So…. When chemists talk about a “mole,” what exactly are they talking about???

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So…. When chemists talk about a “mole,” what exactly are they talking about???
Ummm… I don’t think so... THAT’S IT! Cute… but no… Not quite…

That’s right, in chemistry, a “mole” is a number!
Just like a “dozen” is 12 and a “gross” is 144, a “mole” is simply a word that stands for a number: 6.02 x 1023

So… What is so special about the number 6.02 x 1023?
For one thing, it represents a very, very, VERY large quantity. If written out, it would be: 602,000,000,000,000,000,000,000. That’s 602 million, million, billion! That’s big!

How big you ask… Consider the following examples.
(And come up with your own if you want. They’re really not that hard to do…)

This computer has a remarkably fast processor: 1.83 GHz.
That means, for example, that it is capable of counting to 1,830,000,000 every second! So… at that rate, how long do you think it would take this computer to count to 1 mole???

(That’s a lot of counting!)
At 1,830,000,000 per second, it would take this computer a little over ten million years to count to 1 mole!! (That’s a lot of counting!) By the way, factor label makes this calculation fairly simple: 1sec 1.83x109counts X 1min 60 sec X 1hr 60 min X 1 day 24 hr X 1 year 365 day X 6.02 x 1023counts = 1.04 x 107 years = 10.4 million years

A grain of rice is pretty small.
It takes about fifty grains of rice to fill a tiny one-milliliter box! Even still, one mole of rice grains would be HUGE… Take a guess: How much space would be taken up by one mole of rice grains?

Area that rice has to cover
One mole of rice – that’s 6.02 x 1023 grains – would cover the entire planet Earth… over 23 meters deep! And… in case you were wondering how this was calculated: Volume of all that rice 1 mL 50 grains X 1 m3 1 x 106 mL X 6.02 x 1023 grains = 1.2 x 1016 m3 Earth’s radius = 6400 km = 6.4 x 106 m Surface area of a sphere = 4Pr2 Area that rice has to cover Surface area of Earth = 4P(6.4 x 106m)2 = 5.1 x 1014 m2 Height = Volume Area 1.2 x 1016 m3 5.1 x 1014 m2 = = m

And… in case you are still not impressed…
If you had a mole of pennies you would indeed be rich. Let’s say you wanted to buy kite string at the outrageous cost of 1 million dollars per inch… you would get your money's worth.

In fact, after stretching your string around the Earth one million times, and to the Moon and back twenty-five times, you could take the string you have left over and sell it back at only one penny an inch.

And this would get you enough money to buy every man, woman and child in the US a $5000 motorcycle and enough gasoline to run it non-stop at 55 mph for a year!!

And… after those purchases, you would still have enough money left over to give every man, woman, and child in the whole world $4365!!! (WOW)

But what is so special about the number: 6.02 x 1023?
It relates back to the unit “amu.” Recall that “amu” stands for “atomic mass unit.” 1 amu is roughly the mass of a proton (or a neutron).

But exactly how big is an “amu
But exactly how big is an “amu?” And how does it relate to the mass unit we are more familiar with – the gram? That’s where 6.02 x 1023 comes in: 6.02 x 1023 amu = 1 g.

Still, why is that such a special number that it gets its own name?
2.54 cm = 1 in, but there is no special name given to “2.54.” The special thing about the mole (6.02 x 1023) is that it provides a simple way for us to “count out” atoms and molecules by weighing them.

That’s so important, it’s worth repeating:
The mole provides a simple way for us to “count out” atoms and molecules by weighing them.

To understand how this works, let’s take a look at the periodic table:
Take Cu for example We can see that the average mass for a Cu atom is amu. 29 Cu 63.55 29 Cu 63.55

So how much would a mole of Cu weigh?
Well, if one atom of Cu weighs amu, then one mole of Cu (that’s 6.02x1023 atoms of Cu) should weigh:(6.02 x 1023)x(63.55 amu). And what would that be in grams? Let’s use factor label: 29 Cu 63.55 1 g 6.02 x 1023 amu = g (6.02 x 1023)x(63.55 amu) X Now watch how this simplifies: First, the amu’s cancel… And so do the 6.02 x 1023’s… What does that leave us with…? That means we didn’t really have to do any calculation at all: The mass of one mole of Cu is simply Cu’s atomic mass expressed in grams.

