Question

6.0 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 138 g/mol, is burned completely in excess oxygen, and the mass of the products carefully measured:
carbon dioxide - 13.39 g
water - 2.35 g

Use this information to find the molecular formula of X.

Answer 1

Answer:

The molecular formula of X is given as $C_7 H_6 O_3$

Explanation:

$Moles C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol} \\\\moles \mathrm{C}= moles \mathrm{CO}_{2}=0.304 \mathrm{mol}$

$mass C= moles \times molar mass =0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g \\\\moles \mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol} \\\\moles \mathrm{H}=2 \times moles \mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol} \\\\Mass \mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$

mass O = Total mass of the compound - (mass of C + mass of H)

=6.0 g - ( 3.65 + 0.262 ) g

=2.09 g

$moles O=\frac{2.09 g}{16 g \text { per mole }}=0.131 \mathrm{mol}$

Least moles is for O that is 0.131mol and dividing all by the least we get

$\begin{aligned} C &=\frac{0.304}{0.131}=2.3 \\\\ H &=\frac{0.260}{0.131}=2 \\\\ O &=\frac{0.131}{0.131}=1 \end{aligned}$

Since 2.3 is a fraction it has to be converted to a whole number so we multiply all the answers by 3

$\\ C 2.3 \times 3=7 \\\\ H 2 \times 3=6 \\\\ O 1 \times 3=3$

So the empirical formula is $C_7 H_6 O_3$

Empirical formula mass

$=(7 \times 12) +(6\times1.008)+(3\times16)=138.048g$

$n=\frac{\text { molar mass }}{\text { empirical formula mass }}=\frac{138}{138.048}=1$

Molecular formula =n × empirical formula

$=1 \times C_7 H_6 O_3$

Compound X  = $C_7 H_6 O_3$  is the Answer