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mystix
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[SOLVED] Empirical Formula
Hi, I have a example question in my textbook that I don't fully understand. If someone could please help me out I would really appreciate it! Thanks!
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In this example, 15.5 g of a compound containing carbon, hydrogen and oxygen is completely burned in air; 22.0 g of CO2 and 13.5 g of water vapour are produced. Find the empirical formula.
Moles of CO2 = 22.0 g / 44.0 g / mol = 0.500 mol
Moles of H2O = 13.5 g / 18.0 g / mol = 0.750 mol
moles of C atoms = 0.500 mol and moles of H atoms = 0.750 mol x 2 = 1.50 mol
We must now find the number of of moles of oxygen atoms.
(I don't understand how you can get mols of C atoms and H atoms, but not mols of O atoms at this point.)
Mass of carbon = 0.500 mol x 12.0 g / mol = 6.00 g
Mass of hydrogen = 1.50 mol x 1.00 g / mol = 1.50 g
mass of oxygen = 15.5g - (6.00 + 1.50) g = 8.0 g
moles of oxygen atoms = 8.0 g / 16.0 g / mol = 0.50 mol
Hence mole ratio is
C = 0.500 mol
H = 1.50 mol
O = 0.50 mol
Multiply ratios by 1.5 to get whole numbers
C = 1 mol
H = 3 mol
O = 1 mol
Therefore, the empirical formula = CH3O
Hi, I have a example question in my textbook that I don't fully understand. If someone could please help me out I would really appreciate it! Thanks!
----------------------------------------------------------------------------
In this example, 15.5 g of a compound containing carbon, hydrogen and oxygen is completely burned in air; 22.0 g of CO2 and 13.5 g of water vapour are produced. Find the empirical formula.
Moles of CO2 = 22.0 g / 44.0 g / mol = 0.500 mol
Moles of H2O = 13.5 g / 18.0 g / mol = 0.750 mol
moles of C atoms = 0.500 mol and moles of H atoms = 0.750 mol x 2 = 1.50 mol
We must now find the number of of moles of oxygen atoms.
(I don't understand how you can get mols of C atoms and H atoms, but not mols of O atoms at this point.)
Mass of carbon = 0.500 mol x 12.0 g / mol = 6.00 g
Mass of hydrogen = 1.50 mol x 1.00 g / mol = 1.50 g
mass of oxygen = 15.5g - (6.00 + 1.50) g = 8.0 g
moles of oxygen atoms = 8.0 g / 16.0 g / mol = 0.50 mol
Hence mole ratio is
C = 0.500 mol
H = 1.50 mol
O = 0.50 mol
Multiply ratios by 1.5 to get whole numbers
C = 1 mol
H = 3 mol
O = 1 mol
Therefore, the empirical formula = CH3O