A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.8635 g of CO2 and 0.1767 g of H2O. What is the empirical formula of the compound?

Question

i got C3H3O2 IS IT RIGHT?

Answer 1

mass of C in sample = mass of C in CO2
mass of C in sample = (0.8635 g CO2 / 44 g/mol)(1 mol C / 1 mol CO2)(12 g/mol) = 0.2355 g

mass of H in sample = mass of H in H2O
mass of H in sample = (0.1767 g H2O / 18 g/mol)(2 mol H / 1 mol H2O)(1 g/mol) = 0.0196 g

mass of O in sample = total mass of sample - mass of C - mass of H
mass of O in sample = 0.4647 g - 0.2355 g - 0.0196 g = 0.2096 g
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Calculate for moles of C, H & O from the calculated masses:

moles of C = 0.2355 g / 12 g/mol = 0.0196 mol
moles of H = 0.0196 g / 1 g/mol = 0.0196 mol
moles of O = 0.2096 g / 16 g/mol = 0.0131 mol
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Divide this moles with the lowest value which is 0.0131 mol

C = 0.0196 / 0.0131 = 1.5
H = 0.0196 / 0.0131 = 1.5
O = 0.0131 / 0.0131 = 1
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Multiply by 2 to make the ratios a whole number:

C = 1.5 x 2 = 3
H = 1.5 x 2 = 3
O = 1 x 2 = 2

Empirical Formula is C3H3O2

Answer 2

Looks correct to me.

Answer 3

I didn't punch in your numbers, but the setups are correct, and if the CH ratio are 1.5 and O is 2, then you are correct.