Question

Compound X contains only carbon, hydrogen, nitrogen, and chlorine. When 1.00 g of X is dissolved in water and allowed to react with excess silver nitrate, AgNO3, all the chlorine in X reacts and 1.95 g of solid AgCl is formed. When 1.00 g of X undergoes complete combustion, 0.900 g of CO2 and 0.735 g of H2O are formed. What is the empirical formula of X?

Answer 1

Answer:

$C_{3}H_{12}N_{2}Cl_{2}$

Explanation:

Lets say the formula of the compound is:

$C_{a}H_{b}N_{c}Cl_{d}$

Experiment 1:

With 1.00 g of the compound you get 1.95 g of AgCl

The molar mass of AgCl is 143.32 g/mol.

The moles of AgCl in 1.95g are:

$moles = \frac{1.95}{143.32} = 0.0136$

Also the moles of Cl are 0.0136 because 1 mol of AgCl has 1 mol of Cl

Experiment 2:

With 1.00 g of the compound you get 0.900 of CO2 and 0.735 g of H2O

The molar mass of CO2 is 44 g/mol

The molar mass of H2O is 18 g/mol

$moles\ CO_{2} = \frac{0.900}{44} = 0.0205 \\moles\ H_{2}O = \frac{0.735}{18} = 0.0408$

moles of C are 0.0205

moles of H are 0.0816 (2 times the moles of H2O)

With the given information so far you get that in 1.00 g of the compound there are 0.0136 moles of Cl, 0.0205 moles of C and 0.0816 moles of H.

In mass you have:

Mass Cl = 0.0136 * 35.5 = 0.4828 g

Mass C = 0.0205 * 12 = 0.246 g

Mass H = 0.0816 * 1 = 0.0816 g

Total mass = 0.4828 + 0.246 + 0.0816 + mass N

1.00 = 0.8104 + Mass N

Mass N = 0.1896

$moles\ N =\frac{0.1896}{14} =0.0135$

Now you divide all moles for the least

Cl: 0.0136 / 0.0135 = 1.0007

C: 0.0205 / 0.0135 = 1.52

H: 0.0816 / 0.0135 = 6.04

N: 0.0135 / 0.0135 = 1

multiplying by 2 to get an integer for C

Cl = 2

C = 3

H = 12

N = 2

$C_{3}H_{12}N_{2}Cl_{2}$