Question

Question #f2c30

Answer 1

The molar mass of the coumpound is 280 g/mol and its molecular formula is `C_20H_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.

`DeltaT_"f" = i * K_f * b`, where

`DeltaT_"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_"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

`(CH_2)_n = "280 g/mol"`

This will get you

`n = (280cancel("g/mol"))/((12.0 + 2.02)cancel("g/mol")) = 280/14.02 = 19.97 ~= 20`

Therefore, your compound's molecular formula is

`(CH_2)_(20) = color(green)(C_20H_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.