Question

A 100 mg sample of compound CxHyOz (where x,y, and z are subscripts) was burned to form 199.8 mg CO2 and 81.8 mg water. Find the empirical formula

Answer 1

First, try writing out the chemical equation for the reaction.

Answer 2

CxHyOz >> xCO2+ y/2H2O
since the number of moles in both side of the equation must be equal
x moles of C in CxHyOz will give x moles of CO2
y moles of H in CxHyOz will give y/2 moles of H2O (because there's 2 moles of H in one mole of H2O)

so back to the calculations:
m(CO2)=0.1998g m(H2O)=0.0818g
n(CO2)=m/M n(H2O)=m/M
= 0.1998/(12.01+2x16) =0.0818/(2x1.008+16)
=4.540x10^-3 mol = 4.54x10^-3mol

so n(C in CxHyOz ) = n(CO2)
=4.540x10^-3 mol
n(H in CxHyOz ) = 2n(H20)
= 9.081x10^-3 mol
n(O in CxHyOz)=2n(CO2) + n(H2O)
=1.36x10^-2 mol
for
so ratio in C:H:O
= 4.540x10^-3 : 9.081x10^-3 : 1.36x10^-2
= 1 : 2 : 3

the empirical formula for CxHyOz is CH2O3 （H2CO3, the substance is carbonic acid)