Question

An organic compound containing only C,H and O was subjected to combustion analysis. A sample weighing 0.4758g yielded 0.765 g CO2 and 0.313 g H2O. What is the empirical formula of the compound?

Answer 1

well by using the combustion analysis method we can sort it out As
for carbon % = mass of CO2/mass of org.comp x 12.00/44.00 x100
for hydrogen= mass of H2O/mass of org.comp x 2.016/18 x100

now %age oxygen = 100 - (%age of C + % of H) get
so now simply putting the values............. we get
C = 43.84% H= 7.36% so O=100-(43.84+7.36)=48.8%

now moles calculation..............

simply div. %age/atomic mass(molarmass) so we get
C=3.65 moles
H=7.24.....
O=3.05moles
the last step is to obtain the atomic ratio so just div. the moles of elements by the least number of produced,,,,,,,,,,,,, (3.05 in present case)
for C =3.65/3.05=1.2
H=7.24/3.05= 2.3
O=3.05/3.05=1 so the empirical formula is CH2O ??? is that rite Ma'm @Shaekitchen