0.290g of an organic compound containing C, H and O gave on combustion 0.270g of water and 0.66g of CO2. What is emperical formula of the compound.
Dear friend
First calculate % of c, h and o by using following formula.
% of c=(12/44)(mass of carbon dioxide / mass of organic compound)*100=62.07%
% of h= (2/18)(mass of water/mass of organic compound)*100=10.34.
% of o=100-(10.34+62.07)=27.59
Next you need to follow the following steps
Calculate moles of each elements as you know about mass from the above percentage values.
No. Of mples of
C=5.1725. h=10.34.o=1.72
Divide with number of moles by least number value i.e, 1.72 you get simple whole number ratio.
C=3.h=6.0=1
Empirical formula is C3H6O
Hope u understand and this answer will help u
First calculate % of c, h and o by using following formula.
% of c=(12/44)(mass of carbon dioxide / mass of organic compound)*100=62.07%
% of h= (2/18)(mass of water/mass of organic compound)*100=10.34.
% of o=100-(10.34+62.07)=27.59
Next you need to follow the following steps
Calculate moles of each elements as you know about mass from the above percentage values.
No. Of mples of
C=5.1725. h=10.34.o=1.72
Divide with number of moles by least number value i.e, 1.72 you get simple whole number ratio.
C=3.h=6.0=1
Empirical formula is C3H6O
Hope u understand and this answer will help u