Question

Butyric acid contains only $ {\text{C, H and O}} $ . A $ {\text{4}}{\text{.24 mg}} $ sample of butyric acid is completely burned. It gives $ {\text{8}}{\text{.45 mg}} $ of carbon dioxide $ \left( {{\text{C}}{{\text{O}}_2}} \right) $ and $ {\text{3}}{\text{.46 mg}} $ of water. What is the mass percentage of each element in butyric acid?
If the elemental composition of butyric acid is found to be $ 54.2\% $ C, $ 9.2\% $ H and $ 36.6\% $ O, determine the empirical formula.
The molecular mass of butyric acid was determined by experiment to be 88. What is the molecular formula?

Answer 1

Hint: The percentage is calculated by dividing the mass of each element present with the given mass. The empirical formula can be calculated by taking the ratio of percentage given. The exact molecular formula will be obtained by multiplying empirical formula with 2.

Complete step by step solution:
We know in carbon dioxide there is 1 carbon atom present. The molar mass of carbon is 12 and the molar mass of carbon dioxide is 44. This means that 44 gram of carbon dioxide has 12 g of carbon.
According to this $ {\text{8}}{\text{.45 mg}} $ gram of carbon dioxide will contain $ \dfrac{{12}}{{44}} \times {\text{8}}{\text{.45 mg}} $ gram of carbon. The total sample of butyric acid is $ {\text{4}}{\text{.24 mg}} $ so the percentage of carbon present will be:
 $ \dfrac{{12}}{{44}} \times \dfrac{{{\text{8}}{\text{.45}}}}{{4.24}} \times 100{\text{ mg}} = 54.3\% $
In water there are 2 hydrogen atoms present. The molar mass of hydrogen is 1 and the molar mass of water is 18. This means that 18 gram of water has 2 g of hydrogen.
According to this $ {\text{3}}{\text{.46 mg}} $ gram of carbon dioxide will contain $ \dfrac{2}{{18}} \times 3.46{\text{ mg}} $ gram of carbon. The total sample of butyric acid is $ {\text{4}}{\text{.24 mg}} $ so the percentage of carbon present will be:
 $ \dfrac{2}{{18}} \times \dfrac{{3.46}}{{4.24}}{\text{ mg}} \times 100 = 9\% $
The total percentage is always hundred and hence the percentage of oxygen will be:
 $ 100 - 54.3\% - 9\% = 36.7\% $
Part B
We will get the empirical formula by taking the whole number ratio of the given percentage composition. The ratio of C : O : H is $ 54.2\% $ : $ 36.6\% $ : $ 9.2\% $
The smallest whole number ratio is 2 : 4 : 1
So the empirical formula is $ {{\text{C}}_2}{{\text{H}}_4}{\text{O}} $
Empirical formula just represents the smallest ratio and not the exact number of atoms present. We will divide the exact molecular mass with the molecular mass of the empirical formula to calculate n.
 $ {\text{n}} = \dfrac{{88}}{{44}} = 2 $
We will multiply the empirical formula with 2 to get the complete molecular formula as: $ {{\text{C}}_4}{{\text{H}}_8}{{\text{O}}_2} $ .

Note:
The butyric acid that is botanic acid is a carboxylic acid. It is derived from the word butter because it is also present in the butter. It is colourless and has unpleasant odour.