Question

the percent composition by mass of a compound is 81.53 % c, 4.89 % h, and 13.59 % n. what is the molecular formula of this compound if its molar mass is 206.24 g/mol?

o c13h22n2

o c14h10n2

o c13h8n3

o c12h6n4

o c15h12n

Explanation:

Step1: Assume 100g of the compound

The masses of C, H, and N are 81.53g, 4.89g, and 13.59g respectively.

Step2: Calculate the number of moles of each element

The molar - mass of C is 12.01g/mol, of H is 1.01g/mol, and of N is 14.01g/mol.
The number of moles of C: $n_{C}=\frac{81.53g}{12.01g/mol}\approx6.79mol$
The number of moles of H: $n_{H}=\frac{4.89g}{1.01g/mol}\approx4.84mol$
The number of moles of N: $n_{N}=\frac{13.59g}{14.01g/mol}\approx0.97mol$

Step3: Find the mole - ratio of the elements

Divide each number of moles by the smallest number of moles (0.97mol).
For C: $\frac{6.79mol}{0.97mol}\approx7$
For H: $\frac{4.84mol}{0.97mol}\approx5$
For N: $\frac{0.97mol}{0.97mol}=1$
The empirical formula is $C_{7}H_{5}N$.

Step4: Calculate the empirical - formula mass

The empirical - formula mass of $C_{7}H_{5}N$ is $7\times12.01g/mol + 5\times1.01g/mol+1\times14.01g/mol=84.07g/mol + 5.05g/mol+14.01g/mol = 103.13g/mol$.

Step5: Find the multiple (n)

$n=\frac{\text{Molar mass}}{\text{Empirical - formula mass}}=\frac{206.24g/mol}{103.13g/mol}=2$.

Step6: Determine the molecular formula

Multiply the sub - scripts in the empirical formula by n.
The molecular formula is $C_{14}H_{10}N_{2}$.

Answer:

B. $C_{14}H_{10}N_{2}$