Question

Combustion of a 0.9835-g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900 g of co2 and 1.070 g of h2o. what is the empirical formula of the compound?

Answer 1

Find grams of C and H, using molar masses: 
12.0 g C  x  (1.900 g CO2 /  44.0 g CO2)  = 0.5182 grams C 
2.02 g H  x  (1.070 g H2O /18 g H2O) = 0.1201 grams H 

Since all the C and H in CO2 and H2O will come from the sample except oxygen, because we need to provide oxygen for sample to burn.

0.9835 g sample - 0.5182 g O - 0.1201 g H = 0.3452 grams O 


find moles, using molar masses: 
0.5182 grams C / 12.0 g/mol C = 0.0432 moles C 
0.1201 grams H / 1.01 g/mol H = 0.119 moles H 
0.3452 grams O / 16.0 g/mol O = 0.0216 moles O 

0.0216 is lesser, so use it to normalize to find the ratio.
find ratios: 
0.0432 moles C / 0.0216 = 2 moles C 
0.119 moles H / 0.0216 = 5.5 moles H 
0.0216 moles O / 0.0216 = 1 mole O 


so, the ratio is C2 H{5.5} O1
double the ration to eliminate decimals.
 
gives your answer is $C_4 H_{11} O_2$