May the mole be with you
Michael Faraday, an English scientist, discovered the two laws of electrolysis in 1834 that describe the relationship between chemical change and electricity. Just as Newton’s laws of motion laid the foundation for classical mechanics, Faraday’s laws of electrolysis led to the development of a range of concepts and equations in the field of electrochemistry.
Perrin needed to prove that NA is a constant in the following equation that he derived:
The challenge was that N’, N and m could not be measured directly. He circumvented this issue by replacing the gas in the cylinder with visible particles suspended in a liquid. Drawing from the principles of Brownian motion, Perrin assumed that collisions between liquid molecules and the visible particles would result in a distribution similar to that of gas molecules in the cylinder. He also assumed that the motion of the particles obeyed the ideal gas law.
Perrin used a monodisperse colloid of a gum called gamboge, consisting of thousands of gamboge spheres in a water cylinder. He studied the distribution of the spheres with a microscope and adjusted eq9 to account for the upthrust of water on the gamboge spheres. For a single gamboge sphere (“particle”),
upthrust = weight of liquid displaced = ml g = dVl g (10)
where ml is the mass of the liquid displaced, d is the density of the liquid and Vl is the volume of the liquid displaced.
Furthermore, the volume of liquid displaced is equal to the volume of the particle, Vp :
Vl = Vp (11)
Substituting eq11 in eq10 yields
upthrust = dVp g (12)
Since Vp = m/D, where m is the mass of a particle and D is the density of the particle, eq 12 becomes:
The effective weight of the particle is therefore the difference between the weight of the particle and the upthrust on the particle:
Replacing the weight mg of a gas molecule in eq9 with the effective weight of a suspended particle gives
The suspended particles must be heavier than the liquid molecules for the particles to exert a downward pressure on the liquid, producing an upthrust that results in a lower effective weight for the particles. This means that d < D for eq14 to be valid. Hence, the choice of particle material is important.
Perrin meticulously prepared emulsions containing particles that were equal in size. He calculated the average mass of a particle by weighing a specified number of particles, determined its density using various methods (including the specific gravity bottle method), and counted the number of suspended particles per unit volume at different heights using a microscope.
After repeating the experiment with different particle materials (e.g. mastic), sizes, masses, liquids and temperatures, he found that the value of NA remained fairly constant, reporting numbers ranging from 6.5 x 1023 to 7.2 x 1023. He further conducted experiments using methods based on radioactivity, blackbody radiation and the motion of ions in liquids, obtaining very similar results for the value of NA. Perrin concluded that the results justified the hypotheses that had guided him, including Avogadro’s law, and named the constant the Avogadro constant, in honour of Avogadro.
The accuracy of the value of the Avogadro constant was subsequently improved by other scientists, one of whom was Robert Millikan.
Perrin developed the concept to analyse the vertical distribution of a volume of molecules of an ideal gas, with total mass of mc, in a cylinder with a cross-sectional area A and small height h at temperature T. Firstly, he considered an infinitesimal layer of the gas with mass dmc and thickness dh in the cylinder.
This layer of gas experiences an upward force Fu = pA from the pressure exerted by the gas it, and a downward force Fd = gdmc + (p+dp)A due to the gravitational force and the pressure exerted by the gas above it. At equilibrium, Fu = Fd and we get:
At this juncture, Perrin, with reference to the works of previous scientists (Avogadro, Dalton, etc.) on relative mass of molecules and introduced an important definition:
A gramme-molecule, M, is the mass of a gas that occupies the same volume as 2g of hydrogen gas at the same temperature and pressure*
* Perrin’s definition of the gramme-molecule is somewhat similar to the 1967 definition of the mole. This definition, together with Avogadro’s law, implies that a gramme-molecule of gas X and a gramme-molecule of gas Y have the same number of molecules.
