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.

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01-03-2021
Chemistry

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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.

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Answer: The molecular formula will be [tex]H_{16}NOCl[/tex]

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 =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.80g}{1g/mole}=5.80moles[/tex]

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

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

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

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

For H = [tex]\frac{5.80}{1.44}=4[/tex]

For N = [tex]\frac{1.44}{1.44}=1[/tex]

For O = [tex]\frac{1.44}{1.44}=1[/tex]

For Cl = [tex]\frac{1.44}{1.44}=1[/tex]

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

Hence the empirical formula is [tex]H_4NOCl[/tex]

The empirical weight of [tex]H_4NOCl[/tex] = 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.

[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{278}{69.5}=4[/tex]

The molecular formula will be=[tex]4\times H_4NOCl=H_{16}NOCl[/tex]

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