Question

A laboratory analysis of a 100 g sample finds it is composed of 1.8 g hydrogen, 56.1 g sulfur, and 42.1 g oxygen. What is its empirical formula? Give your answer in the form H#S#O#, where the number following the element’s symbol corresponds to the subscript in the formula. (Don’t include a 1 subscript explicitly.) For example, the formula CHO would be entered as CH2O.

Answer 1

Answer: The empirical formula is $H_2S_2O_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mas of H = 1.8 g

Mass of S = 56.1 g

Mass of O = 42.1 g

Step 1 : convert given masses into moles.

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.8g}{1g/mole}=1.8moles$

Mass of S =$\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{56.1g}{32g/mole}=1.8moles$

Moles of O=$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{42.1g}{16g/mole}=2.6moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For H = $\frac{1.8}{1.8}=1$

For S = $\frac{1.8}{1.8}=1$

For O =$\frac{2.6}{1.8}=1.5$

Converting to whole number ratios

The ratio of H: S: O= 2: 2: 3

Hence the empirical formula is $H_2S_2O_3$