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Chapter 3: Problem 127

In using a mass spectrometer, a chemist sees a peak at a mass of \(30.0106 .\) Of the choices \({ }^{12} \mathrm{C}_{2}{ }^{1} \mathrm{H}_{6}{ }_{6}{ }^{12} \mathrm{C}^{1} \mathrm{H}_{2}{ }^{16} \mathrm{O}\), and \({ }^{14} \mathrm{~N}^{16} \mathrm{O}\), which is responsible for this peak? Pertinent masses are \({ }^{1} \mathrm{H}\), \(1.007825 ;{ }^{16} \mathrm{O}, 15.994915 ;\) and \({ }^{14} \mathrm{~N}, 14.003074 .\)

Short Answer

Expert verified
The molecule responsible for the mass spectrometer peak at 30.0106 is \({ }^{14}\mathrm{N}^{16}\mathrm{O}\), as its molecular mass (30.008) is closest to the observed mass peak.

Step by step solution

01

Identify the given information

The mass spectrometer peak is at a mass of 30.0106. Possible choices are: 1. \({ }^{12}C_{2}{ }^{1}H_{6}{ }_{6}{ }^{12}C^{1}H_{2}{ }^{16}O\) 2. \({ }^{14}\mathrm{N}^{16}\mathrm{O}\) Pertinent masses are: - \({ }^{1}\mathrm{H}\): 1.007825 - \({ }^{16}\mathrm{O}\): 15.994915 - \({ }^{14}\mathrm{N}\): 14.003074
02

Calculate the molecular mass of each choice

We need to find the molecular mass of each choice by adding up the atomic masses present in the molecule. This can be calculated as follows: For choice 1: \({ }^{12}C_{2}{ }^{1}H_{6}{ }_{6}{ }^{12}C^{1}H_{2}{ }^{16}O\) Molecular mass = (2 脳 (12.0000)) + (6 脳 1.007825) + (6 脳 12.0000)+ (2 脳 1.007825) + (1 脳 15.994915) For choice 2: \({ }^{14}\mathrm{N}^{16}\mathrm{O}\) Molecular mass = (1 脳 14.003074) + (1 脳 15.994915)
03

Compare calculated molecular mass with the observed mass peak

Now we need to compare the calculated molecular mass for each choice with the observed mass peak from the mass spectrometer to determine which molecule is responsible for the mass peak. For choice 1: Calculated molecular mass = (2 脳 (12.0000)) + (6 脳 1.007825) + (6 脳 12.0000)+ (2 脳 1.007825) + (1 脳 15.994915) 鈮� 30.016 For choice 2: Calculated molecular mass = (1 脳 14.003074) + (1 脳 15.994915) 鈮� 30.008
04

Identify the molecule responsible for the peak

Upon comparing the calculated molecular masses with the observed mass peak (30.0106), we can conclude that choice 2, i.e., \({ }^{14}\mathrm{N}^{16}\mathrm{O}\), is the molecule responsible for the peak as its molecular mass (30.008) is the closest to the observed mass peak.

