Question

A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of H2O. Determine the masses of C and H in the sample and the percentages of these elements in this hydrocarbon.

Answer 1

Answer: The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=0.3986g$

Mass of $H_2O=0.0578g$

Mass of sample = 0.1153 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

• For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.3986 g of carbon dioxide, $\frac{12}{44}\times 0.3986=0.1087g$ of carbon will be contained.

• For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.0578 g of water, $\frac{2}{18}\times 0.0578=0.0066g$ of hydrogen will be contained.

To calculate the percentage composition of a substance in sample, we use the equation:

$\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100$      ......(1)

• For Carbon:

Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{0.1087g}{0.1153g}\times 100=94.27\%$

• For Hydrogen:

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

$\%\text{ composition of hydrogen}=\frac{0.0066g}{0.1153g}\times 100=5.72\%$

Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.