Module 4- Unit 1
Determining Composition
John Pollard
University of Arizona
As we have discovered, the particulate model of matter
postulates that differences between substances result from
differences in the composition and structure of the
particles that make them up. It is therefore important for
us to be able to determine both the composition and
structure of the particles that make up substances. To start, we will focus on how one can
experimentally determine the mass of relative mass of particles with a technique called
mass spectrometry.
In the common approach to mass spectrometry, substances of interest are vaporized and
then broken up into ions or charged particles.
These ions are then accelerated and separated
using a magnetic field according to their
respective mass to charge ratios. The mass
spectrometry experiment takes advantage of
our understanding of the electrical nature of
matter. To explain the existence of ions, we
look to our model of the internal structure of
atoms. In this model, atoms are made up of
positively charged components called protons,
negatively charged particles called electrons
and neutrally charged particles called
neutrons. The protons and neutrons are
packed together at the nucleus of the atom and
the electrons exist around the nucleus and extend out a great distance with reference to
the size of nucleus. An atoms nucleus is approximately 10-14 meters in diameter and the
total diameter of the atom as defined by the regions the electrons occupy is approximately
10-10 meters. As an analogy, if a baseball represented the nucleus of an atom, the
electrons would occupy a region with a diameter of about 750 meters! Despite the size
difference, nearly the entire mass of an atom is made up by the nucleus with electrons
being about 1800 times lighter than both protons and neutrons (which are about the same
mass).
On the periodic table, each atom is characterized by the number of protons in the nucleus,
otherwise referred to as the atomic number. In a neutral atom, the number of protons and
electrons are equal to each other. If an atom has a negative charge, then the electrons
outnumber the protons. If an atom is positively charged, the atom has a deficiency of
electrons compared to the number of protons. The number of neutrons that an atom
possesses does not affect the charge but does affect the mass. Most elements have atoms
of different mass referred to as isotopes. The existence of isotopes is clearly seen in the
mass spectrum of an element like krypton.
The mass spectrum of a substance plots the mass to charge ratio (m/z) to the relative
abundance of the ion detected. The relative abundances of all ions detected are with
reference to the ion in highest abundance (100 on the Rel. Abundance scale). First, a
sample of Kr atoms are ionized (an electron is removed from each) to produce +1 ions.
These ions are then separated by their mass to charge ratio. Since they all have the same
charge, the m/z ratio reveals the relative masses of each ion. For Kr it can be seen that
there are six different isotopes with masses 78, 80, 82, 83, 84 and 86 amu respectively.
Since every Kr atom or ion has 36 protons, the mass differences are due to the number of
neutrons present in each isotope. For the isotope of mass 82, there are 36 protons and 46
neutrons. For the isotope of mass 86 there are 36 protons and 50 neutrons. Krypton’s
most abundant isotope is Kr-84.
If you look at the periodic table, you will notice that there is a relative
atomic mass reported for Kr that is not exactly equal to any of the
isotope masses. This is because the number reported is the weighted
average of the masses of all the isotopes. To calculate this value one
multiplies the % abundance (not the relative abundance from the mass
spectrum) by the mass for each isotope then sum these values. For Kr it would be:
(77.9 x 0.0035) + (79.9 x 0.0225) + (81.9 x 0.116) + (82.9 x 0.115) + (83.9 x 0.57) + (85.9 x 0.173) = 83.8
Krypton’s atomic number is 36 meaning all atoms of krypton have 36 protons.
Therefore, the remaining mass is made up by neutrons. For krypton, there are six stable
isotopes having masses of 78, 80, 82, 83, 84 and 86 amu respectively. These are all
isotopes of krypton and can be seen in the mass spectrum.
Mass spectrometry can also be used to determine the mass and obtain structural
information for molecules. What is different with molecules is that upon ionization, they
often break into several fragments. It is the patterns of these fragments that can be used
to elucidate structural information. As an example, let’s look at the mass spectrum of
methane (CH4).
