Chemistry help?

Convert masses of CO2 and H2O into moles and then into moles C and moles H:

10.9 g CO2 / 44 g/mol = 0.248 mol CO2 X (1 mol C/1 mol CO2) = 0.248 mol C

3.73 g H2O / 18 g/mol = 0.207 mol H2O X (2 mol H/1 mol H2O) = 0.414 mol H

Divide each by the smaller:

0.248 mol C / 0.248 = 1 C

0.414 mol H / 0.248 = 1.67 H

Multiply those both by 3 to make the second a whole number:

1C X 3 = 3 C

1.67 H X 3 = 5 H

So, the empirical formula is C3H5

Initially follow the same process to get moles C and moles H:

7.19 g CO2 / 44 g/mol X (1 mol C/1 mol CO2) = 0.163 mol C

2.94 g H2O / 18 g/mol = 0.163 mol H2O X (2 H/1 H2O) = 0.327 mol H

Convert moles C and H to masses and subtract from mass of sample to get mass O, and convert that to moles:

0.163 mol C X 12 g/mol = 1.95 g C

0.327 mol H X 1 g/mol = 0.33 g H

Mass O = 3.60 g - (1.95 + 0.33) = 1.32 g O

moles O = 1.32 g / 16 = 0.082 mol O

So, you have"

0.163 mol C

0.327 mol H

0.082 mol O

Divide each by the smallest:

0.163 C / 0.082 = 2C

0.327 / 0.082 = 4 H

0.082/0.082 = 1 O

So, the empirical formula is C2H4O

10.9 g CO2 x 1mole CO2//44.0 g x 1 moles C/1 mole CO2 = 0..2477 moles C

3.73 g H2O x 1mole H2O/18.0 g x 2mole H/1 mole H2O = 0.4144

0.2477 moles C ; 0.4144 moles H

divide the above moles by ; 0.2477

C1H1.67 to obtain whole number multiply by 3

C3H5

7.19 g CO2 x 1mole CO2/44g x 1 mole C/1mole CO2 = 0.163 moles C

2.94 g H2O x 1mole H2O/ 18.0 g x 2 moles H/1mole H2O = 0.3266 moles H

0.163 moles C x 12.0 g C/1mole C = 1.956 g C

0.3266 moles H x 1.0 g H/1mole H = 0.3266 g H

grams of oxygen = 3.60 g – 1.956 g C - 0.3266 g H = 1.317 g O

1.317 g O x 1 mole O/16.0 g = 0.0823 moles O

0.163 moles C; 0.3266 moles H, 0.0823 moles O

Divide the moles by 0.0823

Empirical formula is C2H4O1