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Avogado's Number is so large many students have trouble comprehending its size. Consequently, a small sidelight of chemistry instruction has developed for writing analogies to help express how large this number actually is.

Before looking over the following examples, here's a nice YouTube video about the mole and Avogadro's Number. Have fun and please come back to the ChemTeam when you're done exploring.

**1) Avogadro's Number compared to the Population of the Earth:**

We will take the population of the earth to be six billion (6 x 10^{9} people). We compare to Avogadro's Number like this:

6.022 x 10^{23}divided by 6 x 10^{9}= approx. 1 x 10^{14}

In other words, it would take about 100 trillion Earth populations to sum up to Avogadro's number.

If we were to take a value of 7 billion (approximate population in 2012), it would take about 86 trillion Earth populations to sum up to Avogadro's Number.

**2) Avogadro's Number as a Balancing Act:**

At the very moment of the Big Bang, you began putting H atoms on a balance and now, 19 billion years later, the balance has reached 1.008 grams. Since you know this to be Avogadro Number of atoms, you stop and decide to calculate how many atoms *per second* you had to have placed.

1.9 x 10^{10}yrs x 365.25 days/yr x 24 hrs/day x 3600 sec/hr = 6.0 x 10^{17}seconds to reach one mole6.022 x 10

^{23}atoms/mole divided by 6.0 x 10^{17}seconds/mole = approx. 1 x 10^{6}atoms/second

So, after placing one million H atoms on a balance every second for 19 billion years, you get Avogadro Number of H atoms (approximately).

**3) Avogadro's Number in Outer Space:**

If all the matter in the universe were spread evenly throughout the entire universe, there would be approx. 1 x 10¯^{6} nucleons per cm^{3}. We could do several things with that. For example:

a) What volume (in cm^{3}) of space would hold Avogadro Number of nucleons?6.022 x 10b) How many Earths would equal this volume of space (take Earth's radius to be 6380 km)?^{23}nucleons/mole divided by 1 x 10¯^{6}nucleons/cm^{3}= 6.022 x 10^{29}cm^{3}/mole

**4) Avogadro Number of Coins:**

Take a common coin of your country and stack up 30 of them. Measure the height in cm and divide by 30. You now have the average height of one coin in centimeters.

a) How high in cm is a stack of Avogadro Number of that coin?

b) How many light years is this? (Light travels 3.00 x 10^{8}km per second)

c) How many "round-trips" is this to the moon? (Go there and back = one round-trip. The Earth-Moon distance (measured center-to-center is a bit more than 384,000 km.)

Another way to express this type of problem: If you placed one mole of pills (coins, etc.) with a diameter of 1.00 cm side by side, how many trips around the Sun's equator can you make?

**Solution:**

1) Convert diameter of Sun from km to cm:

(1.392 x 102) Calculate circumference of sun:^{6}km x (10^{5}cm / 1 km) = 1.392 x 10^{11}cmI looked up the diameter of the Sun online.

c = πdc = (3.14159) (1.392 x 10

^{11}cm)c = 4.3731 x 10

^{11}cm

3) Calculate trips around the Sun:

Since each pill = 1.00 cm, one mole of them covers 6.022 x 10^{23}cm6.022 x 10

^{23}cm / 4.3731 x 10^{11}cm1.377 x 10

^{12}times around the Sun.

**5) Avogadro Number of Pieces of Paper:**

If you had a mole of sheets of paper stacked on top of each other, how many round trips to the Moon could you make? (Hint: a stack of 100 sheets of ordinary printer paper is about 1.0 cm.)

Video: A Solution to Problem #5

6) The area of the ChemTeam's home state of California is 403932.8 km^{2}. Suppose you had 6.022 x 10^{23} sheets of paper, each with dimensions 30 cm x 30 cm. (a) How many times could you cover California completely with paper? (b) Suppose each sheet of paper is 1 mm thick. How high would the paper be stacked?

7) If you drove 6.022 x10^{23} days at a speed of 100 km/h, how far would you travel?

8) If you spent 6.022 x 10^{23} dollars at an average rate of 1.00 dollar/s, how long in years would the money last? (Assume that every year has 365 days.)