n. The molecular formula is $(CH)_6 = C_6H_6$.

Answer:

C. $C_6H_6$