Question

You add 0.255 g of an orange, crystalline compound whose empirical formula is `C_10H_8Fe` to 11.12 g of benzene. The boiling point of the benzene rises from 80.10 C to 80.26 C. What are the molar mass and molecular formula of the compound?

Answer 1

The molecular formula is `"C"_20"H"_16"Fe"_2`, and the molar mass is 368.03 g.

Explanation:

The formula for boiling point elevation is

`color(blue)(bar(ul(|color(white)(a/a) ΔT_"b" = K_"b"mcolor(white)(a/a)|)))" "`

where

`ΔT_"b"` = the increase in the boiling point
`K_"b"` = the molal boiling point elevation constant of the solvent
`m` = the molality of the solution

We can rearrange the formula to get

`m = (ΔT_"f")/K_"f"`

In your problem,

`ΔT_"f" = T_2 - T_1 = "80.26 °C - 80.10 °C" = "0.16 °C"`
`K_"b" = "2.53 °C·kg·mol"^"-1"`

∴ `m = (0.16 color(red)(cancel(color(black)("°C"))))/(2.53 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "0.0632 mol·kg"^"-1"`

Now, `m = "0.0632 mol"/(1 color(red)(cancel(color(black)("kg")))) = "0.255 g"/("0.011 12" color(red)(cancel(color(black)("kg"))))`

Divide both sides of the equilibrium by 0.0632.

∴ `"1 mol" = "0.255 g"/"0.011 12" × 1/0.0632 = "363 g"`

The molar mass is 363 g/mol, so the molecular mass is 363 u.

The empirical formula mass of `"C"_10"H"_8"Fe"` is 184.02 u.

The molecular mass must be an integral multiple of the empirical formula mass.

`"MM"/"EFM"= n`

`n = "MM"/"EFM" = (363 color(red)(cancel(color(black)("u"))))/(184.02 color(red)(cancel(color(black)("u")))) = 1.97 ≈ 2`

The molecular formula must be twice the empirical formula.

`"MF" = ("C"_10"H"_8"Fe")_2 = "C"_20"H"_16"Fe"_2`

Its molar mass is 368.03 g.