Question

Boron consists of two isotopes, ¹⁰B and ¹¹B. Chlorine also has two isotopes, ³⁵CI and ³⁷CI.Consider the mass spectrum of BCI₃. How many peaks would be present, and what approximate mass would each peak correspond to in the BCI₃ mass spectrum?

Short answer

There are8 peaks would be present.

Step-by-step solution

Definition

The molar mass of a materialis defined as the mass of the substancecontained in one moleof the substance, and it is equal to the total massof the component atoms. It may be stated mathematically as follows:
Molar mass =$\frac{Weighting}{numberofmoles}$

Determine the possible combination

The 8 different possible combinationsare below,

1. ¹⁰B³⁵CI
2. ¹⁰B³⁷CI³⁵CI₂
3. ¹⁰B³⁵CI³⁷CI₂
4. ¹⁰B³⁷CI₃
5. ¹⁰B³⁵CI₃
6. ¹¹B³⁵CI³⁷CI₂
7. ¹¹B³⁷CI³⁵CI₂
8. ¹¹B³⁷CI₃

Here are the 8 possible combinations, from that 8 peaks would be present.

Determine the Atomic weight

Atomic weight ofboron-10is10.013 Da

Atomic weight ofboron-11is11.010 Da

Atomic weight ofChlorine-35is34.969 Da

Atomic weight ofChlorine-37is 36.966 Da

Determine the mass of each option

$$\begin{gathered} {}^{10}B{}^{35}C{I}_3\to m \\ =10.013Da+3\times 34.969Da \\ =114.92Da \\ {}^{10}B{}^{37}CI{}^{35}C{I}_2\to m \\ =10.013Da+36.966Da+2\times 34.969Da \\ =116.92Da \\ {}^{10}B{}^{35}CI{}^{37}C{I}_2\to m \\ =118.91Da \\ {}^{10}B{}^{37}C{I}_3\to m \\ =120.911Da \\ {}^{11}B{}^{35}C{I}_3\to m \\ =115.32Da \\ {}^{11}B{}^{35}CI{}^{37}C{I}_2\to m \\ =119.91Da \\ {}^{11}B{}^{37}CI{}^{35}C{I}_2\to m \\ =117.91Da \\ {}^{10}B{}^{37}C{I}_3\to m \\ =121.91Da \end{gathered}$$