With Perrin’s definition, the physical state of the gas in the cylinder can be described using the ideal gas law (eq2), where the amount of gas n is expressed in multiples of gramme-molecule M:
Substituting eq4 in eq3 by eliminating dmc/A yields
The expression for the distribution of the gas in the entire cylinder can be obtained by integrating both sides of eq5 (see above diagram):
From eq2, we have p’ = n’RT/V’ and p = nRT/V. Substituting these two equations in eq6 gives
where N’ = n’/V’ and N = n/V; that is, the number densities of the gas at the upper and lower levels of the cylinder.
Furthermore, the gramme-molecule of the gas, M, is equal to the number of molecules in a gramme-molecule, NA, multiplied by the mass of a molecule of the gas, m:
Substituting eq8 in eq7 results in
If Avogadro’s hypothesis – that ‘equal volumes of all gases at the same temperature and pressure have the same number of molecules’ – is true, NA in eq9 must be constant for different ideal gases. To determine the value of NA, Perrin would need to (i) count the number of molecules per unit volume at the upper and lower levels of the cylinder (of fixed height, h, at a constant temperature, T) for a particular gas, (ii) repeat the count for other gases with different m, and (iii) calculate the average value of NA. However, a problem arose.
Jean Perrin, a French scientist, was one of the earlier researchers who attempted to determine the Avogadro constant.
Prior to Perrin’s work in 1909, Amedeo Avogadro, an Italian scientist, published papers between 1811 and 1841 and suggested that
Equal volumes of all gases at the same temperature and pressure have the same number of molecules
This became known as Avogadro’s law. It implies that for a given mass of an ideal gas at constant temperature and pressure, the volume V of the ideal gas is directly proportional to its amount n:
V = kn (1)
where k is the proportionality constant.
When eq1 is incorporated into the combined gas law, which was developed many years earlier, we have the ideal gas law:
pV = nRT (2)
At that time, n was known as ‘amount of gas’ rather than ‘number of moles’, as the mole concept had not yet been developed. Since the amount of gas can be in measured in different ways, the gas constant R had different units back then.
Avogadro also investigated the relative mass of different gases; for example, he deduced from gas density data that the relative molecular weight of nitrogen and hydrogen is in the ratio of 13.2 : 1 and that the ratio of oxygen molecules and hydrogen molecules in water is 0.5 : 1.
Lastly, the research of botanist Robert Brown in 1872, which involved the random motion of particles suspended in a liquid or a gas due to their collisions with the surrounding molecules, also played an important role in Perrin’s calculations.
Mass is the quantity of matter that a physical body contains. In daily life, we measure almost everything using the concept of inertia mass, which is a measure of a body’s resistance to acceleration. Such measurements are made in relation to the kilogramme, which is now defined by the Planck constant. Prior to Nov 2018, the kilogramme was defined by the mass of a platinum alloy cylinder called the International Prototype Kilogramme (IPK), which is stored in France. The IPK contains octillions of atoms and has inevitably gained or lost mass over time through oxidation. Therefore, it cannot be used to accurately measure the mass of atoms.
To circumvent this problem, scientists decided to measure the mass of an atom or isotope relative to that of another atom or isotope, just as the mass of everyday objects was measured in relation to the IPK. The question then arises: which isotope should be chosen as the reference mass, and why?
The unified atomic mass unit, defined as one twelfth the mass of a carbon-12 atom, serves as a standard reference for quantifying the masses of atoms and molecules, facilitating comparisons across the diverse elements of the periodic table.
12C was chosen in 1961 as the standard reference because it has a high relative isotopic abundance, which makes it easier to isolate for measurement. Another reason for the choice of 12C is that it was already used as a reference standard in mass spectrometry prior to its selection.
Thus, 12C was arbitrarily assigned a value of exactly twelve unified atomic mass units or 12 u. Everything seems in order after this definition, but how is the carbon-12 unified atomic mass unit scale relevant to a chemist who is more familiar with calculating and measuring the inertia mass of macroscopic amounts of chemicals in the laboratory?
The following are common mass-related definitions that students encounter when they study Chemistry:-
Relative isotopic mass
The relative isotopic mass is the ratio of the mass of an isotope in unified atomic mass unit to one unified atomic mass unit. It is a dimensionless quantity, for example, the relative isotopic mass of 1H is 1.007825.
