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Example 3.3: Deriving Moles from Grams for an Element

Deriving Moles from Grams for an Element

According to nutritional guidelines from the \(\ce{US}\) Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?

Solution

The mass of K is provided, and the corresponding amount of K in moles is requested.

\(m_{\mathrm{\ce{K}}}\) \(= 4.7\ \mathrm{g}\)


\(n_{\mathrm{\ce{K}}}\) = ?


Referring to the periodic table, the molar mass of K is 39.10 g/mol. The given mass of K (4.7 g) is a bit more than one-tenth of the mass of one mole (39.10 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol.
The molar amount of a substance may be calculated by dividing its mass \(\ce{(g)}\) by its molar mass (g/mol):

\(n_{\mathrm{\ce{K}}}\) \(= \dfrac{m_{\mathrm{\ce{K}}}}{M_{\mathrm{\ce{K}}}}\)

\(\ \ \ =\dfrac{4.7\ \mathrm{g}}{39.0983\ \frac{\mathrm{g}}{\mathrm{mol}}}\)

\(\ \ \ =0.120\ \mathrm{mol}\)



The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”
The calculated chemical amount of K (0.12 mol) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.