Chemistry · General chemistry I · Worked example

Find isotope abundances from an atomic mass

Boron has two stable isotopes, boron-10 $(10.013 u)$ and boron-11 $(11.009 u)$ , and an atomic mass of $10.81 u$ . Estimate the natural abundance of each isotope.

Name the unknown

Let $x$ be the fraction of boron-10. Boron has only these two isotopes, so the fraction of boron-11 is $1 - x$ .

Write the weighted average

Each isotope’s mass times its fraction, added, must give the atomic mass from the periodic table.

10.013 x + 11.009 (1 - x) = 10.81

Expand the bracket

Multiply $11.009$ into both terms of $1 - x$ . Distributing is allowed because it does not change the value of the left side.

10.013 x + 11.009 - 11.009 x = 10.81

\begin{array}{c} 10.013 x + 11.009 - 11.009 x \\ = 10.81 \end{array}

Collect the $x$ terms

$10.013 x - 11.009 x$ $= - 0.996 x$ . The coefficient is negative because boron-10 is the lighter isotope, so more of it pulls the average down.

11.009 - 0.996 x = 10.81

Isolate the $x$ term

Subtract $10.81$ from both sides and add $0.996 x$ to both. Doing the same thing to both sides keeps the equation true. $11.009 - 10.81 = 0.199$ .

0.996 x = 0.199

Divide

$0.199 \div 0.996 = 0.1998$ , which is $0.20$ to the two figures the data support.

x \approx 0.20

Report what the data support

Boron is about $20 %$ boron-10 and $80 %$ boron-11. Two figures are all the data justify. $10.81$ has two decimal places, so the difference $11.009 - 10.81 = 0.199$ is good only to two decimal places, $0.20$ , and $x$ keeps two significant figures. The Periodic Table lists $19.65 %$ and $80.35 %$ as representative values, and it marks them as variable, because boron’s isotope mix differs a little from one natural source to another.

Result

About $20 %$ boron-10 and $80 %$ boron-11.

Your turn

1. One like the example

Chlorine’s atomic mass is $35.45 u$ , and its isotopes have masses $34.969 u$ and $36.966 u$ . Which isotope is more abundant, and about what fraction of chlorine atoms is it?

Show the answer and explanation

Chlorine-35, about $76 %$ .

$35.45$ is nearer to $34.969$ than to $36.966$ , so chlorine-35 is the more abundant isotope. With $x$ as its fraction, $34.969 x + 36.966 (1 - x)$ $= 35.45$ gives $1.997 x = 1.516$ , so $x \approx 0.76$ .

\begin{array}{l} 34.969 x + 36.966 (1 - x) = 35.45 \\ 1.997 x = 1.516 \\ x \approx 0.76 \end{array}

2. A variation

Gallium has two stable isotopes, gallium-69 $(68.926 u)$ and gallium-71 $(70.925 u)$ , and an atomic mass of $69.72 u$ . What fraction of gallium atoms is gallium-69?

Show the answer and explanation

About $0.60$ , or $60 %$ .

With $x$ as the fraction of gallium-69, $68.926 x + 70.925 (1 - x)$ $= 69.72$ . Expanding and collecting gives $70.925 - 1.999 x = 69.72$ , so $1.999 x = 1.205$ and $x = 0.60$ . The average is nearer gallium-69, which agrees.

\begin{array}{l} 68.926 x + 70.925 (1 - x) = 69.72 \\ 70.925 - 1.999 x = 69.72 \\ 1.999 x = 1.205 \\ x \approx 0.60 \end{array}

3. Somewhere new

Neon is $90.48 %$ neon-20 $(19.992 u)$ and $0.27 %$ neon-21 $(20.994 u)$ . The rest is neon-22 $(21.991 u)$ . Find the abundance of neon-22 and the atomic mass of neon.

Show the answer and explanation

$9.25 %$ , and $20.18 u$ .

With three isotopes the missing abundance comes from the total. The fractions add to $1$ , so neon-22 is $1 - 0.9048 - 0.0027$ $= 0.0925$ , or $9.25 %$ . Then $0.9048 \times 19.992$ $+ 0.0027 \times 20.994$ $+ 0.0925 \times 21.991$ $= 20.180$ , or $20.18 u$ .

\begin{array}{l} f = 1 - 0.9048 - 0.0027 \\ f = 0.0925 \\ m = 0.9048 \cdot 19.992 + 0.0027 \cdot 20.994 + 0.0925 \cdot 21.991 \\ m \approx 20.18 \end{array}

\begin{array}{c} f = 1 - 0.9048 - 0.0027 \\ f = 0.0925 \\ m = 0.9048 \cdot 19.992 \\ + 0.0027 \cdot 20.994 \\ + 0.0925 \cdot 21.991 \\ m \approx 20.18 \end{array}

Keep exploring

In the Math box, change the last row to $x \approx 0.80$ . The row turns to ✗: $0.80$ is the fraction of boron-11, the isotope the average sits closer to.

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Name the unknown

Write the weighted average

Expand the bracket

Collect the xx terms

Isolate the xx term

Divide

Report what the data support

Result

Your turn

1. One like the example

Chlorine-35, about 76⁢%76%.

2. A variation

About 0.600.60, or 60⁢%60%.

3. Somewhere new

9.25⁢%9.25%, and 20.18 u20.18 u.

Keep exploring

Study it, then try it

Collect the $x$ terms

Isolate the $x$ term

Chlorine-35, about $76 %$ .

About $0.60$ , or $60 %$ .

$9.25 %$ , and $20.18 u$ .