Answer
a) $C_9H_9N$
b) $131.178\ g/mol$
Work Step by Step
Atomic weights: C - 12.011, H - 1.008, N - 14.007
a) Calculate the mass of each element in 100.00 g of this compound:
$m_C=100.00\ g\cdot\frac{82.40\ g_C}{100.00\ g}=82.40\ g_C$
$m_H=100.00\ g\cdot\frac{6.92\ g_H}{100.00\ g}=6.92\ g_H$
$m_N=100.00\ g\cdot\frac{(100.00-82.40-6.92)\ g_N}{100.00\ g}=10.68\ g_N$
Calculate the number of moles of each element in this sample:
$n=\frac mM$
$n_C=\frac{82.40\ g}{12.011\ g/mol}=6.860\ mol$
$n_H=\frac{6.92\ g}{1.008\ g/mol}=6.87\ mol$
$n_N=\frac{10.68\ g}{14.007\ g/mol}=0.7625\ mol$
Find the ratio of the number of moles of each element by the one with the least number of moles, Nitrogen:
$\frac{n_C}{n_N}=\frac{6.860\ mol}{0.7625\ mol}\approx9$
$\frac{n_H}{n_N}=\frac{6.87\ mol}{0.7625\ mol}\approx9$
The formula of this compound should follow these proportions, so the formula is:
$C_9H_9N$
b) Calculate the molar mass of this compound, given the formula obtained in part (a):
$M=9\cdot 12.011+9\cdot1.008+14.007=131.178\ g/mol$
