Menthol is composed of C, H, and O. A 0.1005-g sample of menthol is combusted in oxygen, producing 0.2829 g of CO2 and 0.1159 g

Question

3 years ago

13

of H2O. What is the empirical formula of menthol?

1 answer:

ira [324] · Answer 1 · 2022-03-08T03:30:04+03:00

Answer: The empirical formula for the given organic compound is $CH_2O$

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 $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

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

So, in 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

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

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.013g$ of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.0066moles$

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0064 moles.

For Carbon = $\frac{0.0064}{0.0064}=1$

For Hydrogen = $\frac{0.013}{0.0064}=2.03\approx 2$

For Oxygen = $\frac{0.0066}{0.0064}=1.03\approx 1$

The ratio of C : H : O = 1 : 2 : 1

Hence, the empirical formula for the given compound is $C_1H_{2}O_1=CH_2O$