Example 3.4: Deriving Grams from Moles for an Element

Deriving Grams from Moles for an Element

A liter of air contains \(9.2\times 10^{-4 }\)mol argon. What is the mass of Ar in a liter of air?

Solution

\(n_{\mathrm{\ce{Ar}}}\) \(= 9.2\times 10^{-4}\ \mathrm{mol}\)


\(m_{\mathrm{\ce{Ar}}}\) = ?


The chemical amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10^-3) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):

The molar mass M is defined as

\(M = \dfrac{m}{n}\)     

Solving for m gives

\(m = n \cdot M\)     

\(m_{\mathrm{\ce{Ar}}}\) \(= n_{\mathrm{\ce{Ar}}} \cdot M_{\mathrm{\ce{Ar}}}\)

\(\ \ \ =9.2\times 10^{-4}\ \mathrm{mol} \cdot 39.948\ \frac{\mathrm{g}}{\mathrm{mol}}\)

\(\ \ \ =0.0368\ \mathrm{g}\)


The result is in agreement with our expectations, around 0.04 g Ar.