Question

what is the mass in grams of a sample of $v_2o_3$ that contains $1.03\times10^{12}$ atoms of o?

Explanation:

Step1: Calculate the number of moles of \(O\) atoms

Use Avogadro's number \(N_A = 6.022\times 10^{23}\text{ atoms/mol}\).
The number of moles of \(O\) atoms, \(n_O=\frac{1.03\times 10^{12}\text{ atoms}}{6.022\times 10^{23}\text{ atoms/mol}}\)

$$n_O=\frac{1.03}{6.022}\times10^{- 11}\text{ mol}\approx1.71\times 10^{-12}\text{ mol}$$

Step2: Relate moles of \(O\) to moles of \(V_2O_3\)

From the formula \(V_2O_3\), the ratio of \(n_{V_2O_3}:n_O = 1:3\). So \(n_{V_2O_3}=\frac{n_O}{3}\)

$$n_{V_2O_3}=\frac{1.71\times 10^{-12}\text{ mol}}{3}=5.7\times 10^{-13}\text{ mol}$$

Step3: Calculate the molar mass of \(V_2O_3\)

The molar mass of \(V\) is \(M_V = 50.94\text{ g/mol}\) and of \(O\) is \(M_O=16.00\text{ g/mol}\).
\(M_{V_2O_3}=2\times50.94 + 3\times16.00=101.88+48.00 = 149.88\text{ g/mol}\)

Step4: Calculate the mass of \(V_2O_3\)

Use the formula \(m = n\times M\).
\(m_{V_2O_3}=n_{V_2O_3}\times M_{V_2O_3}\)

$$m_{V_2O_3}=5.7\times 10^{-13}\text{ mol}\times149.88\text{ g/mol}\approx8.54\times 10^{-11}\text{ g}$$

Answer:

\(8.54\times 10^{-11}\text{ g}\)