The percentages of carbon, hydrogen, and oxygen in vitamin c were determined by burning a sample weighing 2.00mg, the masses of carbon dioxide and water formed are 3.00mg and 0.81mg respectively. from this information, find the percentage of each element and the empirical formula of vitamin c.
CxHyOz + O2 ==> CO2 + H2O
2 mg........................3 mg......0.81 mg
mass C = 3 x 12/44 = 0.818 mg C
% C = 0.818/2)*100 = 40.9%
g H = 0.81 x 2*1/18 = 0.09 mg
% H = 0.09/2)*100 = 4.5%
%O = 100 - 40.9 - 4.5 = 54.6%
Take 100 g sample to have 40.9 g C, 4.5 g H, 54.6 g O.
mols C = 40.9/12 = 3.41
mols H = 4.5/1 = 4.5
mols O = 54.6/16 = 3.41
Now you want to find the ratio of each to each other with no whole number being less than 1.The easy way to do this is to divide the smallest number by itself and divide the other numbers by the same small number; therefore,
mols C = 3.41/3.41 = 1.00
mols H = 4.5/3.31 = 1.32
mols O = 3.41/3.41 = 1.00
To get whole numbers that 1.32 would be a whole number if we multiplied it by 3 so
mols C = 1 x 3 = 3.00
mols H = 1.32 x 3 = 3.96 rounds to 4.00
mols O = 1 x 3 = 3
So the empirical formula is C3H4O3. That may NOT be the molecular formula.