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In using a mass spectrometer, a chemist sees a peak at a mass of 30.0106 . Of the choices \(^{12} \mathrm{C}_{2}^{1} \mathrm{H}_{6},^{12} \mathrm{C}^{1} \mathrm{H}_{2}^{16} \mathrm{O},\) 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 peak at a mass of 30.0106 in the mass spectrometer is \(^{12}C^1H_2^{16}O\).

Step by step solution

01

Choice 1: \(^{12}C_2^1H_6\)

Calculate the mass of \(^{12}C_2^1H_6\) by multiplying the mass of \(^{12}C\) by 2 and the mass of \(^{1}H\) by 6 and adding the results: \[Mass (^{12}C_2^1H_6) = 2(12.000) + 6(1.007825)\]
02

Choice 2: \(^{12}C^1H_2^{16}O\)

Calculate the mass of \(^{12}C^1H_2^{16}O\) by adding the mass of \(^{12}C\), multiplying the mass of \(^{1}H\) by 2, and adding the mass of \(^{16}O\): \[Mass (^{12}C^1H_2^{16}O) = 12.000 + 2(1.007825) + 15.994915\]
03

Choice 3: \(^{14}N^{16}O\)

Calculate the mass of \(^{14}N^{16}O\) by adding the mass of \(^{14}N\) and \(^{16}O\): \[Mass (^{14}N^{16}O) = 14.003074 + 15.994915\] #Step 2: Comparing calculated masses to the given peak# Compare the calculated masses from Step 1 to the given peak mass of 30.0106 to determine which is responsible for the peak.
04

Choice 1: \(^{12}C_2^1H_6\)

Calculate the mass of \(^{12}C_2^1H_6\): \[Mass (^{12}C_2^1H_6) = 2(12.000) + 6(1.007825) = 24.000 + 6.04695 = 30.04695\]
05

Choice 2: \(^{12}C^1H_2^{16}O\)

Calculate the mass of \(^{12}C^1H_2^{16}O\): \[Mass (^{12}C^1H_2^{16}O) = 12.000 + 2(1.007825) + 15.994915 = 12.000 + 2.01565 + 15.994915 = 30.010565\]
06

Choice 3: \(^{14}N^{16}O\)

Calculate the mass of \(^{14}N^{16}O\): \[Mass (^{14}N^{16}O) = 14.003074 + 15.994915 = 29.997989\] #Step 3: Determine the responsible molecule# Based on the calculations in Step 2, the mass of \(^{12}C^1H_2^{16}O\), 30.010565, is the closest to the given peak at a mass of 30.0106.
07

Conclusion

The molecule responsible for the peak at a mass of 30.0106 in the mass spectrometer is \(^{12}C^1H_2^{16}O\).

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Key Concepts

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

Isotopes
Isotopes are atoms of the same element that have the same number of protons but differ in the number of neutrons. This difference in neutron numbers means isotopes display varied atomic masses. Despite their mass differences, isotopes possess nearly identical chemical properties because chemical reactions primarily involve electron interactions and isotopes of an element share the same electronic configurations.
  • For example, carbon has several isotopes, including the common \(^{12}C\), and the less abundant \(^{13}C\) and \(^{14}C\).
  • Hydrogen's isotopes include protium \(^1H\), deuterium \(^2H\), and tritium \(^3H\).
Knowledge of isotopes is essential in mass spectrometry, as each isotope will appear as a distinct peak on a mass spectrum due to its unique mass.
Molecular Mass
Molecular mass, often referred to as molecular weight, is the sum of the atomic masses of all atoms in a molecule. This can be found by adding together the standard atomic weights of the constituent atoms, taking into account the number of each type of atom present. Molecular mass is a crucial concept when working with compounds in a variety of contexts, including determining the results of mass spectrometry analyses.
  • For instance, the molecular mass of water (\(H_2O\)) is calculated by taking the sum of twice the atomic mass of hydrogen (\(1.007825\)) plus the atomic mass of oxygen (\(15.999\)).
  • In the exercise, molecular masses help to identify which chemical species corresponds to a given mass spectrum peak.
Understanding how to calculate molecular mass allows chemists to pinpoint which molecules present on a spectrometer correspond to what element compositions.
Mass Calculation
Hard calculations are reduced to a simple mathematical task when determining the mass of molecular compounds. When calculating molecular mass, it involves multiplying the atomic mass of each element by the number of atoms of that element present in the molecule, and then adding these values together. This task is fundamental when interpreting mass spectra.
  • To find the mass of \(^ {12}C_2^1H_6\), for example, you calculate: \[2(12.000) + 6(1.007825) = 30.04695\]
  • In contrast, for \(^ {14}N^{16}O\), you sum: \(14.003074 + 15.994915 = 29.997989\)
By performing these calculations, one can compare the calculated molecular mass to experimental data, in this case, the mass spectrometer's output. Such calculations are critical for identifying compounds responsible for specific peaks, as shown by identifying \(^ {12}C^1H_2^{16}O\) in the example, based on the match of calculated mass to observed peak mass.

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Most popular questions from this chapter

One of the components that make up common table sugar is fructose, a compound that contains only carbon, hydrogen, and oxygen. Complete combustion of 1.50 \(\mathrm{g}\) of fructose produced 2.20 \(\mathrm{g}\) of carbon dioxide and 0.900 \(\mathrm{g}\) of water. What is the empirical formula of fructose?

In the spring of \(1984,\) concern arose over the presence of ethylene dibromide, or EDB, in grains and cereals. EDB has the molecular formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2}\) and until 1984 was commonly used as a plant fumigant. The federal limit for \(\mathrm{EDB}\) in finished cereal products is 30.0 parts per billion (ppb), where \(1.0 \mathrm{ppb}=\) \(1.0 \times 10^{-9} \mathrm{g}\) of \(\mathrm{EDB}\) for every 1.0 \(\mathrm{g}\) of sample. How many molecules of EDB are in 1.0 lb of flour if 30.0 ppb of EDB is present?

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen contains \(2.59 \times 10^{23}\) atoms of hydrogen and is 17.3\(\%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and 65 g/mol, what amount (moles) of compound is present, and what is the mass of the sample?

Some bismuth tablets, a medication used to treat upset stomachs, contain 262 \(\mathrm{mg}\) of bismuth subsalicylate, \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4},\) per tablet. Assuming two tablets are digested, calculate the mass of bismuth consumed.

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react 52.9 g of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) with excess red phosphorus, what mass of tetraphosphorus decaoxide \(\left(\mathrm{P}_{4} \mathrm{O}_{10}\right)\) could be produced? \(\mathrm{KClO}_{3}(s)+\mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{KCl}(s) \quad\) (unbalanced)

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