Relative atomic mass
The relative atomic mass of an element is the weighted average of the relative isotopic masses of all its isotopes. It is represented by the symbol Ar and is the value usually found next to the symbol of an element in a periodic table. For example, the relative atomic mass of copper, Ar(Cu), is (62.929601 x 0.6917) + (64.927794 x 0.3083) ≈ 63.546.
| Relative isotopic mass | % Abundance | |
| 63Cu | 62.929601 | 69.17 |
| 65Cu | 64.927794 | 30.83 |
The terms ‘atomic weight’ and ‘standard atomic weight’ are sometimes used in place of relative atomic mass. Strictly speaking, standard atomic weight is only equivalent to relative atomic mass when studying the masses of elements on earth.
Relative molecular mass
The relative molecular mass of a covalent bonded molecule is the sum of the relative atomic masses of all atoms making up the molecule. It is represented by the symbol Mr . For example, the relative molecular mass of carbon dioxide, Mr(CO2), is 12.011 + (15.999 x 2) = 44.009.
| Relative atomic mass | |
| C | 12.011 |
| O | 15.999 |
For an ionic compound, the term relative formula mass is used instead of relative molecular mass.
Atomic mass
The mass of an atom (i.e. the mass of an isotope and not an average mass of all isotopes of an element). It is defined in unified atomic mass unit, u, and can be converted to the basic SI mass unit, kg. When the atomic mass of an isotope is stated in u, it has exactly the same numerical value as the isotope’s relative isotopic mass. For example,
| Relative isotopic mass | Atomic mass, u | Atomic mass, kg | |
| 2H | 2.014104 | 2.014104 | 3.34450 x 10-27 |
The relationship between the unified atomic mass unit, u, and the basic SI mass unit, kg, is:
where is NA is the Avogadro constant. The value 0.001 kgmol-1 is called the molar mass constant and is usually written as 1 gmol-1. With the new definition of the Avogadro constant as exactly 6.02214076 x 1023 mol-1, the value of the molar mass constant deviates very slightly from the exact value of 1 gmol-1 and has to be determined through future experiments.
Molecular mass
The molecular mass is the mass of a molecule. It is defined in unified atomic mass unit, u, and can be converted to the basic SI mass unit, kg. Since an atom in a molecule is a specific isotope of an element, different molecules of the same compound may have different molecular mass. For example, the molecular mass of nitrogen gas can be
| Molecular mass, u | |
| 14N14N | 14.003074 + 14.003074 = 28.006148 |
| 14N15N | 14.003074 + 15.000109 = 29.003183 |
| 15N15N | 15.000109 + 15.000109 = 30.000218 |
For ionic compounds, the term formula mass is used instead of molecular mass.
Molar mass
The molar mass of a substance is the mass of the substance per mole. It is represented by the symbol M and has the SI unit of kgmol-1. However, most molar masses are expressed in gmol-1 for practical purpose. For an isotope, it is its atomic mass in gmol-1. For an element, it is its average atomic mass in gmol-1. For a covalent and an ionic compound, it is the average molecular mass in gmol-1 and average formula mass in gmol-1 respectively. For example,
| M, gmol-1 | |
| 35Cl | 34.97 |
| Cl | 35.45 |
| N2 | 28.014 |
| CuSO4 | 159.602 |
How do you weigh an atom?
As mentioned in the earlier sections, the mass of an atom is measured on the unified atomic mass unit scale. This is carried out in a mass spectrometer with carbon-12 as a standard reference. For example, mass spectrometric data for the ratio of the mass-to-charge ratio (u/z) of 2H to that of 12C is 0.167842. Thus, the mass of 2H on the carbon-12 unified atomic mass unit scale is:
Using eq2, the mass of one 2H is 2.014104 x 1.660539 x 10-27 = 3.3450 x 10-27 kg.
The conversion from u to kg is based on eq2, which is dependent on the uncertainty in the Avogadro constant prior to Nov 2018. However, the molar mass of 2H, which is equal to 2.0141 g, is not. This is shown by substituting eq1 in eq6,
Multiplying both sides by NA,
where the RHS of eq7 is independent of NA.
