Download Introduction to Avogadro`s Number, Relative Atomic Mass and the

Introduction to Avogadro’s Number, Relative Atomic Mass and the Mole
A lot of students have problems with this part of chemistry at first. Fortunately, once you
understand it , you will find it quickly becomes second nature to think about quantitative chemistry
(quantity-chemistry - i.e. the numerical amounts of substances as masses or volumes we are working
with) this way. Try this approach. First, a little historical background, which you can skip if you want.
Ever since the time of John Dalton (1800 or so), it has been known that atoms of different chemical
elements combine to form compounds in simple whole number ratios. For example two atoms of
hydrogen combine with one atom of oxygen to form a molecule of water. That is a 2:1 ratio. Now no
one in the seventeenth century knew what an atom looked like, how big it was or how much it might
weigh. They could only do experiments on quantities like litres (or cubic metres) of gases and grams
(or kilograms) of solid and liquid materials and weigh how much product they got. They had no way
of knowing how many atoms in total might be involved. Dalton’s idea was that atoms of each
element had a characteristic weight, or as we now say, an atomic mass. Then Amadeo Avogadro ‘s
hypothesis that equal volumes of all gases under the same conditions of temperature and pressure
contain the same number of molecules when combined with Gay-Lussac, Berzelius and others
discoveries that the volumes of reacting gases are in simple whole number ratios gave the
opportunity to assign a scale of atomic masses to the different elements. Since at this stage of
chemistry, only the ratios in which elements react was known, the scale could only be of relative
atomic masses, not absolute atomic masses.
To illustrate this, imagine having two sets of weighing scales and two plastic bins, like flip-top bins
you have for the kitchen, without the flip-top part. Also, you have two large sets of identically-sized
balls, one set made of a lightweight plastic (like polystyrene, say) and the other set made of lead.
Start by putting one empty bin on each weighing scales and doing a zero-reset. Then fill one bin to
the brim with plastic balls and the other with lead balls. The balls are all of equal size and let us
assume they are jiggled about during filling of the bins so they pack in neatly leaving no big gaps
anywhere, then there will be the same number of plastic balls in one bin as there are lead balls in
the other.
Bins
Polystyrene
Balls
2.0
Kg
Lead Balls
Scales
22.0
Kg
Suppose the weight shown on the scales under the bin containing lead balls is eleven times greater
than the weight shown on the other scales. Then we can say that each lead ball is eleven times as
heavy as (or has eleven times the mass of) each plastic ball. So if we decided to invent our own mass
unit, a ‘relative ball mass’ such that a plastic ball was 1 on this scale, then each lead ball would have
a relative ball mass of 11. Note that you have not had to count any of the balls, nor weigh an
individual ball, nor know exactly how many balls are in each bin to be able to do this.
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Of course, if you had suitable and accurate enough scales, you could weigh one of each of the types
of ball to compare them, but with only kitchen or perhaps warehouse scales, it is hard to measure a
small mass accurately and so much easier to make your measurements on bulk quantities as
described above. This puts us in the same situation as those seventeenth and eighteenth century
chemists who had chemical balances precise enough to work in fractions of a gram, which were
nevertheless far too crude to be able to weigh individual atoms.
Suppose now that you were going to make a large modern art structure by connecting each plastic
ball to a lead ball with a short length of dowel (a lightweight stick) to make a pair, the pairs to be laid
out in some sort of pattern over the floor or suspended throughout the gallery – in the Tate Modern,
say. Since we have equal numbers of both kinds of ball available in our two bins, they could all be
used in this construction, with none left over. The balls thus combine in a simple whole number ratio
(1:1) to make an equal number of ‘lead-and plastic-ball’ pairs which we can call ‘ball-molecules’.
These are rather like simple diatomic molecules, such as hydrogen chloride (HCl) or carbon
monoxide (CO). Using the ‘relative ball mass’ scale we invented earlier, each ball-molecule has a
relative ball mass of 12 (given by 11 + 1, if we neglect the weight of the connecting dowel). If we
needed to scale up this artwork in size (to fill a larger hall, say), we still do not need to count
individual balls. We know the mass ratio of the two types of ball which fully combine (a mass ratio of
plastic ball : lead ball - 1:11) and the mass ratio of a plastic ball to a ball-molecule (single plastic ball
: ball-molecule - 1:12). This allows us to scale up the masses of our ‘reactants’ (single balls)
accordingly to obtain a desired amount of ‘product’ (ball-molecules), though with several bin-fulls to
work with perhaps this is starting to become a team project rather than just one person’s creation.
