Question

Looking at the bottom of her tea cup, Prof. Trelawney suddenly has a revelation! She knows one of the previous exam questions! She bursts into the class and and yells: Analysis of a sample is finds it contains 15.0% of species X, by mass, 75.0% of species Y, and 10.0% of species Z, by mass. The molar mass of species X is 45.0 g/mol, the molar mass of species Y is 150. g/mol, and the molar mass of species Z is 30.0 g/mol. If the molar mass of the sample is 1200 g/mol, what is the molecular formula of the sample? Prof. Vitarelli frowns at her disapprovingly

Answer 1

Answer : The molecular formula of the sample is, $X_4Y_6Z_4$

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 X = 15.0 g

Mass of Y = 75.0 g

Mass of Z = 10.0 g

Molar mass of X = 45.0 g/mole

Molar mass of Y = 150  g/mole

Molar mass of Z = 30.0 g/mole

Step 1 : convert given masses into moles.

Moles of X = $\frac{\text{ given mass of X}}{\text{ molar mass of X}}= \frac{15.0g}{45.0g/mole}=0.33moles$

Moles of Y = $\frac{\text{ given mass of Y}}{\text{ molar mass of Y}}= \frac{75.0g}{150g/mole}=0.5moles$

Moles of Z = $\frac{\text{ given mass of Z}}{\text{ molar mass of Z}}= \frac{10.0g}{30.0g/mole}=0.33moles$

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

For X = $\frac{0.33}{0.33}=1$

For Y = $\frac{0.5}{0.33}=1.5$

For Z = $\frac{0.33}{0.33}=1$

The ratio of X : Y : Z = 1 : 1.5 : 1

To make in the whole number we multiply the ratio by 2, we get:

The ratio of X : Y : Z = 2 : 3 : 2

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

The Empirical formula = $X_2Y_3Z_9$

The empirical formula weight = 2(45.0) + 3(150) + 2(30.0) = 600 gram/eq

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

Formula used :

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

$n=\frac{1200}{600}=2$

Molecular formula = $(X_2Y_3Z_2)_n=(X_2Y_3Z_2)_2=X_4Y_6Z_4$

Therefore, the molecular of the sample is, $X_4Y_6Z_4$