[MagicMoriarty]

“0.493 g of a compound with Mr = 86 containing carbon, hydrogen and oxygen only was burned in excess oxygen. It produced 1.261 g of carbon dioxide and 0.516 g of water. Calculate the empirical and molecular formula of the compound.“

I tried this question but I don’t know how to get the answer. I attached my attempt.

[sjb]

Consider typing the answer out to help searching. There is no guarantee that the oxygen in the water and the carbon dioxide is solely from the initial compound.

[chenbeier]

Do another approach
With the given mass of CO₂ and H₂O calculate the Mole and also the mass of C and H. The sum of C and H you substract from die given mass of the substance gives the mass of oxygen. Calculate also the moles.
Now you have 3 numbers of moles, set Oxygen to 1 and calculate H and C. Compare die result with the given molare weight.

[MNIO]

the unbalanced equation looks something like this
  __ CxHyOz + __ O2 ---> __ CO2 + __ H2O

notice 2 things
  (1) all the C in CO2 came from CxHyOz... there's no C in O2 on the left is there?
  (2) all the H in H2O came from CxHyOz... same reason.. no H in oxygen gas.

now let's map out a solution
  (1) convert mass CO2 to mass C.. that's also the mass of C in CxHyOz
  (2) convert mass H2O to mass H.. that's also the mass of H in CxHyOz
  (3) mass O in CxHyOz = mass CxHyOz - mass C - mass H
  (4) convert mass to moles... (empirical and molecular formulas are MOLE not mass ratios)
  (5) simply mole ratios to get empirical formula
  (6) divide molar mass by empirical unit mass to get # of empirical units per mole
  (7) write molecular formula based on (6) and (5)

something like this
  1.261g CO2 * (12.011g C / 44.01g CO2) = ___g C
  0.516g H2O * (2.016g H / 18.02g H2O) = ___g H
  grams O = 0.493g CxHyOz - __g C - __g H
then
  mole C = __g C * (1 mol / 12.011g) = ___ mol
  mole H = __g H * (1 mol / 1.008g) = ___ mol
  mole O = __g ? * (1 mol / __g) = ___ mol
then simplify by dividing all mole values by the smallest value of mole C, mole H, mole O
  mole C = __ mol / __ mol =
  mole H = __ mol / __ mol =
  mole O = __ mol / __ mol =
note that we're dividing all mole #'s by the same value!
then
  empirical formula = simplified mole ratio
then
  empirical unit mass = __C * 12.011 + __H * 1.008 + __O * 16.00 = __
then
  # empirical units per molecule = molar mass / empirical unit mass = 84 / __ = ?
then
  molecular formula = __ * empirical formula = C_H_O_

**********
fill in the blanks

[benzene135]

Here's my work:
0.493 g Total, Molar mass= 86 amu
1.261 g CO₂ (1 mol CO₂ / 44.0096 g CO₂)(1 mol C / 1 mol CO₂)(12.0108 g C / 1 mol C) = 0.3441 g C
0.516 g H₂O (1 mol H₂O / 18.0153 g H₂O)(2 mol H / 1 mol H₂O)(1.00794 g H / 1 mol H) = 0.0577 g H
0.493 g total - 0.3441 g C - 0.0577 g H = 0.0912 g O
0.3441 g C (1 mol C / 12.0108 g C) = 0.02865 mol C
0.0577 g H (1 mol H / 1.00794 g H) = 0.0572 mol H
0.0912 g O (1 mol O / 15.9994 g O) = 0.0057 mol O
Empirical Formula: C_0.02865H_0.0572O_0.0057 <- Divide by smallest number (0.0057)
Empirical Formula: C₅H₁₀O
To calculate the molecular formula, divide the molar mass of the unknown compound by the molar mass of the empirical compound.
Molar Mass of the Empirical Compound: (5)(12.0108) + (10)(1.00794) + 15.9994 = 86.1328
86 amu / 86.1328 amu = approx. 1.
Therefore, the molecular formula is: C₅H₁₀O