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`

Step-by-step 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}}= (5.80g)/(1g/mole)=5.80moles`

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

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

Moles of Cl =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (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 =
`(5.80)/(1.44)=4`

For N =
`(1.44)/(1.44)=1`

For O =
`(1.44)/(1.44)=1`

For Cl =
`(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}}=(278)/(69.5)=4`

The molecular formula will be=
`4* H_4NOCl=H_(16)NOCl`