Question

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.8635 g of CO2 and 0.1767 g of H2O. What is the empirical formula of the compound?

Answer 1

Answer: The empirical formula is $C_3H_3O_2$.

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 $CO_2$ = 0.8635

Mass of $H_2O$= 0.1767 g

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 0.8635 g of carbon dioxide, =$\frac{12}{44}\times 0.8635=0.2355g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.1767 g of water, =$\frac{2}{18}\times 0.1767=0.0196g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.4647) - (0.2355+0.0196) = 0.2096 g

Mass of C = 0.2355 g

Mass of H = 0.0196 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{0.2355g}{12g/mole}=0.0196moles$

Moles of H=$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.0196g}{1g/mole}=0.0196moles$

Moles of O=$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.2096g}{16g/mole}=0.0131moles$

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

For C =$\frac{0.0196}{0.0131}=1.5$

For H =$\frac{0.0196}{0.0131}=1.5$

For O = $\frac{0.0131}{0.0131}=1$

The ratio of C : H: O = 1.5 : 1 .5 : 1

Converting to simple whole number ratio:

The ratio of C : H: O = 3 : 3: 2

Hence the empirical formula is $C_3H_3O_2$.