Module 5ii: What is a mole?
Look at the web site:
http://intro.chem.okstate.edu/ChemSource/Moles/mole15.htm
A Mole is a number!
It is Avagadro’s number!
What is Avagadro’s
Number?
Why do the chemist worldwide have National Mole Day:
6:02 am to 6:02 pm on
October 23^{rd} each year?
Answer:
Mathematical Connections to a mole:
On the above web site there are many
analogies to illustrate a mole:

Analogies
to illustrate the size of
Avogadro's number

f.
One mole of marshmallows would cover
the USA to a depth of 105,000 km (6500 miles). NOTE: The volume of a
marshmallow is estimated as 16 cm^{3} (1.0 in^{3 }).
The area of the USA is 9.32 x 10^{ 6} km^{2} or 3.6 x 10^{ 6}
mi^{ 2}.
g.
If an Avogadro number of pennies
were distributed evenly among the 4.9 x 10^{9} human inhabitants of
earth, each man, woman, and child would have enough money to spend a million
dollars every hourday and nightand still have over half of it unspent at
death.
h.
One guacamole is the amount of taco
chip dip that can be made from an Avogadro number of avocadosplus appropriate
quantities of tomatoes, onions, and chili. A train stretching to the North Star
and back 21/2 times would be required to transport one guacamole. NOTE:
This assumes that the volume of one standard avocado (pit removed) is 278 cm^{
3} and that other ingredients make up 25% of total volume. The average
coal car has a capacity of 110 kL (4000 ft^{ 3 )} and
is 16 m (53 ft) long. The North Star is 680 light years distant.
i.
Suppose the Greek god Zeus, after
observing the Big Bang 15 billion years ago, became bored and decided to count
one mole of atoms. Zeus is omnipotent. He can count very fast (one million
atoms per second) and, of course, never sleeps. He has currently completed over
threefourths of the task, and will be finished in just another four billion
years.
j.
One mole of moles (animaltype),
placed head to tail, would stretch 11 million light years and weigh 9/10 as much as the moon. NOTE: Each mole is
assumed to be 17 cm long with a mass of 100 g. Speed of light = 3.0 x 10^{ 8
}m/s. Mass of moon = 6.7 x 10^{22} kg..
k.
One mole of marbles, each 2 cm in
diameter, would form a mountain 116 times higher than Mount Everest. The base
of the marble mountain would be slightly larger than the area of the USA. NOTE:
Marbles are assumed to have hexagonal closest packing and the mountain has a
cone of angle 30 degrees;. Area of USA = 9.32 x 10^{6}
km^{2}.
More Analogies & Calculations for you
to do:
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.
https://www.youtube.com/watch?v=1R7NiIum2TI
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 mole
6.022 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 10^{23} nucleons/mole divided by 1 x 10¯^{6} nucleons/cm^{3} = 6.022 x 10^{29} cm^{3}/mole
b) How many Earths would equal this volume of space (take Earth's radius to be 6380 km)?
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 "roundtrips" is this to the moon? (Go there and back =
one roundtrip. The EarthMoon distance (measured centertocenter 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 10^{6} km x (10^{5} cm / 1 km) = 1.392 x 10^{11} cm
I looked up the diameter of the Sun online.
2) Calculate circumference of sun:
c = πd
c = (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} cm
6.022 x 10^{23} cm / 4.3731 x 10^{11} cm
1.377 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.)
You Solve:
6) The area of the 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?
You Solve:
7) If you drove 6.022 x10^{23} days at a speed of 100 km/h, how far would you travel?
You Solve:
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.)
You Solve:
Now it is your turn!
Find 10 more analogies to illustrate a mole using
the web and write each in your lab note book or on this report form for a take
home experiment for 10 points. Site each web site with the complete URL as I
have done above. Using the Word Doc on Blackboard you may cut and paste your 10
analogies.
Can you also find illustrations of a mole?
https://www.youtube.com/watch?v=FAh9vhueFVo
Turn in the copy below when you
take M5ii exam.
Title: What is a Mole?
All Roads in Chemistry Lead to a Mole!
Supplemental Online Lab 10 Points
Title: What is a Mole?
1.
A mole is a number!
2.
A count of objects (entities)!
3. It is Avagadro’s Number.
What is Avagadro’s
Number?
6.023 x 10^{23 }entities (Atoms,
Ions, Molecules, electrons, protons)
Objective:
To explain to a person the concept of Avagadro’s number into some kind of analogy measurement
that the lay person can understand.
Analogy #1:
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Analogy #2:
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Analogy #3:
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Analogy #4:
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Analogy #5:
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Analogy #6:
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Analogy #7:
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Analogy #8:
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Analogy #9:
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Analogy #10:
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Now for extra credit (up to 5 points):
You create an analogy that you can use to explain to someone.
You
must show your work with dimensional analysis, like in the RollMole analogy.
(Please do not use the web)