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Combustion of 32.44 g of a compound containing only carbon, hydrogen, and oxygen produces 37.54 gCO2 and 15.37 gH2O. Empirical formula?
2 Answers
- Roger the MoleLv 72 years agoFavorite Answer
(37.54 g CO2) / (44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) = 0.8529949 mol C
(0.8529949 mol C) × (12.01078 g C/mol) = 10.2451 g C
(15.37 g H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) = 1.70633 mol H
(1.70633 mol H) × (1.007947 g H/mol) = 1.71989 g H
(32.44 g total) - (10.2451 g C) - (1.71989 g H) = 20.47501 g O
(20.47501 g O) / (15.99943 g O/mol) = 1.27973 mol O
Divide by the smallest number of moles:
(0.8529949 mol C) / 0.8529949 mol = 1.000
(1.70633 mol H) / 0.8529949 mol = 2.000
(1.27973 mol O) / 0.8529949 mol = 1.500
In order to achieve integer coefficients, multiply by 2 then round to the nearest whole numbers to find the empirical formula:
C2H4O3
