Question

combustion of a 0.9835g of a compound sample

Answer 1

If the compound contains only Carbon, hydrogen and oxygen and it produced 1.9g of CO2 and1.07g of H2O, then the empirical formula becomes:

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.

 The formula is C₄H₁₁O₂