Question

practice problem 2: (find the molar mass)
compound\tmolar mass (g/mol)
na₂o\t
cu(no₃)₂\t
calcium chloride (cacl2)\t
hydrogen gas (h₂)\t
iron (ii) oxide (feo)\t
iron (iii) oxide (fe₂o₃)\t

Explanation:

To find the molar mass of each compound, we use the atomic masses of the constituent elements (from the periodic table: Na = 22.99 g/mol, O = 16.00 g/mol, Cu = 63.55 g/mol, N = 14.01 g/mol, Ca = 40.08 g/mol, Cl = 35.45 g/mol, H = 1.008 g/mol, Fe = 55.85 g/mol). Molar mass is the sum of the atomic masses of all atoms in the formula unit.

1. $\boldsymbol{\ce{Na2O}}$

• 2 Na atoms: $2 \times 22.99 = 45.98$ g/mol
• 1 O atom: $1 \times 16.00 = 16.00$ g/mol
• Total: $45.98 + 16.00 = 61.98$ g/mol

2. $\boldsymbol{\ce{Cu(NO3)2}}$

• 1 Cu atom: $1 \times 63.55 = 63.55$ g/mol
• 2 N atoms: $2 \times 14.01 = 28.02$ g/mol
• 6 O atoms: $6 \times 16.00 = 96.00$ g/mol
• Total: $63.55 + 28.02 + 96.00 = 187.57$ g/mol

3. $\boldsymbol{\ce{CaCl2}}$

• 1 Ca atom: $1 \times 40.08 = 40.08$ g/mol
• 2 Cl atoms: $2 \times 35.45 = 70.90$ g/mol
• Total: $40.08 + 70.90 = 110.98$ g/mol

4. $\boldsymbol{\ce{H2}}$

• 2 H atoms: $2 \times 1.008 = 2.016$ g/mol

5. $\boldsymbol{\ce{FeO}}$

• 1 Fe atom: $1 \times 55.85 = 55.85$ g/mol
• 1 O atom: $1 \times 16.00 = 16.00$ g/mol
• Total: $55.85 + 16.00 = 71.85$ g/mol

6. $\boldsymbol{\ce{Fe2O3}}$

• 2 Fe atoms: $2 \times 55.85 = 111.70$ g/mol
• 3 O atoms: $3 \times 16.00 = 48.00$ g/mol
• Total: $111.70 + 48.00 = 159.70$ g/mol

Final Table of Molar Masses:

Compound	Molar Mass (g/mol)
$\ce{Cu(NO3)2}$	187.57
Calcium chloride ($\ce{CaCl2}$)	110.98
Hydrogen gas ($\ce{H2}$)	2.016
Iron (II) oxide ($\ce{FeO}$)	71.85
Iron (III) oxide ($\ce{Fe2O3}$)	159.70

Answer:

To find the molar mass of each compound, we use the atomic masses of the constituent elements (from the periodic table: Na = 22.99 g/mol, O = 16.00 g/mol, Cu = 63.55 g/mol, N = 14.01 g/mol, Ca = 40.08 g/mol, Cl = 35.45 g/mol, H = 1.008 g/mol, Fe = 55.85 g/mol). Molar mass is the sum of the atomic masses of all atoms in the formula unit.

1. $\boldsymbol{\ce{Na2O}}$

• 2 Na atoms: $2 \times 22.99 = 45.98$ g/mol
• 1 O atom: $1 \times 16.00 = 16.00$ g/mol
• Total: $45.98 + 16.00 = 61.98$ g/mol

2. $\boldsymbol{\ce{Cu(NO3)2}}$

• 1 Cu atom: $1 \times 63.55 = 63.55$ g/mol
• 2 N atoms: $2 \times 14.01 = 28.02$ g/mol
• 6 O atoms: $6 \times 16.00 = 96.00$ g/mol
• Total: $63.55 + 28.02 + 96.00 = 187.57$ g/mol

3. $\boldsymbol{\ce{CaCl2}}$

• 1 Ca atom: $1 \times 40.08 = 40.08$ g/mol
• 2 Cl atoms: $2 \times 35.45 = 70.90$ g/mol
• Total: $40.08 + 70.90 = 110.98$ g/mol

4. $\boldsymbol{\ce{H2}}$

• 2 H atoms: $2 \times 1.008 = 2.016$ g/mol

5. $\boldsymbol{\ce{FeO}}$

• 1 Fe atom: $1 \times 55.85 = 55.85$ g/mol
• 1 O atom: $1 \times 16.00 = 16.00$ g/mol
• Total: $55.85 + 16.00 = 71.85$ g/mol

6. $\boldsymbol{\ce{Fe2O3}}$

• 2 Fe atoms: $2 \times 55.85 = 111.70$ g/mol
• 3 O atoms: $3 \times 16.00 = 48.00$ g/mol
• Total: $111.70 + 48.00 = 159.70$ g/mol

Final Table of Molar Masses:

Compound	Molar Mass (g/mol)
$\ce{Cu(NO3)2}$	187.57
Calcium chloride ($\ce{CaCl2}$)	110.98
Hydrogen gas ($\ce{H2}$)	2.016
Iron (II) oxide ($\ce{FeO}$)	71.85
Iron (III) oxide ($\ce{Fe2O3}$)	159.70