Question

4. Vitamin C, M= 176.12 g/mol, is a compound of composed of carbon, hydrogen, and oxygen. Vitamin C is found in many natural sources especially citrus fruits. When a 1.000-g sample of vitamin C is burned in a combustion apparatus 1.50 g of CO2 and 0.410 g of H2O are produced. What is the empirical formula of Vitamin C? What is the molecular formula of Vitamin C?

Answer 1

I assume M stands for molar mass.
CxHyOz + O2 --> CO2 + H2O
Convert g CO2 to g C and g H2O to g H.
1.50 g CO2 x (M C/M CO2) = 1.50 x 12/44 = 0.409
0.410 g H2O x (2*1/18) = 0.0455
g O = 1.00 - g C - g H = 1.00 -0.409 - 0.0455 = 0.545 g O
Convert to mols.
mols C = 0.409/12 = 0.0341
mols H = 0.0455/1 = 0.0455
mols O = 0.545/16 = 0.0341
Now you want to find the ratio of C to H to O with no number being less than 1.0. The easy way to start is to divide everything by the smallest number.
C = 0.0341/0.0341 = 1.00
H = 0.0410/0.0341 = 1.33
O = 0.0341/0.0341 = 1.00
These numbers are not small whole numbers so we multiply by integers to obtain, when rounded, small whole numbers.
For example, multiply by 2 to get
C = 0.0341/0.0341 = 1.00 x 2 = 2
H = 0.0455/0.0341 = 1.33 x 2 = 2.66
O = 0.0341/0.0341 = 1.00 x 2 = 2.00
Doesn't work so try 3.
C = 0.0341/0.0341 = 1.00 x 3 = 3.00
H = 0.0455/0.0341 = 1.33 x 3 = 3.99
O = 0.0341/0.0341 = 1.00 x 3 = 3.00
So the empirical formula is C3H4O3 and empirical mass is 88
The molecular formula is always an integer x empirical mass so
empirical mass x integer = 176.12 and round to whole number.
88 x integer = 176.12
integer 176.12/88 = 2.001 which rounds to 2.00 as a whole number so the molecular formula is (C3H4O3)2 = C6H8O6.

Answer 2

I thought you needed to add 0.409 +0.0455 and then subtract them from 1.00? Why did you subtract all of them?