Question

lithium has only two naturally occurring isotopes. the mass of lithium - 6 is 6.01512 amu and the mass of lithium - 7 is 7.01601 amu.
part a
use the atomic mass of lithium to calculate the relative abundances of the two isotopes.
express your answers using one decimal place. enter your answers separated by a comma.
lithium - 6 abundance, lithium - 7 abundance =
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Explanation:

Step 1: Recall the atomic mass formula

The atomic mass of an element is the weighted average of the masses of its isotopes. Let the abundance of lithium - 6 be \(x\) (in decimal form) and the abundance of lithium - 7 be \(1 - x\) (since the sum of abundances of all isotopes is 1). The atomic mass of lithium is approximately \(6.941\) amu. The mass of lithium - 6 (\(m_6\)) is \(6.01512\) amu and the mass of lithium - 7 (\(m_7\)) is \(7.01601\) amu. The formula for atomic mass (\(M\)) is \(M=x\times m_6+(1 - x)\times m_7\).

Step 2: Substitute the values into the formula

Substitute \(M = 6.941\), \(m_6=6.01512\) and \(m_7 = 7.01601\) into the formula:
\[

$$\begin{align*} 6.941&=x\times6.01512+(1 - x)\times7.01601\\ 6.941&=6.01512x+7.01601-7.01601x\\ 6.941 - 7.01601&=6.01512x-7.01601x\\ - 0.07501&=- 1.00089x\\ x&=\frac{0.07501}{1.00089}\approx0.075 \end{align*}$$

\]

Step 3: Convert abundances to percentages

The abundance of lithium - 6 in percentage is \(x\times100 = 0.075\times100=7.5\%\)
The abundance of lithium - 7 is \((1 - x)\times100=(1 - 0.075)\times100 = 92.5\%\)

Answer:

7.5, 92.5