And does this neat little trick work for every element in the table?
Let’s try neon. We can see that the average mass for a neon atom is amu. 10 Ne 20.18 10 Ne 20.18

How convenient is that! So how much would a mole of Ne weigh?
If one atom of Ne weighs amu, then one mole of Ne (that’s 6.02x1023 atoms of Ne) should weigh:(6.02 x 1023)x(20.18 amu). And what would that be in grams? Let’s use factor label: 10 Ne 20.18 1 g 6.02 x 1023 amu = g (6.02 x 1023)x(20.18 amu) X Again, watch how this simplifies: First, the amu’s cancel… And so do the 6.02 x 1023’s… What does that leave us with…? So the “20.18” tells us two things about neon. First, it tells us that one atom of Ne weighs amus. Second, it tells us that one mole of Ne weighs grams. How convenient is that!

So now, take out a periodic table and try answering the following questions.
As you figure out the answers, record them on the mole tutorial work sheet.

1. How much would one mole of He weigh?
1 amu 2 amu amu 6.02x1023 amu 1 g 2 g g 6.02x1023 g 1 atom 2 atoms atoms 6.02x1023 atoms

1 amu 2 amu amu 6.02x1023 amu 1 g 2 g g 6.02x1023 g 1 atom 2 atoms atoms 6.02x1023 atoms Sorry, try again

1 amu 2 amu amu 6.02x1023 amu 1 g 2 g g 6.02x1023 g 1 atom 2 atoms atoms 6.02x1023 atoms CORRECT! One mole of He would weigh g Write that answer on your worksheet, and then try the next question.

2. How much would one atom of He weigh?

1 amu 2 amu amu 6.02x1023 amu 1 g 2 g g 6.02x1023 g 1 atom 2 atoms atoms 6.02x1023 atoms CORRECT! One atom of He would weigh amu. Write that answer on your worksheet, and then try the next question.

3. How much would one atom of Ca weigh?
1 amu 12 amu 40.08 amu 6.02x1023 amu 1 g 12 g 40.08 g 6.02x1023 g 1 atom 12 atoms 40.08 atoms 6.02x1023 atoms

1 amu 12 amu 40.08 amu 6.02x1023 amu 1 g 12 g 40.08 g 6.02x1023 g 1 atom 12 atoms 40.08 atoms 6.02x1023 atoms Sorry, try again

1 amu 12 amu 40.08 amu 6.02x1023 amu 1 g 12 g 40.08 g 6.02x1023 g 1 atom 12 atoms 40.08 atoms 6.02x1023 atoms CORRECT! One atom of Ca would weigh amu. Write that answer on your worksheet, and then try the next question.

4. How much would one mole of Ca weigh?
1 amu 12 amu 40.08 amu 6.02x1023 amu 1 g 12 g 40.08 g 6.02x1023 g 1 atom 12 atoms 40.08 atoms 6.02x1023 atoms

1 amu 12 amu 40.08 amu 6.02x1023 amu 1 g 12 g 40.08 g 6.02x1023 g 1 atom 12 atoms 40.08 atoms 6.02x1023 atoms Sorry, try again

1 amu 12 amu 40.08 amu 6.02x1023 amu 1 g 12 g 40.08 g 6.02x1023 g 1 atom 12 atoms 40.08 atoms 6.02x1023 atoms CORRECT! One mole of Ca would weigh g. Write that answer on your worksheet, and then try the next question.

5. How many atoms would be present in one mole of He?

5. How many atoms would be present in one mole of He weigh?
1 amu 2 amu amu 6.02x1023 amu 1 g 2 g g 6.02x1023 g 1 atom 2 atoms atoms 6.02x1023 atoms CORRECT! In one mole of He there would be 6.02x1023 atoms present. Write that answer on your worksheet, and then try the next question.

6. How many atoms would be present in one mole of Ca?

1 amu 2 amu amu 6.02x1023 amu 1 g 2 g g 6.02x1023 g 1 atom 2 atoms atoms 6.02x1023 atoms CORRECT! In one mole of Ca there would be 6.02x1023 atoms present. Write that answer on your worksheet, and now let’s return to the tutorial.

So what have we learned so far?
1) One atom of any element weighs its atomic mass in amu’s. (thus one atom of C weighs amu) 2) One mole of any element weighs its atomic mass in grams. (thus one mole of C weighs g) 3) One mole of any element contains 6.02 x 1023 atoms. (thus one mole of C contains 6.02 x 1023 atoms)

Does the mole apply to them too?
But what about compounds like water (H2O), carbon dioxide (CO2) and propane (C3H8)… Does the mole apply to them too? ABSOLUTELY! But there is a difference… When we talk about 1 mole of an element like C, we are talking about 6.02x1023 atoms of C… That’s 6.02x1023 of these: But when we talk about 1 mole of a compound like H2O, for example, we are talking about 6.02x1023 molecules of H2O… C C C C C C C C C C C C C C C C C C C C C C C C C C C C C O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H O H

To figure out how much a single molecule of water weighs, you would simply look at its chemical formula: H2O. That’s two atoms of H and one atom of O. So the total mass would just be 2(1.0 amu) + 1(16.0 amu) = 18.0 amu. So if a molecule of water weighs 18.0 amu, how much would one mole of water weigh? If you were thinking 18.0 g, then good job! How much would one molecule of CF4 weigh? That’s one C and four F’s: 1(12.0 amu) + 4(19.0 amu) = 88.0 amu And one mole of CF4 would weigh 88.0 g

Again, using a periodic table (and perhaps a calculator if you need one), try answering the following questions? As you figure out the answers, record them on the mole tutorial work sheet.

