Question

Methyl salicylate is a common active ingredient in liniments such as ben-gay. It is also known as oil of wintergreen. It is made up of carbon, hydrogen, and oxygen atoms. When a sample of methyl salicylate weighing 5.287 g is burned in excess oxygen, 12.24 g of carbon dioxide and 2.505 g of water are formed. What is the simplist formula for oil of wintergreen

Answer 1

Answer: The empirical formula is $C_3H_3O$.

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 $CO_2$ = 12.24 g

Mass of $H_2O$ = 2.505 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 12.24 g of carbon dioxide, =$\frac{12}{44}\times 12.24=3.338g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.505 g of water, =$\frac{2}{18}\times 2.505=0.278g$ of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671  g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles$

Moles of H=$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles$

Moles of O=$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles$

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

For C =$\frac{0.278}{0.104}=3$

For H =$\frac{0.278}{0.104}=3$

For O =$\frac{0.104}{0.104}=1$

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is $C_3H_3O$.