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Upon combustion, a 1.3109 g sample of a compound containing only C, H, & O produces 3.2007 gCO2 and 1.3102 gH2O. Find empirical formula.?
1 Answer
- Roger the MoleLv 72 years ago
(3.2007 g CO2) / (44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) = 0.072727248 mol C
(0.072727248 mol C) × (12.01078 g C/mol) = 0.873511 g C
(1.3102 g H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) = 0.145454 mol H
(0.145454 mol H) × (1.007947 g H/mol) = 0.14661 g H
(1.3109 g total) - (0.873511 g C) - (0.14661 g H) = 0.290779 g O
(0.290779 g O) / (15.99943 g O/mol) = 0.01817434 mol O
Divide by the smallest number of moles:
(0.072727248 mol C) / 0.01817434 mol = 4.0016
(0.145454 mol H) / 0.01817434 mol = 8.003
(0.01817434 mol O) / 0.01817434 mol = 1.0000
Round to the nearest whole numbers to find the empirical formula:
C4H8O