7. How much would one mole of CO2 weigh?
22.0 amu 28.0 amu 40.0 amu 44.0 amu 22.0 g 28.0 g 40.0 g 44.0 g

7. How much would one mole of CO2 weigh?
22.0 amu 28.0 amu 40.0 amu 44.0 amu 22.0 g 28.0 g 40.0 g 44.0 g Sorry, try again

CORRECT! One mole of CO2 would weigh 44.0 g.
7. How much would one mole of CO2 weigh? 22.0 amu 28.0 amu 40.0 amu 44.0 amu 22.0 g 28.0 g 40.0 g 44.0 g CORRECT! One mole of CO2 would weigh 44.0 g. Write that answer on your worksheet, and then try the next question.

8. What would one molecule of C2H6 weigh?

CORRECT! One molecule of C2H6 would weigh 30.0 amu.
8. What would one molecule of C2H6 weigh? 13.0 amu 18.0 amu 30.0 amu 74.0 amu 13.0 g 18.0 g 30.0 g 74.0 g CORRECT! One molecule of C2H6 would weigh 30.0 amu. Write that answer on your worksheet, and then try the next question.

9. What would one mole of (NH4)2S weigh?

CORRECT! One mole of (NH4)2S would weigh 68.1 g.
9. What would one mole of (NH4)2S weigh? 38.0 amu 50.1 amu 54.1 amu 68.1 amu 38.0 g 50.1 g 54.1 g 68.1 g CORRECT! One mole of (NH4)2S would weigh 68.1 g. Write that answer on your worksheet, and then try the next question.

10. What would one molecule of NF3 weigh?

CORRECT! One molecule of NF3 would weigh 71.0 amu.
10. What would one molecule of NF3 weigh? 33.0 amu 36.0 amu 61.0 amu 71.0 amu 33.0 g 36.0 g 61.0 g 71.0 g CORRECT! One molecule of NF3 would weigh 71.0 amu. Write that answer on your worksheet, and now let’s return to the tutorial.

There is one more important bit of information we need to consider:
Seven elements on the periodic table Are not normally found as individual atoms like this: Instead, they always bonded together into pairs like this: 1 H 1.008 7 N 14.01 8 O 16.00 9 F 19.00 17 Cl 35.45 35 Br 79.90 53 I 126.9

Bromine, for example, would not just be “Br.”
These seven elements are said to be “diatomic.” And when we write their symbols, we include a subscript “2” after them. Bromine, for example, would not just be “Br.” Instead, it would be “Br2.” Likewise, oxygen would be “O2.” Br Br O O To help remember these seven diatomic elements, just think of the word: Br I N Cl H O F (pronounced “Brinklehof.”) 1 H 1.008 7 N 14.01 8 O 16.00 9 F 19.00 17 Cl 35.45 35 Br 79.90 53 I 126.9

The reason that this is important to mention here is that when we talk of “a mole of oxygen,” for example, we are not talking about 6.02x1023 individual O atoms like this Instead, we are talking about 6.02x1023 diatomic O2 molecules, like this O O O O O O O … O O O O O O O … 1 H 1.008 7 N 14.01 8 O 16.00 9 F 19.00 17 Cl 35.45 35 Br 79.90 53 I 126.9

That means that one mole of oxygen would not weigh 16.00 g.
It would weigh twice that: g. 7 N 14.01 8 O 16.00 Likewise, one mole of nitrogen would not weigh g. It would weigh twice that: g. It’s important to keep this in mind when dealing with these seven elements. 1 H 1.008 7 N 14.01 8 O 16.00 9 F 19.00 17 Cl 35.45 35 Br 79.90 53 I 126.9

would be comprised of 6.022 x 1023 S8 molecules and weigh 256.6 g. S
In addition, P is tetratomic (it usually comes in clusters of four atoms bonded thusly ). And sulfur is octatomic (it usually comes in rings of eight bonded in a zig-zag ring ). Thus a mole of sulfur would be comprised of x 1023 S8 molecules and weigh g. 16 S 32.07 15 P 30.97 16 S 32.07

Now try problems 11-20 on the mole tutorial worksheet.
The answers are listed in the margin, so check them off as you do them to make sure you are on the right track. (When you finish with question #20, come back to the tutorial.)

