Empirical formula : C4H10O
Explanation
mass CO2 = 5.049 g
moles CO2 = (mass CO2) / (molar mass CO2)
moles CO2 = (5.049 g) / (44.01 g/mol)
moles CO2 = 0.1147 mol
moles C = moles CO2
moles C = 0.1147 mol
mass C = (moles C) * (molar mass C)
mass C = (0.1147 mol) * (12.01 g/mol)
mass C = 1.378 g
mass H2O = 2.584 g
moles H2O = (mass H2O) / (molar mass H2O)
moles H2O = (2.584 g) / (18.01528 g/mol)
moles H2O = 0.1434 mol
moles H = 2 * (moles H2O)
moles H = 2 * (0.1434 mol)
moles H = 0.287 mol
mass H = (moles H) * (molar mass H)
mass H = (0.287 mol) * (1.008 g/mol)
mass H = 0.289 g
mass O = (total mass) - (mass C + mass H)
mass O = (2.126 g) - (1.378 g + 0.289 g)
mass O = 0.4588 g
moles O = (mass O) / (molar mass O)
moles O = (0.4588 g) / (16.0 g/mol)
moles O = 0.028677 mol
moles C / moles O = (0.1147 mol) / (0.028677 mol)
moles C / moles O = 4
moles H / moles O = (0.287 mol) / (0.028677 mol)
moles H / moles O = 10
There are 4 atoms of C and 10 atoms of H per one atom of O
Empirical formula : C4H10O
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