# What is the molecular weight of acetic acid if a solution that contains 30.0 grams of acetic acid per kilogram of water freezes at -0.93°C. Do these results agree with the assumption that acetic acid has the formula CH3CO2H?

Oct 8, 2014

The molecular mass from this experiment is 60 u. This agrees with the value of 60.05 u calculated from the formula CH₃CO₂H.

1. Calculate the molality of the solution.

ΔT_"f" = iK_"f"m, so

m = (ΔT_"f")/( iK_"f")

${T}_{\text{f" = T_"f"^° -ΔT_"f}}$ = 0.00 °C – (-0.93) °C= 0.93 °C
${K}_{\text{f" = "1.86 °C·kg·mol}}^{-} 1$

Assume that $i$ = 1. Then

m = (ΔT_"f")/( iK_"f") = "0.93 °C"/("1 ×1.86 °C·kg·mol"^-1) = 0.50 mol/kg

2. Calculate the molar mass.

Molar mass = $\text{30.0 g"/"0.50 mol}$ = 60 g/mol

So the molecular mass is 60 u.

You can do the above calculation in one step by rearranging the equation to read

Molar mass = (iK_"f")/(ΔT_"f") × "mass of [solute](http://socratic.org/chemistry/solutions-and-their-behavior/solute)"/"kilograms of [solvent](http://socratic.org/chemistry/solutions-and-their-behavior/solvent)"

3. Compare with the known value.

The molecular mass of CH₃COOH is 60.05 u.

The experimental and the theoretical values agree within experimental uncertainty.

Here's a video on calculating the molar mass from freezing point depression.