Asked by Grace
2.4g of a compound of carbon, hydrogen and oxygen gave on combustion 3.52g of carbon diioxide and 1.44g of water. The relative atomic mass of the compound was found to be 60. What is the molecular formula of the compound
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Mass of CO2 = 3.52 g → moles CO2 = 3.52/44 = 0.08 mol → moles C = 0.08 mol → mass C = 0.08×12 = 0.96 g.
Mass of H2O = 1.44 g → moles H2O = 1.44/18 = 0.08 mol → moles H = 0.16 mol → mass H = 0.16×1 = 0.16 g.
Mass O = 2.4 − (0.96 + 0.16) = 1.28 g → moles O = 1.28/16 = 0.08 mol.
Mole ratio C:H:O = 0.08:0.16:0.08 = 1:2:1 → empirical formula CH2O (formula mass = 30). Relative molecular mass = 60 → multiplier = 60/30 = 2. Molecular formula = C2H4O2.
Mass of H2O = 1.44 g → moles H2O = 1.44/18 = 0.08 mol → moles H = 0.16 mol → mass H = 0.16×1 = 0.16 g.
Mass O = 2.4 − (0.96 + 0.16) = 1.28 g → moles O = 1.28/16 = 0.08 mol.
Mole ratio C:H:O = 0.08:0.16:0.08 = 1:2:1 → empirical formula CH2O (formula mass = 30). Relative molecular mass = 60 → multiplier = 60/30 = 2. Molecular formula = C2H4O2.
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