Question

3.9g of an organic compound on combustion gives 13.2g of CO2 and 2.7g OfH2O. The empirical formula of the compound is??

Answer 1

Answer: Empirical formula of the compound is $CH$.

Explanation:-

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

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ =13.2g

Mass of $H_2O$=2.7g

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 13.2 g of carbon dioxide, =$\frac{12}{44}\times 13.2=3.6g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.7 g of water, =$\frac{2}{18}\times 2.7=0.3g$ of hydrogen will be contained.

Mass of oxygen in the compound = (3.9) - (3.6+0.3) = 0 g

Mass of C = 3.6 g

Mass of H = 0.3 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{11.33g}{12g/mole}=0.3moles$

Moles of H=$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.3g}{1g/mole}=0.3moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =$\frac{0.3}{0.3}=1$

For H =$\frac{0.3}{0.3}=1$

The ratio of C : H = 1: 1

Hence the empirical formula is $CH$.