Chemistry
posted by Anonymous on .
Sorry to bother again. I'm stuck in these two problems as well.
1. A 27.0 g smaple of an unknown hydrocarbon was burned in excess oxygen to form 88.0 g of carbon dioxide and 27.0 g of water. What iss a possible molecular formula of the hydrocarbon?
2. A sample of CaCO3 was reported as being 30% Ca. Assuming no Ca was present in any impurities, the % of CaCO3 in the sample is...
thanks.

#2. % CaCO3 = %Ca x (molar mass CaCO3/molar mass Ca) = ??

I would convert 88.0 g CO2 to percent C and 27.0 g H2O to percent H in the sample. Take 100 g sample, convert g C to mols and g H to mols. Then find the ratio of C to H. Post your work if you get stuck.

Thank you for all your help.
would the answer be CH4?
my ratio is 1 C to 4 H 
I don't find 1C to 4H.
%C = 88.9%
%H = 11.1%
Take 100 g cample which gives us 88.9 g C and 11.1 g H.
mols C in 88.9 g = 88.9/12.01 = ??
mols H in 11.1 g = 11.1/1 = ??
Now take the ratio and you don't get 1 to 4 but check me out on that. You do the arithmetic.
Hint: Divide the smaller number by itself which forces it to be 1.00. Divide the other number by the same small number which will then give you mols one to mols of the other. Don't throw away fractions unless they are less than about 0.15 more than or less than a whole number; i.e., 1 to 1.5 will be 2:3 
ok thank you so much for the help! :)

I STILL DONt GET IT
\ 
88g CO2* (1mol/44g) = 2mol
27g H2O* (1mol/18g)= 1.5mol
Convert the 1.5 mol to a whole number by multiplying by 2 so mol H2O is 3.
Then multiply the mol CO2 and the mol H2O by 2 again and your answer will be C4H6