Moving this into the chemistry sphere, the lightest atom is hydrogen and this is given a relative
atomic mass (given the symbol Ar) of 1 (to the nearest whole number). The relative atomic mass of
fluorine is 19 (also to the nearest whole number). Hydrogen and fluorine react to form the diatomic
molecule, hydrogen fluoride. As a (simplified) chemical equation, this is H + F → HF. Hydrogen
fluoride has a relative molecular mass (since it is a molecular compound, not a single element) of 1 +
19 = 20. This is to say, each HF molecule has 20 times the mass of a hydrogen atom. By the
arguments we used above for lead and plastic balls, we can see that for example, 1g of hydrogen
atoms will combine with 19g of fluorine atoms to give 20g of hydrogen fluoride molecules with no
hydrogen or fluorine left over. Relative molecular mass is given the symbol Mr. You may ask why we
have to work with masses of solids or liquids in chemistry, whereas in the bins of balls situation we
can compare volumes. Well, atoms of different elements are certainly not equal in size and because
of their different electron distribution patterns and types of bonding, do not all comfortably sit the
same distances from other atoms of the same or different elements when bonded to them. So the
amount of empty space varies a lot for different elements and compounds and the volume of a
substance, when it is a liquid or solid is not a good measure of the number of atoms it contains. The
mass however, remains an accurate measure of how much of anything we have got. With gases, of
course, the distances between particles (either monatomic atoms or polyatomic molecules) is large
and the particle sizes and weak interaction forces can be ignored to a first approximation, so the
volumes of gases under the same conditions of temperature and pressure is a good measure of the
number of atoms (if monatomic like say, Helium) or molecules (if diatomic or polyatomic, like
hydrogen H2, for example) they contain and in fact experiments on the volume changes during the
reactions of gases were used to establish the ideas explained here – see the historical note above.
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It would be possible to extend our bins of balls model to more complex situations, perhaps building
‘triatomic molecules’ with two plastic balls attached to one lead ball and so on. If the starting
materials were two or more kinds of ball-molecules (rather than single balls) which have to be
disassembled then re-combined to produce several new product ball-molecules, we come much
closer to modelling the sort of chemical reactions on which we tend to do these kinds of calculation.
For example, in the chemical reaction given above, you probably know that both hydrogen and
fluorine gases are themselves diatomic molecules, so the chemical equation really ought to be
written ½H2(g) + ½F2(g) → HF(g). Of course, in terms of mass, this does not change anything: 1g of
hydrogen molecules contains the same number of hydrogen atoms as 1g of dissociated (single,
uncombined) hydrogen atoms since atoms do not change their mass just because they are being
combined with other atoms. When the reactants are compounds rather than elements, then of
course, it does matter how many atoms of each element are in each molecule; to get the relative
molecular masses correct for your calculations and to know the proportions of the atoms obtained
from each molecule of the compound e.g. when water is decomposed by electrolysis, each molecule
of water (H2O) forms two atoms of hydrogen and one atom of oxygen, so 2 × 1 + 16 = 18g of water
forms 2 × 1 = 2g of hydrogen and 16g of oxygen.
Going back to the idea of working with bins of loose single balls, if all of the various type of balls
used to construct the ball-molecules are of equal sizes, your artistic team doesn’t even need to do
any weighing. They can work in the basic unit of a ‘bin-full’. So say you wanted to construct a large
number of ball-molecules each made up of one’ type A’ ball, two ‘type B’ balls and five ‘type C’ balls
(a ball-molecular formula of AB2C5), you could get together one ‘bin-full’ of type A, two ‘bin-full’s of
type B and five ‘bin-full’s of type C and know that you would have exactly the right proportions of
balls to produce your ball-molecule product with no wasted balls. This works without you even
having to know exactly how many balls are in a bin – of course, someone could go to the trouble of
counting them one time and then you could record this special ‘balls-per-bin’ number for future
reference. It might be an interesting fact to know, but is not actually required for the art project in
hand.
What has this got to do with chemistry? Well, the ‘bin-full’ is like the mole – a scaling up of a number
of atoms or molecules to give a workable defined quantity of material and the ‘balls-per-bin’ number
is like Avogadro’s number, how many atoms or molecules that achieves that scaling. For your art
project the ‘balls-per-bin’ is the multiplier which converts a single ball to a ‘bin-full’ of balls; for
chemistry, Avogadro’s number (usually given the symbol L now days, formerly NA) is the multiplier
which converts the numerical value of relative atomic mass of a single atom of an element up to that
same number of grams of that element. For example, one mole (or in other words, one Avogadro’s
number-worth) of hydrogen atoms has a mass of 1g because the relative atomic mass of hydrogen is
1. One mole (Avogadro’s number) of hydrogen molecules has a mass of 2g because the relative
molecular mass of hydrogen is 2 (H2 – so 1 + 1= 2) The only real difference Avogadro’s number has
from our ‘balls-per-bin number ‘ is the sheer size of it: you might get a few hundred balls in a
realistically-sized plastic bin, but Avogadro’s number is enormous, some 6.023 × 1023 (the digits
‘6023’ followed by 20 zeros) which reflects just how minute atoms actually are. Note however, that
it is just a big number and nothing more sinister or difficult to understand than that. How was this
value actually determined, since we can’t actually weigh atoms on a pan balance? Well, that is a
topic you will learn about on your physics course rather than GCSE chemistry.
