# Question f2c30

May 12, 2015

The molar mass of the coumpound is 280 g/mol and its molecular formula is ${C}_{20} {H}_{40}$.

So, you know that when you dissolve 100 mg of your unknown compound in 1 g of benzene, you get a lower freezing point than that of pure benzene.

This means that you can use the freezing-point depression formula to solve for the solution's molality.

$\Delta {T}_{\text{f}} = i \cdot {K}_{f} \cdot b$, where

$\Delta {T}_{\text{f}}$ - the freezing point depression;
$i$ - the van't Hoff factor;
${K}_{f}$ - the cryoscopic constant;
$b$ - the molality of the solution.

For molecular compounds the van't Hoff factor is equal to 1.

Solve for the molality of the solution by

DeltaT_"f" = K_f * b => b = (DeltaT_"f")/K_f

b = (1.75^@cancel("C"))/(4.90^@cancel("C")/"m") = "0.357 m"

Since molality is defined as moles of solute per kilogram of solvent, you can calculate the number of moles of your unknown compound by

b = n_"solute"/m_"solvent" => n_"solute" = b * underbrace(m_"solvent")_(color(blue)("in kilograms!"))

${n}_{\text{solute" = 0.357"moles"/cancel("kg") * 1.00 * 10^(-3)cancel("kg") = 0.357 * 10^(-3)"moles}}$

Determine the compound's molar mass by using its given mass

M_M = m/n = (100 * cancel(10^(-3))"g")/(0.357 * cancel(10^(-3))"moles") = "280 g/mol"#

To determine its molecular formula, use its empirical formula

${\left(C {H}_{2}\right)}_{n} = \text{280 g/mol}$

This will get you

$n = \left(280 \cancel{\text{g/mol"))/((12.0 + 2.02)cancel("g/mol}}\right) = \frac{280}{14.02} = 19.97 \cong 20$

Therefore, your compound's molecular formula is

${\left(C {H}_{2}\right)}_{20} = \textcolor{g r e e n}{{C}_{20} {H}_{40}}$

SIDE NOTE If you go by the number of sig figs you gave for the mass of unknown compound dissolved in benzene, the answer should be 300 g/mol.

However, I'm quite sure that the answer's supposed to be 280, so I'll leave it with two sig figs.