chemistry
posted by Sazon on .
Combustion analysis, Empirical and Molecular formulas, help!?
When 2.686 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 9.224 grams of CO2 and 1.511 grams of H2O were produced.
In a separate experiment, the molar mass of the compound was found to be 128.2 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.
Enter the elements in the order presented in the qustion.

empirical formula is the smallest whole number ratio
molecular formula is the empirical doubled i believe. 
you would divide the grams of each molecule produced by the smallest ratio to find the subscripts.

Start from a balanced equation
CxHy + (x+y/4)O2 > xCO2 + y/2 H2O
so 1 mole of CxHy (molar mass 12x+y)produces
x mole of CO2 and
y/2 mole of H2O
so
2.686 g/(12x+y) moles
produces
9.224/44 =x moles of CO2
so x=0.2096
and produces
1.511/18=y/2 moles of H2O
so y=0.1679
so the ratio is
x:y
0.2096:0.1679
or 5:4
Thus the empirical formula or the simplest formula is
C5H4
for which molar mass is 64 g
we are told in the question that the molar mass is 128.2 g so the molecular formula (ie. the formula for one mole) must be C10H8, that is twice the empirical formula in this case.