Question

A compound is found to contain 37.32 % phosphorus , 16.88 % nitrogen , and 45.79 % fluorine by

Answer 1

Answer: 1. The empirical formula is $PNF_2$  

2. The molecular formula is $PNF_2$

Explanation:

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 P = 37.32 g

Mass of N = 16.88 g

Mass of F = 45.79 g

Step 1 : convert given masses into moles.

Moles of P =$\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{37.32g}{31g/mole}=1.20moles$

Moles of N =$\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.88g}{14g/mole}=1.20moles$

Moles of F =$\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{45.79g}{19g/mole}=2.41moles$

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

For P = $\frac{1.20}{1.20}=1$

For N = $\frac{1.20}{1.20}=1$

For F =$\frac{2.41}{1.20}=2$

The ratio of P: N: F= 1: 1: 2  

Hence the empirical formula is $PNF_2$

The empirical weight of $PNF_2$= 1(31)+1(14)+2(19)= 82.98 g.

The molecular weight = 82.98 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{82.98}{82.98}=1$

The molecular formula will be=$1\times PNF_2=PNF_2$