Question

The combustion of 8.23 mg sample of unknown substance gave 9.62 mg of carbon dioxide and 3.94 mg of water. Another sample, weighing 5.32 mg, gave 13.49 mg of silver chloride in a happen analysis. Determine the percent composition for this organic compound.

Answer 1

1) Find mass and moles of carbon

$n(C)= \frac{9.62\times 10^{-3}}{44.01} =0.000219 moles$

$m(C) =n\times M = 0.000219\times 12.01 = 0.00263 g =2.63 mg$

2) Find mass and moles of hydrogen

$n(H) = n(H_2O)\times 2 =\frac{3.94\times10^{-3}}{18.02} \times 2 = 0.000437 moles$

$m(H) = n\times M = 0.000437\times 1 = 0.437mg$

3) Find mass percent of chlorine in a sample

a) find mass of chlorine if silver nitrate:

$n(AgCl )= \frac{m}{M} = \frac{13.49\times 10^{-3}}{143.32} = 0.0000941 mol$

$n(Cl) = n(AgCl) = 0.0000941 mol$

$m(Cl) = 0.0000941\times35.45 = 0.00334 g = 3.34 mg$

b) find mass percent of chlorine in silver nitrate

$mass\% (Cl)= \frac{3.34}{5.32} =0.627$

4) Find mass and moles of chlorine in the first sample

$m(Cl) = m_{sample}\times mass\%(Cl) = 8.23\times 0.627 = 5.16 mg$

$n(Cl) = \frac{m}{M} = \frac{5.16\times10^{-3}}{35.45} = 0.000146 mol$

5) check up if there is an oxygen in a sample:

$8.23 = m(O) + m(C) + m(H) + m(Cl)$

$8.23 = m(O) +2.63 + 0.437 + 5.16$

$m(O) = 0.003 mg$ (negligible)

6) find mole ratio

$n(C):n(H):n(Cl) = 0.000219:0.000438:0.000146 = 1.5:3:1 = 3:6:2$

The formula is $C_3H_6Cl_2$

7) Percent composition

mass percent $\% (c) = \frac{2.63}{8.23}\times 100\% = 32.0\%$

mass percent $\% (H)= \frac{0.437}{8.23} \times 100\% = 5.31 \%$

mass percent $\% =\frac{5.16}{8.23} \times 100\% = 62.7\%$