(where X comes from the masses in the per. table.)
Perhaps the mole concept is most useful when it comes to doing factor label questions. All you need to remember are these two concepts: atoms or molecules 1 mol = 6.02x1023 1 mol = X g (where X comes from the masses in the per. table.) By the way, “mol” is the abbreviation for “mole!”

Let’s try this problem:
How many moles are there in a sample that contains 39.2 g of Ne? First we write down our given (over 1): Then we set up a factor label fraction: We want to get rid of “g,” so “g” goes on bottom… And we want to get into “mol” so “mol” goes on top. The equality below that relates “mol” to “g” is this one We look up Ne in the per. table and see that 1 mol of Ne weighs g. So we fill these in here and here. 39.2 g 1 1 20.18 mol X g 1 mol = 6.02x1023 atoms or molecules 1 mol = X g (where X comes from the masses in the per. table.)

Let’s try this problem:
How many moles are there in a sample that contains 39.2 g of Ne? Now the “g”s cancel: And we’re ready to do the calculations. We enter “39.2” into the calculator… And since “20.18” is on bottom, we divide by it. That gives us an answer of… Since our given (39.2 g) only has three sig figs, we round our answer off to three sig figs: 39.2 g 1 1 20.18 mol X = 1.94 mol g And that’s the final answer. 1 mol = 6.02x1023 atoms or molecules 1 mol = X g (where X comes from the masses in the per. table.)

Let’s try another one: How many atoms are there in a sample that contains 3.7 moles of K? First we write down our given (over 1): Then we set up a factor label fraction: We want to get rid of “mol,” so “mol” goes on bottom… And we want to get into “atoms” so they go on top. The equality that relates “mol” to “atoms” is this one There’s no need to look at the periodic table this time. We simply fill in these numbers here and here. 3.7 mol 1 6.02x1023 1 atoms X mol 1 mol = 6.02x1023 atoms or molecules 1 mol = X g (where X comes from the masses in the per. table.)

Let’s try another one: How many atoms are there in a sample that contains 3.7 moles of K? Now the “mol”s cancel: And we’re ready to do the calculations. We enter “3.7” into the calculator… And since “6.02 EE23” is on top, we multiply by it. That gives us an answer of… Since our given (3.7 mol) only has two sig figs, we round our answer off to two sig figs: 3.7 mol 1 6.02x1023 1 atoms X = 2.2 274 EE24 atoms mol And that’s the final answer. 1 mol = 6.02x1023 atoms or molecules 1 mol = X g (where X comes from the masses in the per. table.)

Sometimes problems require two factor label steps:
How many molecules are there in 2.78 g of CO2? First we write down our given (over 1): Then we set up a factor label fraction: We want to get rid of “g,” so “g” goes on bottom… And we want to get into molecules But there is no equality that relates “g” & “molecules” Thus, we’ll have to use “mol” as a go-between; so mol goes on top. This is the equality that relates “mol” & “g.” 2.78 g 1 1 44.0 mol Using the periodic table, [12 + 2(16)] X we see 1 mol of CO2 = 44.0 g g So these go on top & bottom. 1 mol = 6.02x1023 atoms or molecules 1 mol = X g (where X comes from the masses in the per. table.)

How many molecules are there in 2.78 g of CO2? So the “g”s cancel, but we don’t stop there… Now we want to get rid of “mol” so it goes on bottom. And we put “molecules” on top. This is the equality that relates “mol” and “molecules.” So we put these two numbers on top and bottom. and we’re ready to calculate the answer: Now “mol”s cancel… 2.78 g 1 1 44.0 mol 6.02x1023 1 molecules X X g mol 1 mol = 6.02x1023 atoms or molecules 1 mol = X g (where X comes from the masses in the per. table.)

How many molecules are there in 2.78 g of CO2? First we enter “2.78” on the calculator… Then, because “44.0” is on bottom, we divide by it. Then, because “6.02 EE23” is on top, we multiply by it. This gives us: Note: since our given (2.78 g) had three sig figs, the answer was rounded to three sig figs as well. 2.78 g 1 1 44.0 mol 6.02x1023 1 molecules X X = 3.80x1022 molecules g mol 1 mol = 6.02x1023 atoms or molecules 1 mol = X g (where X comes from the masses in the per. table.)

Now, hopefully, you have learned about moles, how they relate to grams and to atoms and molecules, and how factor label can be used to convert moles to grams, grams to moles, moles to molecules, etc… At this time, try the Mole Calculations Quia question set.

So…. When chemists talk about a “mole,” what exactly are they talking about???

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So…. When chemists talk about a “mole,” what exactly are they talking about???

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