3g C = 1/4 mole
8g O2 = 1/2 mole
So, you have CO2
A 11.0g sample of compound containing only oxygen and carbon produces 3.0g carbon and 8.0g oxygen. Determine the empirical formula of the compound.
3 answers
CxOy ==> xC + 1/2O2
% C = (3/11)*100 = ? about 27
% O2 = (8/11(*100 = about 73
mols C = about 27/12 = about 2.25
mols O2 = about 73/32 = about 2.28 O2 molecules or about 4.56 mols O atoms.
Now find the ratio of the C atoms to the O ATOMS, round to whole numbers and that will be the empirical formula. The easy way to find the ratio is to divide the smaller number by itself (which makes it 1.00), then divide the other number by the same small number.
% C = (3/11)*100 = ? about 27
% O2 = (8/11(*100 = about 73
mols C = about 27/12 = about 2.25
mols O2 = about 73/32 = about 2.28 O2 molecules or about 4.56 mols O atoms.
Now find the ratio of the C atoms to the O ATOMS, round to whole numbers and that will be the empirical formula. The easy way to find the ratio is to divide the smaller number by itself (which makes it 1.00), then divide the other number by the same small number.
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