BRADY PROBLEM 2.83

Tutorial on the procedure of solving for empirical and molecular formulas from a combustion reaction.

Problem

Citric Acid, the substance that makes lemon juice sour, is composed of only carbon , hydrogen, and oxygen. When a 0.5000 g sample of citric acid was burned, it produced 0.6871 g CO 2 and 0.1874 g H 2 O.

(C + H + O) 0.5000 g

+

O 2 CO 2 0.6871 g

+

H 2 O 0.1874 g

The molecular mass of the compound is 192. What are the empirical and molecular formulas for citric acid?

All the hydrogen has gone to the product of H 2 O All the carbon has gone to the product of CO 2

(C + H + O) 0.5000 g

+

O 2 CO 2 0.6871 g

+

H 2 O 0.1874 g

From these gram quantities, convert to moles of compound, then to moles of elements…

( you may wish to subtotal as you will use the number of moles later )…

then to grams of elements. Starting with CO 2 ….

Convert grams of CO 2 to moles of CO 2 to moles of C then to grams of carbon.

0.6871 g

CO 2 1 mol CO 2 44.0 g 1 mol C 1 mol CO 2 = 0.01562 mol C = 0.01562 mol C 12.011 g 1 mol C = 0.1876 grams C

Convert grams of H 2 O to moles of H 2 O to moles of H then to grams of hydrogen.

0.1874 g

H 2 O 1 mol H 2 O 18.0 g 2 mol H 1 mol H 2 O = 0.02082 mol H = 0.02082 mol H 1.008 g 1 mol H = 0.02099 grams H

Find the number of grams of oxygen in the original sample 0.5000 g <0.02099 g> <0.1876 g> 0.2914 g

original sample size subtract grams hydrogen subtract grams carbon remaining grams are the grams of oxygen in the original sample

Convert these grams to moles of oxygen 0.2914 g O 1 mol O 15.994 g = 0.01821 mol O

Find the lowest mole ratio for the empirical formula

0.1562 mol C/ 0.1562 = 1 0.02082 mol H/ 0.1562 =

1.333

x 6 = x 6 =

0.01821 mol O/ 0.1562 =

1.166

x 6 = 6 8 7

C

6

H

8

O

7

Find the Molecular Formula from the given Molecular Weight

Given molecular weight = 192 g/mol Empirical Formula determined: C 6 H 7 O 7

Calculate the mass of the empirical formula:

C = 12.011 X 6 = 72.061

H = 1.008 x 8 = 8.064

O = 15.9994 x 7 = 111.99

192.12 g/mol

Divide the mass of the empirical into the molecular weight:

192/192.12 = 0.99938 = 1 therefore the empirical formula is the same as the molecular formula