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.
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?
2 answers
How come the 3.00mg the mass of carbon instead of CO2. Please explain
3 answers