5.(2 points) a compound with the empirical formula c₃h₂n, has a molar mass of 312.29 g/mol. determine the…

Explanation:

Step1: Calculate empirical - formula mass

The atomic masses of C = 12.01 g/mol, H = 1.01 g/mol, N = 14.01 g/mol. For the empirical formula (C_3H_2N), the empirical - formula mass (M_{empirical}=(3\times12.01)+(2\times1.01)+14.01) [ \begin{align*} M_{empirical}&=36.03 + 2.02+14.01\ &=52.06\ g/mol \end{align*} ]

Step2: Determine the ratio (n)

The ratio (n=\frac{M_{molecular}}{M_{empirical}}), where (M_{molecular} = 312.29\ g/mol) and (M_{empirical}=52.06\ g/mol) [n=\frac{312.29}{52.06}\approx6]

Step3: Find the molecular formula

Multiply the sub - scripts in the empirical formula by (n). The molecular formula is ((C_3H_2N)6 = C{18}H_{12}N_6)

Answer:

(C_{18}H_{12}N_6)