Hydrocarbon+O2>>> CO2 + H2O
CO2+Ca(OH)2>>CaCO3 + H2O
mass CaCo3=10g
MassC=10/molmassCaCO3*atomicmassC
= 10/100*12=1.2 g of C=.1mol
Mass water=2.25
moles H = 2.25/18 * 2=.25
lowest ratio C:.1 H:.25 or
CH.25 or CH4 is the empirical formula.
a hydrocarbon was burnt completely and the products were bubbled into excess of Ca(OH)2 solution. The resulting mixture was filtered, and the residue dried and weighed. The mass of the hydrocarbon was found to be 10.00g. In a separate experiment, the same amount of hydrocarbon was burnt an the mass of the water produced was 2.25g. Determine the empirical formula of the hydrocarbon
for the sake of my understanding could you explain what this resulting solution is?
3 answers
C2H5
I should point out here that your problem makes no sense to me. True that the CO2 bubbled into Ca(OH)2 produces CaCO3 BUT you say the mass of the hydrocarbon is 10 g. So did you start with 10 g hydrocarbon or did the CaCO3 weigh 10 g. I don't call CaCO3 a hydrocarbon but chemistry may have changed since I was 15. I believe Bob Pursley made an "eye" typo, as he shows C as 0.1 and H as 0.25 but read that as C of 1 and H of 0.25. C of 0.1 and H of 0.25 gives C2H5.
1 answer