Chemistry: Empirical & Molecular Formula

Question

1) A compound containing only C, H and O was subjected to combustion analysis. A sample of 6.800×10-2 g produced 1.359×10-1 g of CO₂ and 5.562×10-2 g of H₂O. Determine the empirical formula of the compound.

 

2) If the molar mass of the compound is 264.318 g/mol, determine the molecular formula of the compound and enter the appropriate subscript after each element. 

Answer 1

To obtain the empirical formula, we want to find moles of C, moles of H and moles of O in the simplest whole number ratio.

Moles of CO2 = 0.1359 g x 1 mole/44 g = 0.003089 moles 

Moles of C = 0.003089 moles CO2 x 1 moles C/mole CO2 = 0.003089 moles C

 

Moles of H2O = 0.005562 g x 1 mole/18 g = 0.000309 moles H2O

Moles of H = 0.000309 moles H2O x 2 moles H/mole H2O = 0.000618 moles H

 

To get moles of O, we need to convert moles of C and moles of H to grams, add them together and subtract that from the original mass of the sample to obtain grams of Oxygen.

 

Grams C = 0.003089 moles C x 12 g/mole = 0.03707 g

Grams H = 0.000618 moles H x 1 g/mole = 0.000618 g 

Adding them we get 0.037688 g

Original sample = 0.06800

Mass of O in original sample = 0.06800 g - 0.03769 g = 0.0303 g O

Moles of O = 0.0303 g x 1 mole/16 g = 0.00189 moles O

 

Recapping:

Moles of C = 0.003089

Moles of H = 0.000618

Moles of O = 0.00189

To get whole number ratios, divide all by the lowest value (0.000618) to get 

C = 0.003089/0.000618) = 4.99 ~ 5

H = 0.000618/0.000618 = 1

O = 0.00189/0.000618 = 3.0

So, the empirical formula is C₅HO₃ 

This has a molar mass of 60.055 + 1.008 + 47.997 = 109.06

264.318/109.06 = 2.4

Multiply this by 5 to get a whole number of 12, meaning there are 5 empirical formulae in 1 molecular formula.  

Molecular formula would be C₂₅H₅O₁₅