When a 1.200g sample of a compound containing carbon, hydrogen, and nitrogen was combusted using exess oxygen, 2.929g of CO2 and 0.8001g of H2O were obtained. Analasys of the compounds showed the molecular mass to be 108.16g/mol. What is the molecular formula of the compound?
Convert grams CO2 and grams H2O to grams C and grams H.
2.929 g CO2 x (atomic mass C/molar mass CO2) = 2.929 x (12.01/44.01) = 0.7993
0.8001 x (2*atomic mass H/molar mass H2O) = 0.8001 x (2*1/18) = 0.0889
mass N = 1.200 - mass CO2 - mass H
mass N = 1.200 - 0.7993 - 0.0889 = 0.3118
Now convert to mols.
mols C = g/atomic mass = 0.7993/12.01 = about 0.0665
mols H = 0.0889/1 = 0.0889
mols C = 0.3118/14 = 0.0222
Now you find the smallest set of whole numbers to fit these three values; the easy way to do that is to divide all by the smallest number. That is 0.0222; therefore,
C = 0.0665/0.0222 = 2.99 = 3
H = 0.0889/0.0222 = 4.00
N = 0.0222/0.0222 = 1.00 so the molecular formula is C3H4N with an empirical formula of 3*12.01 + 4*1.008 + 1*14.007 = approx 54
The molecular formula is a multiple of that so
empirical weight x a multiple = molecular weight.
54 x ? = 108 so the multiple is108/54 = 2. The molecular formula then is (C3H4N)2 or C6H8N2. You're welcome.