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The term ‘molar mass’ which you also mentioned in your question is just the mass of one mole of
something, so for example, since carbon dioxide (CO2) has a relative molecular mass of 12 + 2 × 16 =
44, the molar mass of carbon dioxide is 44g.
When performing calculations in chemistry, unlike in our art project where we have whole numbers
of bins, we generally do not have the luxury of working with an exact whole number of moles, so we
have to do some arithmetic. If you have done the ‘unitary method of solution ‘ in the Ratio and
Proportion ‘ section of your GCSE Maths, you can use that method to convert moles to masses and
vice versa. For example:
Q. 3g of pure carbon are burned in excess oxygen. What mass of carbon dioxide is formed? (Ar for C is
12, Ar for O is 16)
Ans. 12g C is 1 mole
So 1g C is 1/12 mole
So 3g C is 3/12 = ¼ = 0.25 mole
Now C(s) + O2(g) → CO2(g),
So 0.25 moles of carbon form 0.25 moles of carbon dioxide
and
1 mole of CO2 has a mass of 12 + 2 × 16 = 44g
So
0.25 mole of CO2 has a mass of 0.25 × 44 = 11g
Therefore, 3g of C burns to form 11g of CO2
Alternatively, you may have been taught to use a (mathematical) formula or to use a formula
triangle to relate mass, number of moles and relative atomic or molecular masses.
The formula is:
(number of moles) = (mass in grams) ÷ (relative atomic or molecular mass)
which you can rearrange when needed to:
(mass in grams) = (number of moles) × (relative atomic or molecular mass)
These formulas can be used to perform the same kind of calculations as shown in the example
above.
The only other relevant term I have not explained so far is ‘relative formula mass’. You probably
already know that not all compounds are in the form of simple molecules: many form giant
structures, often ionic or covalently-bonded lattices. Let me return to the art project using balls for
one last time. Suppose instead of making simple ball-molecule pairs, you decided to build a giant
cubic lattice of balls using a lot of equal-length dowels, arranging the two kinds of ball in an
alternating pattern in each direction (left-right, near-far and up-down) – you might find it easier to
visualise from a picture of the sodium chloride (common salt) lattice you will find in your chemistry
textbook. Because of the alternating arrangement, assuming you build a big lattice cube you will
need roughly equal numbers of the two kinds of ball. Now in any kind of lattice you can realistically
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build in an art gallery, the numbers probably won’t be quite equal, because your cube will stop
somewhere and the outer layer, if there are any odd side lengths, will have a slight excess of one
kind of ball – try counting up the balls in a cube with three alternating balls on each side to see what
I mean. However, the amount of the excess decreases relative to the total number of balls involved,
so in a really big cube, the effective ball ratio approaches 1:1, so becoming just the same ratio as for
the simple diatomic ball-molecules. Then we would need an equal number of ‘bin-full’s of the two
types of ball to build our lattice. This is the approach used in chemistry: sodium and chlorine atoms
combine in equal ratio to form a sodium chloride lattice (which is a really big lattice, even in a small
grain of salt simply because atoms are so small), so we give the chemical formula as NaCl as if it were
in the form of sodium chloride molecules, (which let me stress, it isn’t). Thus, the relative formula
mass of sodium chloride is simply the sum of the relative atomic masses of sodium and chlorine, just
as for molecular formula calculations. We call it relative formula mass as a reminder that we are not
dealing here with actual molecules. In fact, some examination boards seem now to have dropped
the term ‘relative molecular mass ‘entirely and use the term ‘relative formula mass’ for both cases.
They are still keeping the term ‘relative atomic mass’, though.
As a final note, I made a minor simplification in the above explanations. You recollect that I said that
the hydrogen atom is the basis of the relative atomic mass (Ar) scale. In fact, for technical reasons,
the scale is based on carbon, in particular, a special kind (called an isotope) of carbon atom called
carbon-12. The atomic mass unit is then one-twelfth the mass of a carbon-12 atom. On this scale,
the relative atomic mass of hydrogen is still very close to 1 so the gist of my explanation still applies.
If you look at a Periodic Table or table of relative atomic masses, you will see that none of the
elements have an Ar that is exactly a whole number, not even carbon because naturally occurring
carbon contains other kinds (isotopes) of carbon as well as carbon-12. The quoted relative atomic
masses then are averages of these different mass isotopes, weighted according to their relative
proportions as found in nature. Fortunately, apart for the main isotope, the proportions of most
isotopes with these different masses are usually small so the average relative atomic masses are
near to whole numbers. However, some relative atomic masses are quite a way from being near a
whole number – look at chlorine or copper in your Periodic Table, for example. In problems and
exam questions on this topic, you will be given the Ar values (usually the nearest whole numbers or
rounded to the nearest half in the case of chlorine, say) to use for each chemical element involved in
the calculations.
Bill Bavington
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