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?

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

Heya dearfirst love doesn't matter what the it's just happen so we cannot say that minor in love with someone over 18 years old