Question

there are only two naturally - occurring stable isotopes of carbon, the masses of which are listed in the table below. use whatever data you need from the alexs periodic table to calculate the natural abundance of each isotope and complete the table. be sure your answers have the correct number of significant digits.

isotope	(^{12}c)	(^{13}c)
natural abundance	(%)	(%)

Explanation:

Step1: Let the natural abundance of \(^{12}C\) be \(x\) and of \(^{13}C\) be \(y\).

We know that \(x + y=100\%\) or \(x + y = 1\) (in decimal form). Also, the average - atomic mass of carbon is approximately \(12.011\) amu. The formula for the average - atomic mass (\(A\)) is \(A = m_1x_1+m_2x_2\), where \(m_1\) and \(m_2\) are the masses of the isotopes and \(x_1\) and \(x_2\) are their respective natural abundances.

Step2: Substitute the values into the formula.

The mass of \(^{12}C\) (\(m_1\)) is \(12.000\) amu and of \(^{13}C\) (\(m_2\)) is \(13.003\) amu. So, \(12.011=12.000x + 13.003y\). Since \(y = 1 - x\), we substitute \(y\) in the second equation: \(12.011=12.000x+13.003(1 - x)\).

Step3: Expand and solve for \(x\).

\[

$$\begin{align*} 12.011&=12.000x + 13.003-13.003x\\ 12.011-13.003&=12.000x-13.003x\\ - 0.992&=-1.003x\\ x&=\frac{0.992}{1.003}\approx0.989 = 98.9\% \end{align*}$$

\]

Step4: Solve for \(y\).

Since \(y = 1 - x\), \(y=1 - 0.989 = 0.011=1.1\%\)

Answer:

The natural abundance of \(^{12}C\) is \(98.9\%\) and of \(^{13}C\) is \(1.1\%\)