Phenol commonly known as carbolic acid

Phenol commonly known as carbolic acid, was used by Joseph Lister as an antiseptic for surgery in 1865. Its principal use today is in the manufacture of phenolic resins and plastics. Combustion of 5.23 mg of phenol yields 14.67 mg CO_2 and 3.01 mg H_2O. Phenol contains only C, H, and O. What is the percentage of each element in this substance?

Solution:

We will first find the mass of carbon, and then we will find the mass of Hydrogen. After we find these values, we can find the mass percentages, and then find the Oxygen percentage through the difference.

Let us start first by finding the mass of carbon. The way we will do this is the following. We will multiply the mass of CO_2 by the molar mass of C and the reciprocal molar mass of CO_2 like so:

14.67 \text{mg CO}_2\times \frac{1\text{mol CO}_2}{44.01\text{g CO}_2}\times \frac{12.01\text{g C}}{1\text{mol C}}=4.0033\text{mg C}

Now, we will find the mass of hydrogen by doing the same as before.

3.01\text{mg H}_2\text{O}\times \frac{1\text{mol}\,\text{H}_2\text{O}}{18.02\text{g CO}_2}\times \frac{2\text{ H}}{1\,\text{H}_2\text{O}}\times \frac{1.008\text{ g H}}{1\text{ mol H}}=0.3368\text{mg H}

Now, we can calculate the mass of oxygen by subtracting the total carbon and hydrogen mass from 5.23 mg like so:

Mass O= 5.23 mg- (4.0033 + 0.3368) = 0.8899 mg O

Now that we have all the mass values, we can calculate the percentages.

% C = \frac{4.0033\text{ mg}}{5.23\text{ mg}}\times 100\%=76.54\%

% H = \frac{0.3368\text{ mg}}{5.23\text{ mg}}\times 100\%=6.439\%

%O = \frac{0.8899\text{ mg}}{5.23\text{ mg}}\times 100\%=17.0\%

This question can be found in General Chemistry, 9th edition, chapter 3, question 3.64