Question

When \[0.500g\] of butyric acid is combusted in excess oxygen, and it produces \[0.999g\] of carbon dioxide and \[0.409g\] of water, how do you find the empirical formula of butyric acid?

Answer 1

Hint: From the atomic mass the percentages of carbon, hydrogen and oxygen in carbon dioxide, water and butyric acid can be determined. From the given grams and obtained percentage of carbon dioxide, water and butyric acid the grams of carbon, hydrogen and oxygen can be determined. Convert these grams into moles. From the moles calculate the mole ratio to find the empirical formula.

Complete answer:
Given the weight of butyric acid is \[0.500g\] means it is the combined weight of carbon, hydrogen and oxygen atoms.
Percentage of carbon in carbon dioxide is \[\dfrac{{12}}{{44}} = 27.3\% C\]
Percentage of hydrogen in water is \[\dfrac{2}{{18}} = 11.1\% H\]
The percentage of oxygen is \[100\% - \left( {27.3\% + 11.1\% ) = 61.6\% O} \right)\]
The grams of carbon is \[0.999g \times 27.3\% = 27.3g\]
The grams of hydrogen is \[0.409 \times 11.1\% = 0.045g\]
The grams of oxygen is \[0.500 - \left( {27.3g + 0.0454g} \right) = 0.182g\]
To convert the above grams into moles, divide the amount of weight by molar mass
Moles of carbon is \[C:\dfrac{{27.3}}{{12}} = 0.0227moles\]
Moles of hydrogen is \[H:\dfrac{{0.0454}}{1} = 0.0454moles\]
Moles of oxygen is \[O:\dfrac{{0.0182}}{{16}} = 0.0114moles\]
Divide the moles of carbon, hydrogen and oxygen with the moles of oxygen as oxygen contains least number of moles
The number of carbon atoms will be \[\dfrac{{0.0227}}{{0.0114}} = 2\]
The number of hydrogen atoms will be \[\dfrac{{0.0454}}{{0.0114}} = 4\]
The number of oxygen atoms will be \[\dfrac{{0.0114}}{{0.0114}} = 1\]
Thus, the empirical formula of butyric acid is \[{C_2}{H_4}O\]

Note:
Empirical formula is the shortest mole ratio of all the atoms in butyric acid. For some compounds the empirical formula is the same as the molecular formula. For butyric acid molecular formula is \[{C_4}{H_8}{O_2}\] while empirical formula is \[{C_2}{H_4}O\]