Example 2.5: Calculation of Percent Abundance
Naturally occurring chlorine consists of
\(\ce{^35Cl}\) (mass
34.96885 Da) and
\(\ce{^37Cl}\) (mass
36.96590 Da), with an average mass of
35.453 Da. What is the percent composition of Cl in terms of these two isotopes?
Solution
\(m_{\mathrm{\ce{35}}}\) \(= 34.96885\ \mathrm{Da}\)
\(m_{\mathrm{\ce{37}}}\) \(= 36.96590\ \mathrm{Da}\)
\(m_{\mathrm{average}}\) \(= 35.453\ \mathrm{Da}\)
\(f_{\mathrm{\ce{37}}}\) = ?
\(f_{\mathrm{\ce{35}}}\) = ?
The average mass of chlorine is the fraction that is
\(\ce{^35Cl}\) times the mass of
\(\ce{^35Cl}\) plus the fraction that is
\(\ce{^37Cl}\) times the mass of
\(\ce{^37Cl}\).
\(m_{\mathrm{average}} = f_{\mathrm{\ce{35}}} \cdot m_{\mathrm{\ce{35}}} + f_{\mathrm{\ce{37}}} \cdot m_{\mathrm{\ce{37}}}\) [Equation 1]
The fraction that is
\(\ce{^35Cl}\) and the fraction that is
\(\ce{^37Cl}\) must add up to
1, so the fraction of
\(\ce{^37Cl}\) must equal
1 minus the fraction of
\(\ce{^35Cl}\)
\(f_{\mathrm{\ce{37}}} = 1 - f_{\mathrm{\ce{35}}}\) [Equation 2]
Substituting equation
2 into equation
1 gives
\(m_{\mathrm{average}} = f_{\mathrm{\ce{35}}} \cdot m_{\mathrm{\ce{35}}} + (1 - f_{\mathrm{\ce{35}}}) \cdot m_{\mathrm{\ce{37}}}\)
Now we have an equation with a single unknown
\(f_{\mathrm{\ce{35}}}\), we can solve for it and calculate it:
\(f_{\mathrm{\ce{35},m}} \cdot [\ce{37}] - f_{\mathrm{\ce{35}}} \cdot m_{\mathrm{\ce{35}}} = m_{\mathrm{\ce{37}}} - m_{\mathrm{average}}\) ...terms with \(f_{\mathrm{\ce{35}}}\) on the left
\(f_{\mathrm{\ce{35}}} \cdot (m_{\mathrm{\ce{37}}} - m_{\mathrm{\ce{35}}}) = m_{\mathrm{\ce{37}}} - m_{\mathrm{average}}\) ...isolated \(f_{\mathrm{\ce{35}}}\) on the left
\(f_{\mathrm{\ce{35}}} = \dfrac{m_{\mathrm{\ce{37}}} - m_{\mathrm{average}}}{m_{\mathrm{\ce{37}}} - m_{\mathrm{\ce{35}}}}\) ...solved for \(f_{\mathrm{\ce{35}}}\)
\(\ \ \ =\dfrac{36.96590\ \mathrm{Da} - 35.453\ \mathrm{Da}}{36.96590\ \mathrm{Da} - 34.96885\ \mathrm{Da}} \cdot 100\ \mathrm{％}\)
\(\ \ \ =\dfrac{1.5129\ \mathrm{Da}}{1.997050\ \mathrm{Da}} \cdot 100\ \mathrm{％}\)
\(\ \ \ =0.75757 \cdot 100\ \mathrm{％}\)
\(\ \ \ =75.76\ \mathrm{％}\)
To determine
\(f_{\mathrm{\ce{37}}}\), we can use the fact that the relative ambundances add up to
1 (equation
2)
\(f_{\mathrm{\ce{37}}}\) \(= 1 - f_{\mathrm{\ce{35}}}\)
\(\ \ \ =1 - 75.76\ \mathrm{％}\)
\(\ \ \ =24.24\ \mathrm{％}\)
Therefore, chlorine consists of 75.76%
\(\ce{^35Cl}\) and 24.24%
\(\ce{^37Cl}\).