Empirical formula

There are two methods to solve this problem

Relative atomic mass (H=1, C=12, O=16)
First method,
CxHyOz+(x+y/4-z/2)O2-->xCO2+(y/2)H2O
No. of mole of CO2 produced is 0.1910g/(12+16*2)=0.00434 mole
No. of mole of H2O produced is 0.1172g/(1*2+16)=0.00651 mole
No. of mole of O2 used up is (0.1910g+0.1172g-0.100g)/(16*2)=0.00651 mole
So, the mole ratio of carbon dixode(x) : water(y/2) : oxygen(x+y/4-z/2) is
~ 3 : 3 : 2 respectively
So,x = 2,
y/2 = 3
y = 6
x+y/4 -z/2 = 3
2 + 6/4 -z/2 = 3
z = 1
So, the empirical formula is C2H6O

Second method,
Mass of carbon in CO2 is 0.1910g*(12/(12+16*2))=0.0521g
So, mass of carbon in alcohol is 0.0521g
Mass of hydrogen in H2O is 0.1172g*(1*2/(1*2+16))=0.0130g
So, mass of hydrogen in alcohol is 0.0130g
Mass of oxygen in alcohol is 0.100g-0.0521g-0.0130g=0.0349g
No. of mole of carbon, hydrogen and oxygen is
0.0521/12,0.0130/1,0.0349/16
0.0434,0.0130,0.00218 respectively
So, the mole ratio of carbon : hydrogen : oxygen is
0.0434:0.0130:0.0218
~2:6:1
So, the empirical formula is C2H6O

CxHyOz + (x+y/4-z/2)O2-->xCO2 + (y/2)H2O
No. of mole of CO2 produced = 0.1910 / (12+16x2) = 0.00434 mole
No. of mole of H2O produced = 0.1172 / (1x2 + 16) = 0.00651 mole
No. of mole of O2 used up = (0.1910 + 0.1172 - 0.100) / (16x2) = 0.00651 mole
mole ratio ( oxygen : water : carbon dixode) ~ 3 : 3 : 2
From the mole ratio,
x = 2,
y/2 = 3 => y = 6
x+y/4 -z/2 = 3
2 + 6/4 -z/2 = 3
0.5 = z/2
z = 1
The empirical formula: C2H6O