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Chapter 14: Problem 8

Both \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}\) and \(\mathrm{C}_{7} \mathrm{H}_{14}\) have the same nominal mass, namely 98 . Show how these compounds can be distinguished by the \(m / z\) ratio of their molecular ions in highresolution mass spectrometry.

Short Answer

Expert verified
Answer: The molecular compounds C6H10O and C7H14 can be distinguished using their high-resolution mass spectrometry m/z ratios by comparing their exact masses. For C6H10O, the high-resolution m/z ratio is approximately 98.1434, while for C7H14, the high-resolution m/z ratio is approximately 98.1868. Since these two m/z ratios are different, the compounds can be distinguished.

Step by step solution

01

Calculate the exact mass of each compound

In high-resolution mass spectrometry, the molecular masses are measured with higher accuracy (more decimal places). So, to compute exact masses, we'll need to use accurate atomic masses for carbon (C), hydrogen (H), and oxygen (O). Let's calculate the exact mass for each compound: For C6H10O, we use 6 atoms of carbon, 10 atoms of hydrogen, and 1 atom of oxygen: Exact mass = (6 * 12.011) + (10 * 1.00784) + (15.999) Exact mass = 72.066 + 10.0784 + 15.999 Exact mass ≈ 98.1434 For C7H14, we use 7 atoms of carbon and 14 atoms of hydrogen: Exact mass = (7 * 12.011) + (14 * 1.00784) Exact mass = 84.077 + 14.10976 Exact mass ≈ 98.1868
02

Compare the mass-to-charge (m/z) ratios

Now that we have the exact mass for each compound, we can compare their m/z ratios in high-resolution mass spectrometry to distinguish between them. Since their charge will be the same and not affect the comparison, we just consider their masses: For C6H10O, the high-resolution m/z ratio is approximately 98.1434. For C7H14, the high-resolution m/z ratio is approximately 98.1868. These two m/z ratios are different, and thus, these compounds can be distinguished by their m/z ratios in high-resolution mass spectrometry.

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Most popular questions from this chapter

Draw acceptable Lewis structures for the molecular ion (radical cation) formed from the following molecules when each is bombarded by high-energy electrons in a mass spectrometer. a. b. c. d.

For which compounds containing a heteroatom (an atom other than carbon or hydrogen) does the molecular ion have an even-numbered mass and for which does it have an odd-numbered mass? (a) A chloroalkane with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{Cl}\). (b) A bromoalkane with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{Br}\). (c) An alcohol with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{OH}\). (d) A primary amine with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n-1} \mathrm{NH}_{2}\). (e) A thiol with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{SH}\).

The low-resolution mass spectrum of 2-pentanol shows 15 peaks. Account for the formation of the peaks at \(m / z 73,70,55,45,43\), and, 41 .

The base peak in the mass spectrum of propanone (acetone) occurs at \(m / z 43\). What cation does this peak represent?

Electrospray mass spectrometry is a recently developed technique for looking at large molecules with a mass spectrometer. In this technique, molecular ions, each associated with one or more \(\mathrm{H}^{+}\)ions, are prepared under mild conditions in the mass spectrometer. As an example, a protein (P) with a molecular mass of 11,812 gives clusters of the type \((\mathrm{P}+8 \mathrm{H})^{8+},(\mathrm{P}+7 \mathrm{H})^{7+}\), and \((\mathrm{P}+6 \mathrm{H})^{6+}\). At what mass-to-charge values do these three clusters appear in the mass spectrum?
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