Question

The percentage of CHO in vitiamin C was determined by buring a simple weighing 2.00mg.
The masses of carbon(iv)oxide and water formd are 3.00mg and 0.816mg respectively.
From this, find the percentage of each element and the emperical formular of vitamin C?

Answer 1

mass C = 3.00 mg x (12/44) = 0.818 mg
mass H = 0.816 x (2/18) = 0.0906 mg
mass O = mass sample - mass C - mass H = 2.00 - 0.818 - 0.0906 = 1.09 mg

mols C = 0.818/12 = 0.0682
mols H = 0.0906/1 = 0.0906
mols O = 1.091/16 = 0.0682

Divide by the smallest to obtain smallest ratio of 1 to obtain
C = 0.0682/0.0682 = 1
H = 0.0906/0.0682 = 1.33
O = 0.06820.0682 = 1

Now recalculate to make the ratios even numbers by multiplying H by 3 to obtain
C = 1 x 3 = 3
H = 1.33 x 3 = 3.99 and round to 4
O = 1 x 3 = 3
So empirical formula is C3H4O3.

Answer 2

How come the 3.00mg the mass of carbon instead of CO2. Please explain