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BRADY PROBLEM 2.83 Tutorial on the procedure of solving for empirical and molecular formulas from a combustion reaction.

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Presentation on theme: "BRADY PROBLEM 2.83 Tutorial on the procedure of solving for empirical and molecular formulas from a combustion reaction."— Presentation transcript:

1 BRADY PROBLEM 2.83 Tutorial on the procedure of solving for empirical and molecular formulas from a combustion reaction.

2 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. The molecular mass of the compound is 192. What are the empirical and molecular formulas for citric acid? (C + H + O) 0.5000 g CO 2 0.6871 g + H 2 O 0.1874 g + O2O2

3 (C + H + O) 0.5000 g CO 2 0.6871 g + H 2 O 0.1874 g + O2O2 All the hydrogen has gone to the product of H 2 O All the carbon has gone to the product of CO 2 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 ….

4 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 1 mol C= 0.01562 mol C 44.0 g1 mol CO 2 = 0.01562 mol C12.011 g= 0.1876 grams C 1 mol C

5 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 O1 mol H 2 O2 mol H= 0.02082 mol H 18.0 g1 mol H 2 O = 0.02082 mol H1.008 g= 0.02099 grams H 1 mol H

6 Find the number of grams of oxygen in the original sample 0.5000 g original sample size subtract grams hydrogen subtract grams carbon 0.2914 g remaining grams are the grams of oxygen in the original sample 0.2914 g O1 mol O= 0.01821 mol O 15.994 g Convert these grams to moles of oxygen

7 Find the lowest mole ratio for the empirical formula 0.1562 mol C/ 0.1562 = 1x 6 =6 0.02082 mol H/ 0.1562 = 1.333x 6 =8 0.01821 mol O/ 0.1562 = 1.166x 6 =7 C6H8O7C6H8O7

8 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


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