A sample of pure hydrated aluminum nitrate, \text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}, h… — JAMB UTME

A sample of pure hydrated aluminum nitrate, $\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}$ , has a mass of $75.0\text{ g}$ . What is the mass, in grams, of oxygen contained in this sample? $[\text{Al} = 27, \text{N} = 14, \text{O} = 16, \text{H} = 1]$

Answer: 57.6 g

Answer

57.6

The molar mass of

\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}

375\text{ g/mol}

. A

75.0\text{ g}

sample equals

0.20\text{ mol}

of the hydrated salt. Because each formula unit contains

18

oxygen atoms (

9

from the nitrate groups and

9

from the water molecules),

0.20\text{ mol}

of the salt contains

3.60\text{ mol}

of oxygen atoms. Multiplying

3.60\text{ mol}

by the atomic mass of oxygen (

16\text{ g/mol}

) yields

57.6\text{ g}

Step-by-Step Solution

Calculate the molar mass of hydrated aluminum nitrate,

\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}

375\text{ g/mol}

Sum the atomic masses of all constituent atoms:

\text{Al} = 27

3 \times \text{NO}_3 = 3 \times 62 = 186

9 \times \text{H}_2\text{O} = 9 \times 18 = 162

. Total

= 27 + 186 + 162 = 375\text{ g/mol}

Determine the amount (in moles) of the compound in the

75.0\text{ g}

sample.

0.20\text{ mol}

Moles of compound

= \frac{\text{mass}}{\text{molar mass}} = \frac{75.0\text{ g}}{375\text{ g/mol}} = 0.20\text{ mol}

Determine the total moles of oxygen atoms per mole of the hydrated compound.

18\text{ mol of O}

Each formula unit contains

9

oxygen atoms in the nitrate groups (

3 \times 3

) and

9

oxygen atoms in the water molecules (

9 \times 1

), giving a total of

18

oxygen atoms.

Calculate the total mass of oxygen atoms in the sample.

57.6\text{ g}

Mass of oxygen

= \text{moles of compound} \times 18 \times \text{molar mass of O} = 0.20 \times 18 \times 16 = 3.60 \times 16 = 57.6\text{ g}

Key Concept

Mole Concept and Stoichiometric Mass Composition in Hydrated Compounds

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