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Key Concepts

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

Molecular Mass Calculation
Calculating the molecular mass of a compound is essential in mass spectrometry when trying to identify substances based on their mass. The process involves summing up the atomic masses of each atom present in the molecule.
Given a chemical formula, you can calculate the molecular mass by substituting the atomic masses for each element. For example,
  • the molecular mass for a molecule like extsuperscript{14}N extsuperscript{16}O is calculated as \( (1 \times 14.003074) + (1 \times 15.994915) \approx 30.008 \).
  • For extsuperscript{12}C extsubscript{2} extsuperscript{1}H extsubscript{6} extsubscript{6} extsuperscript{12}C extsuperscript{1}H extsubscript{2} extsuperscript{16}O, the molecular mass will be the sum of each element's atomic mass.
    This involves more steps:\( (2 \times 12.0000) + (6 \times 1.007825) + (6 \times 12.0000)+ (2 \times 1.007825) + (1 \times 15.994915) \approx 30.016 \).
It鈥檚 crucial to have accurate atomic masses for a precise calculation, ensuring the identity of the compound can be determined through comparison.
Isotopes
Isotopes are variants of a particular chemical element that differ in neutron number while retaining the same number of protons.
This difference results in varying atomic masses between isotopes, an important factor when using mass spectrometry.
Each element can have several isotopes, and these differences can be subtle yet crucial for molecular identification.
  • For example, the element carbon has isotopes such as extsuperscript{12}C and extsuperscript{13}C. Their difference is the number of neutrons, impacting the atomic mass.
  • Similarly, nitrogen's isotopes are extsuperscript{14}N and extsuperscript{15}N. In the exercise, we focused on extsuperscript{14}N.
Understanding isotopes is fundamental in mass spectrometry as they can lead to different peaks, helping identify which specific isotope, and hence which element, is present in a sample.
Atomic Masses
Atomic masses are critical values that indicate the average mass of an atom of an element, accounting for its isotopes.
They play a significant role in calculating molecular masses. These numbers are based on the relative abundance of isotopes in nature.
When determining the atomic masses used in calculations, it is vital to use accurate figures such as:
  • Hydrogen extsuperscript{1}H at 1.007825,
  • Oxygen extsuperscript{16}O at 15.994915,
  • and Nitrogen extsuperscript{14}N at 14.003074.
These numbers, though small, lead to large differences in results when multiplied to calculate molecular mass. Accurate atomic masses allow chemists to pinpoint substances swiftly and with certainty when analyzing mass spectrometry readouts.
Peak Identification
In the context of mass spectrometry, peak identification is a crucial step that follows the calculation of molecular masses. Each peak in a mass spectrum corresponds to a particle's mass-to-charge ratio \( \left(\frac{m}{z}\right) \) .
This is a pivotal step in determining the identity of a molecule among possible candidates. By comparing the mass of options identified with their observed molecular masses,
  • like comparing the 30.0106 mass spectrometer reading against calculated masses,
  • or recognizing that extsuperscript{14}N extsuperscript{16}O aligns closely with the peak due to its calculated molecular mass of 30.008,
chemists can identify the compound contributing to the peak. Slight deviations in mass calculations can help determine which isotope mix, or compound, corresponds to the observed peak in the mass spectrometer.

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Most popular questions from this chapter

Aspirin \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)\) is synthesized by reacting salicylic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) with acetic anhydride \(\left(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\right)\). The balanced equa- tion is $$ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3} \longrightarrow \mathrm{C}_{9} \mathrm{H}_{3} \mathrm{O}_{4}+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} $$ a. What mass of acetic anhydride is needed to completely consume \(1.00 \times 10^{2} \mathrm{~g}\) salicylic acid? b. What is the maximum mass of aspirin (the theoretical yield) that could be produced in this reaction?

With the advent of techniques such as scanning tunneling microscopy, it is now possible to "write" with individual atoms by manipulating and arranging atoms on an atomic surface. a. If an image is prepared by manipulating iron atoms and their total mass is \(1.05 \times 10^{-20} \mathrm{~g}\), what number of iron atoms were used? b. If the image is prepared on a platinum surface that is exactly 20 platinum atoms high and 14 platinum atoms wide, what is the mass (grams) of the atomic surface? c. If the atomic surface were changed to ruthenium atoms and the same surface mass as determined in part b is used, what number of ruthenium atoms is needed to construct the surface?

Considering your answer to Exercise 79, which type of formula, empirical or molecular, can be obtained from elemental analysis that gives percent composition?

A \(0.4230-\mathrm{g}\) sample of impure sodium nitrate was heated, converting all the sodium nitrate to \(0.2864 \mathrm{~g}\) of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the oriainal sample.

Glass is a mixture of several compounds, but a major constituent of most glass is calcium silicate, \(\mathrm{CaSiO}_{3-}\) Glass can be etched by treatment with hydrofluoric acid; HF attacks the calcium silicate of the glass, producing gaseous and water-soluble products (which can be removed by washing the glass). For example, the volumetric glassware in chemistry laboratories is often graduated by using this process. Balance the following equation for the reaction of hydrofluoric acid with calcium silicate. $$ \mathrm{CaSiO}_{3}(s)+\mathrm{HF}(a q) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$
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