Answer:
The empirical formula is C2H4O3
The molecular formula is C4H8O6
The molar mass is 152 g/mol
Explanation:
The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.
Step 1: Data given
Mass % of Carbon = 31.57 %
Mass % of H = 5.30 %
Freezing point = -5.20 °C
10.56 grams of the compound dissolved in 25.0 grams of water
Kf water = 1.86 °C kg/mol
Step 2: Calculate moles of Carbon
Suppose 31.57% = 31.57 grams
moles C = mass C / Molar mass C
moles C = 31.57 grams / 12.0 g/mol = 2.63 moles
Step 3: Calculate moles of Hydrogen:
Moles H = 5.30 grams / 1.01 g/mol
moles H = 5.25 moles
Step 4: Calculate moles of Oxygen
Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol
Moles O = 3.95 moles
Step 5: We divide by the smallest number of moles
C: 2.63 / 2.63 = 1 → 2
H: 5.25/2.63 = 2 → 4
O: 3.95/ 2.63 = 1.5 → 3
The empirical formula is C2H4O3
The molar mass of the empirical formula = 76 g/mol
Step 6: Calculate moles solute
Freezing point depression = 5.20 °C = m * 1.86
m = 5.20 / 1.86
m = 2.80 molal = 2.80 moles / kg
2.80 molal * 0.025 kg = 0.07 moles
Step 7: Calculate molar mass
Molar mass = mass / moles
Molar mass = 10.56 grams / 0.07 moles
Molar mass = 151 g/mol
Step 8: Calculate molecular formula
151 / 76 ≈ 2
We have to multiply the empirical formula by 2
2*(C2H4O3) = C4H8O6
The molecular formula is C4H8O6
The molar mass is 152 g/mol
is the density