Question

1.5 g of a hydrocarbon on combustion gave 4.4 g of $C{O}_2$ and 2.7 g of $H_{2}O$. The vapor density of the compound is 15. Calculate the molecular formula.

• A. $C_2{H}_4$
• B. $C_2{H}_2$
• C. $C_2{H}_6$
• D. $C_3{H}_4$

Answer 1

Answer: (C)

$C_2{H}_6$

The first step here is to determine the mass of C in $C{O}_2$ and the mass of _H in $H_{2}O$. This is done by dividing the atomic mass by the molecular mass and then multiplying by the mass of compound produced.

For C: $\frac{12.01g}{44.009g}\times 4.40g$= 1.1999 g

For H: $\frac{1.0079\times 2}{18.0148g\times 2.70g}$ = 0.3021 g

The next step is to convert these masses into moles. This is done by dividing the mass by the relative atomic mass of the element:

C: $\frac{1.1999g}{12.011gmo{l}^{-1}}$ = 0.0999 mol

H: $\frac{0.3021g}{1.0079gmo{l}^{-1}}$ = 0.2997 mol

The final step is to divide each of these two values by the smallest number, in this case, this is the number of moles of carbon:

C: $\frac{0.0999mol}{0.0999mol}$ = 1

H: 0.2997 mol / 0.0999 mol = 3

Empirical formula $C{H}_3$

vapor density = molecular weight/2

molecular weight = vapor density x 2 = 15 x2 = 30

molecular formula = n*(Empirical formula)

n= $\frac{EW}{MW}$ = $\frac{30}{15}$ = 2

molecular formula = $2\times C{H}_3=C_2{H}_6$

answer is C