[iluvlost]

A Compound Contains Only C, H, And O. Combustion Of 10.68mg Of The Compound Yields 16.01mg CO2 And 4.37mg H2O. The Molar Mass Of The Compound Is 17.61g/mol. What Are The Empirical And Molecular Formulas Of The Compound?

.01601 g CO2 * 12g C/ 44 g CO2 = .00437 g C
.00437 g H2O * 1 g H/ 18 g H2O = 2.428 E -4 g H
.01068g - .00437 g C + 2.428 E-4 g H = .006065 g
.00437 g C/ 12.011 g C = 3.64 E-4 mol C
2.428 E-4 g H / 1g H = 2.42 E-4 mol H
.006065 g O/ 16 g O = 3.79 E -4 mol O
3.64 E-4 mol C/ 3.79 E-4 mol O = .960
2.42 E-4 mol H/ 3.79 E-4 mol O = .639
EF C3H4O3

3*12g C + 4 * 1 g H + 3* 16 gO = 88
176.1/88 = 2.0
MF C6H8O6

[Borek]

Looks OK.