45.62mg of toluene is combusted and we get 152.5mg of carbon dioxide and 35.67mg of water vapor. What is the empirical formula of toluene?

Empirical formula is C₅H₆O₄₀₇.

Explanation:

Given dа ta:

Mass of carbondioxide = 152.5 mg

Mass of water vapors = 36.67 mg

Mass of toluene = 45.62 mg

Empirical formula = ?

Solution:

First of all we will convert the mg into g.

Mass of carbondioxide = 152.5/1000 = 0.1525 g

Mass of water vapors = 36.67 / 1000 = 0.0367 g

Mass of toluene = 45.62 / 1000 = 0.0456 g

Percentage of each atoms:

C = 0.1525/0.0456 * 12 / 44 = 0.91

H = 0.0367/0.0456 * 2 / 18 = 0.089

O = 100 - (0.91+0.089)

O = 100 - 0.999 = 99

Number of gram atoms:

Number of gram atoms of C = O. 91/12 = 0.076

Number of gram atoms of H = 0.089/1 = 0.089

Number of gram atoms of O = 99 / 16 = 6.188

Atomic ratio:

C : H : O

0.076/0.076 : 0.089/0.076 : 6.188/0.076

1 : 1.17 : 81.4

C : H : O = 5 (1: 1.7 : 81.4)

C : H : O = 5 : 6 : 407

Empirical formula is C₅H₆O₄₀₇.