Hydroquinone used as a photographic developer

Hydroquinone used as a photographic developer, is 65.4% C, 5.5% H, and 29.1% O, by mass. What is the empirical formula of hydroquinone?


In most cases, when a question asks us to find an empirical formula without giving us a specific sample size, it’s best to assume a sample size of 100 g. It makes the math easy and leads us to a solution quicker.


Thus, we will assume there is a 100 g sample of hydroquinone. We will now multiply by the percentage compositions to figure out the number of grams of each individual element.


Grams of carbon: 100 g \times 65.4% = 65.4 g

Grams of hydrogen: 100 g \times 5.5% = 5.5 g

Grams of oxygen: 100 g \times 29.1% = 29.1 g


Now, we will convert each of these masses to moles by dividing by it’s specific molar mass.

Hydroquinone used as a photographic developer


\text{Mol C} = 65.4\,g\,\text{C}\times \frac{1\text{mol C}}{12.01\, g\,\text{C}}

\text{Mol C} =5.445\text{mol}


\text{Mol H} = 5.5\,g\,\text{H}\times \frac{1\text{mol H}}{1.008\, g\,\text{H}}

\text{Mol H} =5.46\text{mol}


\text{Mol O} = 29.1\,g\,\text{O}\times \frac{1\text{mol O}}{15.999\, g\,\text{O}}

\text{Mol O} =1.819\text{mol}


To figure out the integers of the empirical formula, simply divide each value by the smallest mol number we found, which was 1.819.


C= \frac{5.445}{1.819}\approx 3

(rounded to the closest integer)

H=\frac{5.46}{1.819}\approx 3

O=\frac{1.819}{1.819}= 1


Thus, the empirical formula is C_3H_3O

This question can be found in General Chemistry, 9th edition, chapter 3, question 3.68

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