Questions: Calculate the number of oxygen atoms in a 60.0 g sample of vanadium (V) oxide (V2O5). Be sure your answer has a unit symbol if necessary, and round it to 3 significant digits.

Solution Steps

First, we need to calculate the molar mass of vanadium(V) oxide \( \mathrm{V}_2\mathrm{O}_5 \).

• The atomic mass of vanadium (V) is approximately 50.9415 g/mol.
• The atomic mass of oxygen (O) is approximately 16.00 g/mol.

The molar mass of \( \mathrm{V}_2\mathrm{O}_5 \) is calculated as follows: \[ \text{Molar mass of } \mathrm{V}_2\mathrm{O}_5 = 2 \times 50.9415 + 5 \times 16.00 = 101.883 + 80.00 = 181.883 \, \text{g/mol} \]

Next, we calculate the number of moles of \( \mathrm{V}_2\mathrm{O}_5 \) in a 60.0 g sample using the molar mass.

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{60.0 \, \text{g}}{181.883 \, \text{g/mol}} = 0.3298 \, \text{mol} \]

Each molecule of \( \mathrm{V}_2\mathrm{O}_5 \) contains 5 oxygen atoms. Therefore, the number of moles of oxygen atoms is: \[ \text{Moles of oxygen atoms} = 0.3298 \, \text{mol} \times 5 = 1.649 \, \text{mol} \]

Using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol), we can find the number of oxygen atoms: \[ \text{Number of oxygen atoms} = 1.649 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 9.926 \times 10^{23} \, \text{atoms} \]

Final Answer