This is why the molar mass of another isotope, silicon-28, is used in determining the Avogadro constant in X-ray diffraction experiments. Furthermore, it is simpler to present the mass of an atom or isotope in the form of relative isotopic mass, which is a dimensionless quantity defined as the ratio of the mass of an isotope in unified atomic mass unit to one unified atomic mass unit. The table below lists the relative masses of isotopes of the first few elements in the periodic table:
| Atomic no. (Z) | Mass no. (A) | Symbol | Relative isotopic mass* |
| 1 | 1 | 1H | 1.007825 |
| 2 | 2H | 2.014104 | |
| 2 | 3 | 3He | 3.016029 |
| 4 | 4He | 4.002603 | |
| 3 | 6 | 6Li | 6.015122 |
| 4 | 9 | 9Be | 9.012182 |
| 5 | 10 | 10B | 10.012937 |
| 11 | 11B | 11.009305 | |
| 6 | 12 | 12C | 12.000000 |
| 8 | 16 | 16O | 15.994915 |
| 17 | 17O | 16.999132 | |
| 18 | 18O | 17.999160 | |
| 9 | 19 | 19F | 18.998403 |
| 10 | 20 | 20Ne | 19.992440 |
| 21 | 21Ne | 20.993847 | |
| 22 | 22Ne | 21.991386 |
*With the new definition of the Avogadro constant, the mass of an atom in kg is no longer subject to the uncertainty of the Avogadro constant, but is contingent on the uncertainty in the value of the molar mass constant, since . However, the relative masses of isotopes remain unchanged.
The mole concept plays a fundamental role in chemistry, enabling scientists to quantify substances and facilitating various applications, from stoichiometry in chemical reactions to drug formulation in pharmacology.
Consider the following chemical equation, which is the combustion of methane to give carbon dioxide and water:
It clearly states that 1 molecule of methane reacts with 2 molecules of oxygen to give 1 molecule of carbon dioxide and 2 molecules of water. Since the reactant ratio of 1:2 results in a product ratio of 1:2, we can say that 1 mole of methane reacts with 2 moles of oxygen to yield 1 mole of carbon dioxide and 2 moles of water when considering the equation from a macroscopic perspective. For convenience, you may treat all chemical equations as if they are written from a macroscopic viewpoint, meaning that the numbers in front of chemical formulas represent moles (molar format). Note that the number ‘1’ is not added in front of CH4 and CO2 because it is redundant. The letters in parentheses indicate the physical states of the molecules: (g) for gaseous, (l) for liquid, (s) for solid, and (aq) for aqueous.
With reference to eq3, how many moles of CO2 are produced from 1 mole of CH4 and 1.5 moles of O2?
From eq3, 2 moles of O2 react with 1 mole of CH4 to give 1 mole of CO2 and 2 moles of water. So, 1.5 moles of O2 must react with only 1.5/2 moles of CH4 to give 1.5/2 moles of CO2 and 1.5 moles of water. There will be an excess of 1 – (1.5/2) = 0.5/2 moles of CH4 that is unreacted. Note that O2 is the limiting reactant in this example, i.e. the chemical that is totally consumed when the reaction is complete.
Let’s try another one.
How many oxygen atoms are there in two moles of sulphate ions, SO42-, and how many moles of SO42- are there in 5g of CaSO4?
In one ion of SO42-, there are 1 atom of sulphur and 4 atoms of oxygen (note that the superscript “2-” refers to the charge of the ion). In one mole of SO42-, there are 1 mole of sulphur and 4 moles of oxygen. Using eq1,
number (of atoms) = number of moles x 6.02 x 1023
number (of oxygen atoms) = 2 x 4 x 6.02 x 1023
Therefore, there are 4.82 x 1024 atoms of oxygen in two moles of SO42-. 1 mole of CaSO4 contains 1 mole of SO42-. Using eq2 where , the number of moles of SO42- in 5g of CaSO4 is: