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Chapter 3: Problem 136

Boron consists of two isotopes, 10 \(\mathrm{B}\) and 11 \(\mathrm{B}\) . Chlorine also has two isotopes, 35 \(\mathrm{Cl}\) and 37 \(\mathrm{Cl}\) . Consider the mass spectrum of \(\mathrm{BCl}_{3}\) . How many peaks would be present, and what approximate mass would each peak correspond to in the BCl_ mass spectrum?

Short Answer

Expert verified
There would be 8 peaks in the mass spectrum of BCl鈧�, with corresponding approximate masses of 115, 116, 117, 118, 119, 120, 121, and 122.

Step by step solution

01

List all possible combinations of boron and chlorine isotopes in BCl鈧�

For each boron isotope, there are three possible chlorine isotopes. So, for both boron isotopes, we have: - B鈦铰光伆鈦綜l鈧� - B鈦铰孤光伨Cl鈧� For each BCl鈧�, there are 3 chlorine atoms with two possible isotopes each (35Cl and 37Cl), resulting in 2鲁 = 8 combinations. Therefore, we have: - B鈦铰光伆鈦綜l鈧�: 8 combinations - B鈦铰孤光伨Cl鈧�: 8 combinations Total possible combinations: 8 + 8 = 16
02

Find the number of unique approximate masses

Now, we'll look for unique masses by considering the mass of each combination. The approximate mass of BCl鈧� is the sum of the mass of the boron isotope and the three chlorine isotopes. For B鈦铰光伆鈦綜l鈧�: 1. 10 + 3(35) = 115 2. 10 + 2(35) + 37 = 117 3. 10 + 35 + 2(37) = 119 4. 10 + 3(37) = 121 For B鈦铰孤光伨Cl鈧�: 1. 11 + 3(35) = 116 2. 11 + 2(35) + 37 = 118 3. 11 + 35 + 2(37) = 120 4. 11 + 3(37) = 122 The unique approximate masses are 115, 116, 117, 118, 119, 120, 121, and 122.
03

Determine the number of peaks and their corresponding approximate masses

We have 8 unique approximate masses, which means there will be 8 peaks in the mass spectrum of BCl鈧�. The approximate masses corresponding to these peaks are: - 115 - 116 - 117 - 118 - 119 - 120 - 121 - 122 So, there would be 8 peaks in the mass spectrum of BCl鈧�, with approximate masses of 115, 116, 117, 118, 119, 120, 121, and 122.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isotopes
Isotopes are variants of a chemical element that have the same number of protons but a different number of neutrons. This difference in neutrons leads to variations in atomic mass.
Boron's isotopes, for example, are denoted as 10 \(\text{B}\) and 11 \(\text{B}\). Here, the numbers 10 and 11 represent the atomic mass of each isotope.
  • The atomic number is the same for all isotopes of boron, meaning they all have the same number of protons.
  • The different masses arise from differing numbers of neutrons.
This principle applies to all elements, including chlorine isotopes, which are 35 \(\text{Cl}\) and 37 \(\text{Cl}\). Understanding isotopes is key to interpreting mass spectrometry results, where variations in atomic mass create multiple peaks.
Boron
Boron is a chemical element with the symbol \(\text{B}\) and atomic number 5. It is a metalloid, which means it has properties that are intermediate between metals and non-metals.
The two naturally occurring isotopes of boron are 10 \(\text{B}\) and 11 \(\text{B}\). These isotopes have different neutron counts, contributing to unique physical properties.
  • 10 \(\text{B}\) has 5 protons and 5 neutrons.
  • 11 \(\text{B}\) has 5 protons and 6 neutrons.
These properties make boron useful in various industrial applications, like glass manufacturing, due to its chemical versatility. In mass spectrometry, these isotopes help us understand the distribution of elements and compounds in a sample.
Chlorine
Chlorine, represented by the symbol \(\text{Cl}\), has an atomic number of 17. Like boron, chlorine also has two main isotopes: 35 \(\text{Cl}\) and 37 \(\text{Cl}\).
  • 35 \(\text{Cl}\) contains 17 protons and 18 neutrons.
  • 37 \(\text{Cl}\) contains 17 protons and 20 neutrons.
These isotopes exist in nature with different abundances. The average atomic mass of chlorine is around 35.5 u because of these naturally occurring isotopes.
In applications like mass spectrometry, understanding the ratio and mass of these isotopes allows scientists to predict how chlorine will appear in spectra, leading to more accurate identification of substances.
Mass Spectrometry
Mass spectrometry is a powerful analytical tool used to identify the amount and type of molecules present in a sample. It works by ionizing chemical compounds to generate charged molecules or molecule fragments, measuring their mass-to-charge ratio.
In mass spectrometry, each peak represents a different isotopic composition of a molecule. For \(\text{BCl}_3\), each peak corresponds to a combination of boron and chlorine isotopes.
  • The number of peaks in a mass spectrum reflects the isotopic diversity of the sample.
  • This diversity arises because boron and chlorine each have two stable isotopes, leading to multiple combinations.
These peaks are critical for evaluating how different isotopic configurations contribute to the molecular mass, enabling detailed chemical analysis and identification.

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Most popular questions from this chapter

The most common form of nylon (nylon-6) is 63.68\(\%\) carbon, 12.38\(\%\) nitrogen, 9.80\(\%\) hydrogen, and 14.14\(\%\) oxygen. Calculate the empirical formula for nylon-6.

A binary compound between an unknown element \(\mathrm{E}\) and hydrogen contains 91.27\(\% \mathrm{E}\) and 8.73\(\% \mathrm{H}\) by mass. If the formula of the compound is \(\mathrm{E}_{3} \mathrm{H}_{8},\) calculate the atomic mass of \(\mathrm{E}\)

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When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) In a certain experiment, 20.00 \(\mathrm{g}\) iron metal was reacted with 11.20 \(\mathrm{g}\) oxygen gas. After the experiment, the iron was totally consumed, and 3.24 \(\mathrm{g}\) oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.
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