I assume this is NOT a hydrocarbon. I assume it consists of C, H, and O from the wording. It;s implied but not stated that way.
.......................Y + O2 ==> CO2 + H2O
....................15.2g...........22.0g........13.5
g C = 22.0 x (12/44) = 6
g H = 13.5 x (2/18) = 1.5
g O = 15.2 - 6 - 1.5 = 7.7
mols C = 6/12 = 0.5
mols H = 1.5/1 = 1.5
mols O = 7.7/16 = 0.48
Find the ratio in whole numbers to each other. The easy way to do that is to divide by the smallest number; i.e., 0.48, this way.
0.5/0.48 = about 1 rounded to a whole number.
1.5/0.48 = about 3 rounded to a whole number.
0.48/0.48 = 1
empirical formula is CH3O. Check my work.
For percentages:
%C = (g C/15.2)*100 =
% H = (g H/15.2)*100
etc.Post your work if you get stuck.
15.2g of sample Y were burned and a mass of CO2 in H2O results in 22.0g and 13.5g respectively.what is the empirical formula of Y (ii) calculate the percentage composition of carbon hydrogen and oxygen?
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