A second does not seem a very long time whereas 18 years might. On the exact anniversary of your 18th birthday, will you have lived for longer than a mole of seconds?

A quick calculation will show that you have lived 6574 days, that is (18 x 365) + 4 days if we chuck in 4 leap years.  This works out to be (6574 x 24 x 60 x 60) = 567993600 seconds. 5.68 x 108 is an impressive number of seconds but it is nowhere close to one mole of seconds.

On the associated page there is a slide show in the further resources section where you are asked to guestimate the number of grains of sand on a local beach. Even when you do this for the Sahara desert, which covers an area of approximately 4000 x 1500 km2 , you still end up with less than a mole of grains of sand. I think it is virtually impossible to really comprehend the number 6.02 x 1023, as numbers on that scale are beyond our everyday experience. Yet, just 12 g of carbon contain exactly that number of carbon atoms, so in turn the radius of an atom is also difficult to relate to fully .

You have perhaps come across an old legend which may, or more likely may not, be true. The brahmin Sissa ibn Dahir is said to have invented chaturanga, the forerunner of chess, and he was asked by the king what reward he would like for his invention. His response was that he would like to receive an amount of rice. The exact amount would relate to the total amount arrived at by using a chessboard. It would be equal to the sum of one grain on the first square which is then doubled on every subsequent square until the 64th square is filled. This legend is often used in mathematics to show how quickly an exponential  sequence grows. I have adapted it for use in chemistry. The following question is not likely to be asked in an IB examination, but I hope it will help you to at least begin to grasp both the enormous size of Avogadro’s constant and the incredibly small size of a single atom.
(Since it is not a test and, if you struggle a bit with maths, I've provided you with fully worked model answers should you need them - just click on the 'eye'. I've also included some real-life context in the answer to number iv.)

### Context question on Avogadro's constant

A chess board contains 64 squares. The first square contains one copper atom (Ar = 63.55) and each subsequent square contains twice the number of copper atoms than the previous square.

i. What mass of copper will be present on the 64th square?

On the second square there will be 21 = 2 Cu atoms.

On the third square there will be 22 = 4 Cu atoms

On the fourth square there will be 23 = 8 Cu atoms

On the fifth square there will be 24 = 16 Cu atoms

On the 64th square there will be 263  = 9.223 x 1018 Cu atoms (you can obtain this by taking the antilog of 63 x log102)

1 mole of Cu has a mass of 63.55 g and contains 6.02 x 1023 Cu atoms

Mass of Cu on the 64th square = (9.223 x 1018 / 6.02 x 1023) x 63.55 = 9.74 x 10-4 g

ii. How many squares would be needed before you need to place at least one mole of Cu atoms on a square?

The number of Cu atoms in one mole = 6.02 x 1023

6.02 x 1023 = 279.0

So you would need another board,  as one mole of copper is reached on the eightieth square.

iii. A copper atom has a metallic radius of 1.22 x 10-10 m.
A modern optical microscope can resolve images more than 200 nm in length.
Estimate on which square the copper would become visible under such a microscope if (a) the copper packs in a single layer and (b) the copper packs in a three-dimensional cubic structure.
(Assume each copper atom behaves like a perfect cube and packs so that there are no spaces between the atoms in each row.)

(a) 1 nm = 10-9 m, so 200 nm = 2000 x 10-10 m. The diameter of a Cu atom = 2 x 1.22 x 10-10 m

The number of Cu atoms in one row to reach a length of 200 nm

= 2000 x 10-10  / 2.44 x 10-10 = 819.67

The number of Cu atoms in a layer of copper which is one atom thick with sides of 2000 x 10-10 m = 819.672 = 671859

671859 = 219.36

So by the twentieth square you should still not be able to see the copper, but by the twenty first square it should be visible.

(b) The number of Cu atoms in a cube of copper with sides of 2000 x 10-10 m = 819.673 = 5.507 x 108

5.507 x 108= 229.04

So by the thirtieth square is should still just not be able to be seen, but by the thirty first square the copper should be visible.

iv. If 1 g of gold had been placed on the first square of the chess board what would be the total value of the gold on the board when the addition to the 64th square had been completed?  (The current price of gold is US\$56.44 g-1)

On the first square 1 g is  placed so the total mass = 1 g

On the second square 2 g are added so the total mass = 22 - 1  = 3 g

On the third square 4 g  are added so the total mass = 23 -1  = 7  g

On the 64th square 263 g are added so the total mass = 264 – 1 = 1.8447 x 1019 g

Total value = 56.44 x 1.8447 x 1019 = US\$1.04 x 1021

Let's put this into context.

As of 31 August 2021 the total national debt of the US was 28.43 trillion dollars, i.e. US\$ 2.67 x 1013 so the value of the gold on the chessboard is approximately 40 million (4 x 107) times greater than the US national debt. It is also about 4 billion (4 x 109) times more than the total amount of gold reserves in the world which are estimated to be 244000 metric tonnes (i.e. 2.44 x 1011 g) – and yet this is still nowhere near a mole of gold.

All materials on this website are for the exclusive use of teachers and students at subscribing schools for the period of their subscription. Any unauthorised copying or posting of materials on other websites is an infringement of our copyright and could result in your account being blocked and legal action being taken against you.