Topic 19 – Modern Analytical Techniques II
In order to develop their practical skills, students should be encouraged to carry out a range of practical experiments related to this topic. Possible
experiments include using chromatography to identify unknowns such as metals in a ‘silver’ coin or amino acids in a protein. Mathematical skills
that could be developed in this topic include calculating Rf values, interpreting a variety of different spectra, calculating relative molecular masses,
using the (n+1) rule for proton NMR. Within this topic, students can consider a wider range of instrumental methods used for analysis, such as NMR;
and see how this technique is used in medicine through MRI scans. They can also see a wide range of applications that rely on a combination of
different analytical techniques.
Topic 19A: High-Resolution Mass Spectrometry
1. be able to use data from mass spectra to:
For the following questions, take the relative atomic masses of atoms to be:
H = 1.0078 C = 12.0000 O = 15.9949 N = 14.0031
i suggest possible structures of a simple organic compound given relative molecular masses,
accurate to four decimal places
Draw and name two molecules that could represent C5H10O2 (see below).
a
b - Ethyl propanoate -
Methyl butanoate
ii calculate the accurate relative molecular mass of a compound, given relative atomic masses to
four decimal places, and therefore identify a compound
A compound is found to have an accurate relative formula mass of 46.0417. It is thought to be either
CH3CH2OH or H2NCH2NH2. Calculate the Mr of each compound to 4.d.p. to find out which one it is.
CH3CH2OH = 2(12.0000) + 6(1.0078) + 15.9949 = 46.0417
H2NCH2NH2 = 12.000 + 6(1.0078) + 2(14.0031) = 46.053
Analysis of an organic compound showed that its relative formula mass is 102. High-resolution mass
spectrometry showed it to be 102.0678. What could its molecular formula be?
C5H14N2 = 5(12.0000) + 14(1.0078) + 2(14.0031) = 102.1154
C3H6N2O2 = 3(12.0000) + 6(1.0078) + 2(14.0031) + 2(15.9949) = 102.0428
C5H10O2 = 5(12.0000) + 10(1.0078) + 2(15.9949) = 102.0678
How could HRMS be used to distinguish between propane and ethenol?
Ethenol has the structural formula CH2CHOH. Propane has the structural formula CH3CH2CH3.
In a low-res mass spectrometer, both have a mass of 44. However, in a high-res one, ethenol has a mass
of 44.0261. Propanol has a mass of 44.0624. Therefore, they can be told apart on a mass spectrometer.
In order to develop their practical skills, students should be encouraged to carry out a range of practical experiments related to this topic. Possible
experiments include using chromatography to identify unknowns such as metals in a ‘silver’ coin or amino acids in a protein. Mathematical skills
that could be developed in this topic include calculating Rf values, interpreting a variety of different spectra, calculating relative molecular masses,
using the (n+1) rule for proton NMR. Within this topic, students can consider a wider range of instrumental methods used for analysis, such as NMR;
and see how this technique is used in medicine through MRI scans. They can also see a wide range of applications that rely on a combination of
different analytical techniques.
Topic 19A: High-Resolution Mass Spectrometry
1. be able to use data from mass spectra to:
For the following questions, take the relative atomic masses of atoms to be:
H = 1.0078 C = 12.0000 O = 15.9949 N = 14.0031
i suggest possible structures of a simple organic compound given relative molecular masses,
accurate to four decimal places
Draw and name two molecules that could represent C5H10O2 (see below).
a
b - Ethyl propanoate -
Methyl butanoate
ii calculate the accurate relative molecular mass of a compound, given relative atomic masses to
four decimal places, and therefore identify a compound
A compound is found to have an accurate relative formula mass of 46.0417. It is thought to be either
CH3CH2OH or H2NCH2NH2. Calculate the Mr of each compound to 4.d.p. to find out which one it is.
CH3CH2OH = 2(12.0000) + 6(1.0078) + 15.9949 = 46.0417
H2NCH2NH2 = 12.000 + 6(1.0078) + 2(14.0031) = 46.053
Analysis of an organic compound showed that its relative formula mass is 102. High-resolution mass
spectrometry showed it to be 102.0678. What could its molecular formula be?
C5H14N2 = 5(12.0000) + 14(1.0078) + 2(14.0031) = 102.1154
C3H6N2O2 = 3(12.0000) + 6(1.0078) + 2(14.0031) + 2(15.9949) = 102.0428
C5H10O2 = 5(12.0000) + 10(1.0078) + 2(15.9949) = 102.0678
How could HRMS be used to distinguish between propane and ethenol?
Ethenol has the structural formula CH2CHOH. Propane has the structural formula CH3CH2CH3.
In a low-res mass spectrometer, both have a mass of 44. However, in a high-res one, ethenol has a mass
of 44.0261. Propanol has a mass of 44.0624. Therefore, they can be told apart on a mass spectrometer.
