Question

Pheromones are compounds secreted by the females of many insect species to attract males. One of these compounds contains 80.78 percent C, 13.56 percent H, and 5.66 percent O. A solution of 1.00 g of this pheromone in 8.50 g of benzene freezes at 3.37°C. What is the molecular formula of the compound?

Answer 1

Answer : The molecular formula of a compound is,
`C_(19)H_(38)O`

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 80.78 g

Mass of H = 13.56 g

Mass of O = 5.66 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (80.78g)/(12g/mole)=6.732moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (13.56g)/(1g/mole)=13.56moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (5.66g)/(16g/mole)=0.3538moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =
`(6.732)/(0.3538)=19.0\approx 19`

For H =
`(13.56)/(0.3538)=38.3\approx 38`

For O =
`(0.3538)/(0.3538)=1`

The ratio of C : H : O = 19 : 38 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula =
`C_(19)H_(38)O_1=C_(19)H_(38)O`

The empirical formula weight = 19(12) + 38(1) + 1(16) = 282 gram/eq

Now we have to calculate the molar mass of pheromone.

Formula used :

`\Delta T_f=i* K_f* m\\\\T^o-T_s=i* K_f*\frac{\text{Mass of pheromone}}{\text{Molar mass of pheromone}* \text{Mass of benzene in Kg}}`

where,

`\Delta T_f` = change in freezing point

`\Delta T_s` = freezing point of solution =
`3.37^oC`

`\Delta T^o` = freezing point of benzene =
`5.50^oC`

i = Van't Hoff factor = 1 (for non-electrolyte)

`K_f` = freezing point constant for benzene =
`5.12^oC/m`

m = molality

Now put all the given values in this formula, we get

`(5.50-3.37)^oC=1* (5.12^oC/m)* \frac{1.00g}{\text{Mass of pheromone}* 0.00850kg}`

`\text{Mass of pheromone}=282.8g/mol`

Now we have to calculate the molecular formula of the compound.

Formula used :

`n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}`

`n=(282.8)/(282)=1`

Molecular formula =
`(C_(19)H_(38)O)_n=(C_(19)H_(38)O)_1=C_(19)H_(38)O`

Therefore, the molecular of the compound is,
`C_(19)H_(38)O`