"Published in Newark, California, USA"
A mixture of 0.660 grams of camphor and 0.050 grams of an organic solute freezes at 157°C. If the solute contains 10.5% H by weight, determine the molecular formula of the solute if the freezing point of camphor is 178.4°C and the value of Kf is 38.
Solution:
The freezing point constant is defined as the number of degrees the freezing point will be lowered per mole of solute per 1000 g or 1 kg of solvent present. This can be written as
where m is the molality of a solution.
The freezing point depression is defined as the product of the freezing point constant of a solvent and the molality of a solution.
Hence, the molality of a solution is
If the weight of a solvent which is camphor is given, then we can solve for the moles of an organic solute as follows
The molecular weight of an organic solute is
In 1 gmole of an organic solute, if the percent of H is 10.5% by weight, then the weight of H is
and the weight of C is
The number of moles of H is
and the number of moles of C is
In order to get the number of atoms of C and H, we need to divide them1. with the smallest number of moles. In this case, C has a number of smallest moles.
The number of H atoms is
and the number of C atoms is
Hence, the empirical formula of an organic solute is CH1.4.
Since CH1.4 cannot be accepted as a molecule, then we have to multiply each number of atoms by 10 in order to eliminate the decimal number at H atoms.
Therefore, the molecular formula of an organic solute is C10H14 which is an aromatic compound.