#### Answer

(a) $PbCl_3$
(b) $C_{10}H_{14}N_2$

#### Work Step by Step

(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: $C_{10}H_{14}N_2$