CH4+
CH3+
CH2+
C+
CH+
13CH +
4
+
H4
It can be seen that for methane, ionization results in the formation of fragments that
represent the sequential loss of H atoms. This is consistent with the known structure of
methane where all four hydrogens are connected to a central carbon atom. The 13C
isotopic form of CH4+ can also be seen ( 13C makes is a stable isotope in 1.1%
abundance).
Elemental Analysis
Sometimes, the data gathered from techniques like mass spectrometry are combined with
other analytical techniques to provide further information about the particle composition
of a substance. Elemental analysis is a technique that can be used to determine the type
and number of atoms of particles. The technique works by determining the percentage of
each element present in a substance by burning the material and measuring the amount of
known substances that result from this combustion. As an example, let’s look at the
analysis of an automobile exhaust gas.
Elemental analysis of this substance reveals that it contains 92.26% C and 7.74% H.
From this, the empirical formula can be determined. The empirical formula of a
substance is the lowest whole number mole ratio of the elements to each other in a
substance. In this case, to find the empirical formula you do the following:
1) Assume you have 100.0 grams of your sample and use the percentages (which are by
mass) to calculate how much of each element exists in the sample.
For C it would b 92.26 g and for H it would be 7.74 g.
2) Since the empirical formula is a mole ratio, convert these gram values to mole values.
(92.26 g)(1molC/12.01g) = 7.68 mol C
(7.74 g)(1molH/1.008g) = 7.68 mol H
3) Write a preliminary formula putting the mole values in directly as subscripts.
C7.68H7.68
Since the empirical formula is the lowest whole number ratio, divide all subscripts by the
lowest value (which in this case is either value). This gives you an empirical formula of
CH meaning that C and H are in equal molar ratio in the substance AND that each
particle or molecule contains an equal number of C and H atoms.
Now let’s combine this information with the mass spectrum of this substance.
M(X) = 78.1 g/mol
The mass spectrum reveals the molar mass of the substance (78.1 g/mol). This is usually
the highest peak in the mass spectrum and is referred to as the mass peak. With this
information, we can now figure out the actual molecular formula of the substance. To do
this we calculate the mass of the empirical formula know that it is some whole number
fraction of the actual molar mass. In this case, the mass of CH is 13.018 g/mol. Taking
the actual mass and dividing it by this number (78.1/13.018) = 6. The subscripts of the
empirical formula are then multiplied by this value to obtain the molecular formula,
C6H6. The molecular formula provides that actual number of atoms per particle or
molecule of the substance (which in this case is benzene).
Benzene
Analysis of Mixtures (GC/MS)
Many times, samples that are to be analyzed are mixtures of many substances. As an
example, the drug testing of urine involves analysis of a very complex mixture of
substances that first need to be separated. GC/MS is probably one of the most common
techniques for this type of analysis. This technique is actually a combination of two
different analytical methods, gas chromatography (GC) and mass spectrometry (MS).
First, the sample is separated by vaporizing it and passing it through a column where
some components move through faster than others. This is due to the varying attractive
forces the components of the mixture have for the material inside the column. Once
separated, the substances are then passed onto a mass spectrum to determine the relative
masses and to potentially gather structural information. Elemental analysis can even be
done on the sample to provide a more complete picture of the composition and structure
of the particles that make it up.
Study Problems
1) Asbestos was used for years as an insulating material in buildings until exposure was
found to cause lung cancer. Asbestos is a mineral containing magnesium, silicon, oxygen
and hydrogen. One form of asbestos, chrysolite, has a composition of 28.03% Mg,
21.60% Si and 1.16% H. Determine the empirical formula of chrysolite.
2) Below is the mass spectrum of a compound containing only Fe, C and H.
When a 0.9437 g sample of this compound is combusted, 2.233 g of CO2 and 0.457 g
H2O is produced. Assuming all the C and H from the compound is in the CO2 and H2O
after combustion, determine the empirical and molecular formula for this substance.
3) A compound that is a common neurotransmitter is subjected to mass spectrometry and
elemental analysis. Elemental analysis shows it to be 68.2% C, 6.86% H, 15.9% N and
9.08% O. The mass spectrum is shown below. Determine the empirical and molecular
formulas.