[gelboy1015]

Calculate each of the following quantities.

(e) A compound contains only carbon, hydrogen, and oxygen. Combustion of 11.75 mg of the compound yields 17.61 mg CO₂ and 4.81 mg H₂O. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound? (Type your answer using the format CO₂ for CO₂.)

empirical formula: ??

molecular formula: ??

thanks in advance 

[gelboy1015]

stumped

[Padfoot]

stumped

Start by calculating how many moles of C and H are in your products.  Mass of O must be the left over mass from 11.75mg.

[gelboy1015]

ok so far i did:

0.01761 g CO₂ * 1 mol C/12.01 g C = 0.001466278 mol C

0.00481 g H₂O * 2 mol H/2.02 g H = 0.004762376 mol H

then i used 11.75 mg for O:

0.01175 g O * 3 mol O/16.00 = 0.002203125 mol O

then i divided by the lowest: 0.001466278

0.004762376/0.001466278 = 3.247935248 mol H

0.002203125/0.001466278 = 1.502528852 mol O

i multiplied the mol O x 2 to get a whole number - 3.005057704
also did this for the others -
H - 3.247935248 * 2 = 6.495870496
C - 1 * 2 = 2

so it ended up with:

C - 2
H - 6
O - 3

so for empirical, i put C₂H₆O₃

is that right?

[Padfoot]

Almost.

0.01761 g CO₂ * 1 mol C/12.01 g C = 0.001466278 mol C

Divide by molar mass of CO₂ - this give moles CO2 which you can convert to moles C.

Same goes for H₂0. 

then i used 11.75 mg for O:

H and C also contribute to this mass, so:
Mass of O= 0.01175g - (mass of H + mass of C)

Other than that it's fine