Problem 1

Chemistry Level 2

When 8.8 g 8.8\text{ g} of an organic acid, with molar mass 88 gmol − 1 88 \text{ gmol}^{-1} , is burnt in excess oxygen, 17.6 g 17.6\text{ g} of carbon dioxide and 7.2 g 7.2\text{ g} of water are produced. Calculate its empirical formula.

C X 3 H X 6 O \ce{C_3H_6O} C X 3 H X 6 O X 2 \ce{C_3H_6O_2} C X 2 H X 4 O \ce{C_2H_4O} C H X 2 O \ce{CH_2O}

Christopher Boo by Christopher Boo , 24, Singapore

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2 solutions

C2H4O is the only acid with molar mass that is a factor of 88

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Mj Bltz
Jun 1, 2015

The C O 2 CO_{2} gives the amount of carbon, and water gives the amount of hydrogen. The amount of oxygen can be obtained by subtracting the total amount of carbon and hydrogen from the mass of sample (8.8 g).

Note atomic/molar masses: C = 12.01 g/mol, H = 1.01 g/mol, C O 2 CO_{2} = 44.01 g/mol, H 2 O H_{2}O = 18.02 g/mol

C: 17.6 g ( 12.01 44.01 ) = 4.8 g = 0.4 m o l 17.6g(\frac{12.01}{44.01}) = 4.8 g = 0.4 mol

H: 7.2 g ( 2.02 18.02 ) = 0.8 g = 0.8 m o l 7.2g(\frac{2.02}{18.02}) = 0.8 g = 0.8 mol

O: 8.8 g − ( 4.8 g + 0.8 g ) = 3.2 g = 0.2 m o l 8.8g - (4.8g + 0.8g) = 3.2 g = 0.2 mol

Dividing each element by the one with the least amount (in mol), we get the empirical formula: C 2 H 4 O C_{2}H_{4}O

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