## Chemistry (7th Edition)

(a) $PbCl_3$ (b) $C_{10}H_{14}N_2$
(a) The molar weight for "P" is equal to 31.0 u. Therefore, the chlorine mass in this compound should be: 137.3 - 31.0 = 106.3 u Since each chlorine has a molecular mass of 35.45: $106.3 u \times \frac{1 chlorine}{35.45 u } = 3$ chlorine. Therefore, the correct formula is : $PbCl_3$ (b) Mass ($C_{10}H_{14}$): 12.01* 10 + 1.008* 14 = 134.21 u 162.2 - 134.21 = 28.0 u Since each nitrogen has a molecular weight of 14.0 u: 28 u $\times \frac{1 N}{14 u} = 2$ nitrogens. Therefore, the correct formula is: $C_{10}H_{14}N_2$