Question

Lithium forms compounds which are used in dry cells and storage batteries and in high-temperature lubricants. It has two naturally occurring isotopes, 6li (isotopic mass = 6.015121 amu) and 7li (isotopic mass = 7.016003 amu). Lithium has an atomic mass of 6.9409 amu. What is the percent abundance of lithium-6?

Answer 1

7.5%

Further explanation

Given:

Lithium has two naturally occurring isotopes:

• ⁶Li (isotopic mass = 6.015121 amu)
• ⁷Li (isotopic mass = 7.016003 amu).

Lithium has an atomic mass of 6.9409 amu.

Question:

What is the percent abundance of Lithium-6?

The Process:

The average atomic mass is the weighted average of atomic masses of isotopes, weighted by their respective fractional abundance.

$\boxed{ \ Average \ atomic \ mass = \sum_{i = 1}^{n} (isotopic \ mass_{(i)})(abundance_{(i)}) \ }$

Let the percent abundance of Lithium-6 as X and Lithium-7 as (1 - X). Keep in mind, lithium has two naturally occurring isotopes only.

                      mass            the percent abundance (%)

• ⁶Li    6.015121 amu                        X
• ⁷Li    7.016003 amu                   (1 - X)

Lithium has an atomic mass of 6.9409 amu.

Let us find out the percent abundance of Lithium-6.

$\boxed{ \ 6.9409 = (6.015121)X + (7.016003)(1 - X) \ }$

$\boxed{ \ 6.9409 = (6.015121)X - (7.016003)X + 7.016003 \ }$

$\boxed{ \ (7.016003)X - (6.015121)X = 7.016003 - 6.9409 \ }$

$\boxed{ \ (1.000882)X = 0.075103 \ }$

$\boxed{ \ X = \frac{0.075103}{1.000882} \ }$

$\boxed{ \ X = 0.075 \ }$ $\ \rightarrow \boxed{ \ X = 7.5 \% \ }$

Thus, the percent abundance of Lithium-6 is 7.5%.

One more thing, what is the abundance of Lithium-7? 1 - 0.075 = 0.925, i.e., 92.5%

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Answer 2

The naturally occurring isotopes of Li are Li-6 of mass 6.015121 amu and Li-7 of mass 7.016003 amu. The atomic mass of Li is 6.9409 amu, the percent abundance can be calculated using the following relation.

Atomic mass=m(Li-6 )×%(Li-6 )+m(Li-7 )×%(Li-7 )

Let the percent abundance of Li-6 be X thus, that of Li-7 will be 1-X, putting the values,

$6.9409 amu=6.015121 amu\times X+7.016003 amu(1-X)$

Or,

$6.9409 =6.015121 X+7.016003-7.016003 X$

Or,

X=0.075

Thus, $1-X=1-0.075=0.925$

Thus, percent abundance of Li-6 is 0.075 or 7.5 % and that of Li-7 is 0.925 or 92.5%.