Verified answer

mass of CO2 = 31.2 mg or 0209*10^-2

0.709 mmoles of CO2

or 0.709 mmoles of C or

8.51 mg of C

mass of H2O = 9.57 mg or

0.532 mmoles of H2O or

1.063 mmoles of H or

1.072 mg of H

mass of compound burnt = 20.9 mg

mass of C+H = 9.58 mg

mass of O = 11.319 mg

mmoles of O = 0.707

molar ratio of C : H :O = 0.709 : 1.06 : 0.70745

or after dividing by the smallest 1.000 : 1.500 : 0.998

multiply by 2 and we get

C:H:O = 2 : 3 : 2

compound is C2H3O2

The amount of CO2 will give you the amount of C in the sample 3.12×10-2 g

The amount of H2O will give you the amount of H in the sample 2 x 9.57×10-3 g (there are 2 H)

The rest is the amount of O in the sample = 2.09×10-2 less the above =

So now you have the amount of C, H, and O

However,... working out your numbers there seems to be some errors... can you please recheck the numbers?

You have to know that the carbon in the CO2 came from the carbon in the original sample. If we can calculate the mass of carbon in 0.0312g CO2, we know how much carbon was in the original sample

Likewise, the H in the H2O came from the H in the original sample. If we can calculate this, we know how much H was in the sample

If we know the mass C and the mass of H we can calculate the mass of O by subtraction:

Let us do this:

Mass of C in 0.0312g CO2

Molar mass CO2 44.009g/mol

Molar mass C = 12.011g/mol

Mass C in 0.0312g CO2 = 12.011/44.009*0.0312 = 0.008515g

Mass of H in 0.00957g H2O

Molar mass H2O = 18.015g/mol

Molar mass H = 1.008 H2 = 2.016

Mass H in 0.00957g H2O = 2.016/18.015*0.00957 = 0.001071g H

Mass O = 0.0209 - ( 0.008515 + 0.001071) = 0.011314g

divide by respective atomic mass:

C = 0.008515 / 12.011 = 0.0007089

H = 0.001071/1.008 = 0.001063

O = 0.011314/15.999 = 0.00070717

Divide trough by smallest result:

C = 1

H = 1.5

O = 1

Multiply by 2 to bring to whole integers:

C = 2

H= 3

O = 2

Empirical formula = C2H3O2

CxHyOz + O2 --> CO2 + H2O

0.0312gCO2

12gC / 44g/mole x100% = 27.3% C in CO2 or

27.3% of 0.0312g = 0.0085gC

0.00957gH2O

2g/18g/mole x 100% = 11.1%H in H2O

11.1% of 0.00957g = 0.0011gH

the rest is O but some of this O is from the compound and the rest is from the O2

0.0209gsample has 0.0085gC and 0.0011gH then ?gO

0.0209g - 0.0085g - 0.0011g = gO = 0.0113g O

now find moles

0.0085gC/12g/mole = 7.08 x 10^-4molesC

0.0011gH/1g/mole = 1.1 x 10^-3 molesH

0.0113gO/16g/mole = 7.06 x 10^-4moles O

molar ratio, divide each moles by the smallest number

0.000708/0.000706 = ~1

0.0011/0.000706 = 1.5

0.000706/0.000706 = 1

x=1, y=1.5 z=1 but we can't have fractions of moles in an empirical formula so

C2H3O2