Question

Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 −mg sample of menthol is combusted, producing 0.2829 mg of CO2 and 0.1159 mg of H2O. What is the empirical formula for menthol? Express your answer as a chemical formula.

Answer 1

Answer: The empirical formula for the given compound is
`C_(10)H_(20)O`

Step-by-step 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=0.2829mg=0.0002829g`

Mass of
`H_2O=0.1159mg=0.0001159g`

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

So, in 0.0002829 g of carbon dioxide,
`(12)/(44)* 0.0002829=0.00007715g` of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.0001159 g of water,
`(2)/(18)* 0.0001159=0.00001288g` of hydrogen will be contained.

Mass of oxygen in the compound = (0.0001005) - (0.00007715 + 0.00001288) = 0.00001047 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon =
`\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=(0.00007715g)/(12g/mole)=6.429* 10^(-6)moles`

Moles of Hydrogen =
`\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=(0.00001288g)/(1g/mole)=0.01288* 10^(-3)moles`

Moles of Oxygen =
`\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=(0.00001047g)/(16g/mole)=6.544* 10^(-7)moles`

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is
`6.544* 10^(-7)moles`.

For Carbon =
`(6.429* 10^(-6))/(6.544* 10^(-7))=9.82\approx 10`

For Hydrogen =
`(0.01288* 10^(-3))/(6.544* 10^(-7))=19.68\approx 20`

For Oxygen =
`(6.544* 10^(-7))/(6.544* 10^(-7))=1`

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is
`C_(10)H_(20)O_1=C_(10)H_(20)O`