Question

there are only two naturally - occurring stable isotopes of lithium, the masses of which are listed in the table below. use whatever data you need from the aleks 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.

Explanation:

Step1: Set up the equation

Let the abundance of \(^{6}\text{Li}\) be \(x\) (in decimal form), then the abundance of \(^{7}\text{Li}\) is \(1 - x\). The atomic mass of lithium (\(A = 6.941\) amu from the periodic table) is given by the formula \(A=m_1x + m_2(1 - x)\), where \(m_1 = 6.0151\) amu and \(m_2=7.0160\) amu. So, \(6.941=6.0151x+7.0160(1 - x)\).

Step2: Expand the equation

Expand \(7.0160(1 - x)\) to get \(7.0160-7.0160x\). The equation becomes \(6.941 = 6.0151x+7.0160 - 7.0160x\).

Step3: Combine like - terms

Combine the \(x\) terms: \(6.941-7.0160=(6.0151 - 7.0160)x\). Calculate \(6.941-7.0160=- 0.075\) and \(6.0151 - 7.0160=-1.0009\). So, \(-0.075=-1.0009x\).

Step4: Solve for \(x\)

Divide both sides by \(-1.0009\): \(x=\frac{0.075}{1.0009}\approx0.075\).

Step5: Calculate the abundances in percentage

The abundance of \(^{6}\text{Li}\) is \(x\times100\% = 7.5\%\). The abundance of \(^{7}\text{Li}\) is \((1 - x)\times100\%=(1 - 0.075)\times100\% = 92.5\%\).

Answer:

The natural abundance of \(^{6}\text{Li}\) is \(7.5\%\) and the natural abundance of \(^{7}\text{Li}\) is \(92.5\%\).