Use the periodic table to calculate the molar mass of each of the following compounds. Each answer must have 2 decimal places.

Ammonia [tex]$\left( \text{NH}_3 \right)$[/tex]: [tex]$\square$[/tex] g/mol

Magnesium hydroxide [tex]$\left( \text{Mg(OH)}_2 \right)$[/tex]: [tex]$\square$[/tex] g/mol

Iron(III) oxide [tex]$\left( \text{Fe}_2\text{O}_3 \right)$[/tex]: [tex]$\square$[/tex] g/mol



Answer :

Sure! Let's calculate the molar masses of the given compounds step by step using the atomic masses from the periodic table.

### 1. Ammonia (NH₃)

To calculate the molar mass of ammonia (NH₃), we need to sum the atomic masses of 1 nitrogen (N) atom and 3 hydrogen (H) atoms.
- Atomic mass of nitrogen (N): 14.01 g/mol
- Atomic mass of hydrogen (H): 1.01 g/mol

The molar mass of NH₃ is:
[tex]\[ \text{Molar mass of NH}_3 = 1 \times (\text{atomic mass of N}) + 3 \times (\text{atomic mass of H}) \][/tex]
[tex]\[ \text{Molar mass of NH}_3 = 1 \times 14.01 + 3 \times 1.01 \][/tex]
[tex]\[ \text{Molar mass of NH}_3 = 14.01 + 3.03 \][/tex]
[tex]\[ \text{Molar mass of NH}_3 = 17.04 \][/tex]

Thus, the molar mass of ammonia (NH₃) is 17.04 g/mol.

### 2. Magnesium hydroxide (Mg(OH)â‚‚)

To calculate the molar mass of magnesium hydroxide (Mg(OH)â‚‚), we need to sum the atomic masses of 1 magnesium (Mg) atom, 2 oxygen (O) atoms, and 2 hydrogen (H) atoms.
- Atomic mass of magnesium (Mg): 24.31 g/mol
- Atomic mass of oxygen (O): 16.00 g/mol
- Atomic mass of hydrogen (H): 1.01 g/mol

The molar mass of Mg(OH)â‚‚ is:
[tex]\[ \text{Molar mass of Mg(OH)}_2 = 1 \times (\text{atomic mass of Mg}) + 2 \times (\text{atomic mass of O} + \text{atomic mass of H}) \][/tex]
[tex]\[ \text{Molar mass of Mg(OH)}_2 = 1 \times 24.31 + 2 \times (16.00 + 1.01) \][/tex]
[tex]\[ \text{Molar mass of Mg(OH)}_2 = 24.31 + 2 \times 17.01 \][/tex]
[tex]\[ \text{Molar mass of Mg(OH)}_2 = 24.31 + 34.02 \][/tex]
[tex]\[ \text{Molar mass of Mg(OH)}_2 = 58.33 \][/tex]

Thus, the molar mass of magnesium hydroxide (Mg(OH)â‚‚) is 58.33 g/mol.

### 3. Iron(III) oxide (Fe₂O₃)

To calculate the molar mass of iron(III) oxide (Fe₂O₃), we need to sum the atomic masses of 2 iron (Fe) atoms and 3 oxygen (O) atoms.
- Atomic mass of iron (Fe): 55.85 g/mol
- Atomic mass of oxygen (O): 16.00 g/mol

The molar mass of Fe₂O₃ is:
[tex]\[ \text{Molar mass of Fe}_2\text{O}_3 = 2 \times (\text{atomic mass of Fe}) + 3 \times (\text{atomic mass of O}) \][/tex]
[tex]\[ \text{Molar mass of Fe}_2\text{O}_3 = 2 \times 55.85 + 3 \times 16.00 \][/tex]
[tex]\[ \text{Molar mass of Fe}_2\text{O}_3 = 111.70 + 48.00 \][/tex]
[tex]\[ \text{Molar mass of Fe}_2\text{O}_3 = 159.70 \][/tex]

Thus, the molar mass of iron(III) oxide (Fe₂O₃) is 159.70 g/mol.

### Summary

- Ammonia (NH₃): 17.04 g/mol
- Magnesium hydroxide (Mg(OH)â‚‚): 58.33 g/mol
- Iron(III) oxide (Fe₂O₃): 159.70 g/mol

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