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
Solving for m gives
\(m_{\mathrm{\ce{Ar}}}\) \(= n_{\mathrm{\ce{Ar}}} \cdot M_{\mathrm{\ce{Ar}}}\)
\(\ \ \ =9.2\times 10^{-4}\ \mathrm{mol} \cdot 39.948\ \frac{\mathrm{g}}{\mathrm{mol}}\)
\(\ \ \ =0.0368\ \mathrm{g}\)
The result is in agreement with our expectations, around
0.04 g Ar.