Question

Hydroxylamine hydrochloride is a powerful reducing agent which is used as a polymerization catalyst. It contains 5.80 mass % H, 20.16 mass % N, 23.02 mass % O, and 51.02 mass % Cl. What is its empirical formula? Determine the molecular formula of the compound with molar mass of 278 g.

Answer 1

Answer: The molecular formula will be $H_{16}NOCl$

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 H = 5.80 g

Mass of N = 20.16 g

Mass of O = 23.02 g

Mass of Cl = 51.02 g

Step 1 : convert given masses into moles.

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.80g}{1g/mole}=5.80moles$

Moles of N =$\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{20.16g}{14g/mole}=1.44moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.02g}{16g/mole}=1.44moles$

Moles of Cl =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{51.02g}{35.5g/mole}=1.44moles$

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

For H = $\frac{5.80}{1.44}=4$

For N = $\frac{1.44}{1.44}=1$

For O = $\frac{1.44}{1.44}=1$

For Cl = $\frac{1.44}{1.44}=1$

The ratio of H: N: O: Cl= 4: 1: 1: 1

Hence the empirical formula is $H_4NOCl$

The empirical weight of $H_4NOCl$ = 4(1)+1(14)+ 1(16) + 1(35.5)= 69.5 g.

The molecular weight = 278 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{278}{69.5}=4$

The molecular formula will be=$4\times H_4NOCl=H_{16}NOCl$