So how can we relate the mass of elements and compounds to the relative mass of their particles and the number of those particles?
Let's suppose you have an opaque bottle which your friend has weighed for you and put in the bottle a number of 2p coins.
How can you find out how many coins are in the jar without opening it?
What would you need to know about the jar and its contents?
You'd need the mass of a typical 2p coin and the mass of the empty jar.
Here's what you'd do:
First you'd weigh the jar and coins and then subtract the mass of the jar to find the mass of the coins themselves.
If you then divide the mass of the coins on their own by the mass of one 2p coin you will have the number of coins in the jar.
Simples!!
And you are saying: why isn't the mole like that?!!
Well, if you were a bank cashier you would use the coin method all the time to count the coins in a bag.
And that's the point I think, the Mole is really about using the numbers in simple problem calculations so that we get savvy at solving them.
Let's get to the mole and see what can do with it.
First the mole is a number of particles.
We have names for numbers of things: a dozen eggs (12), a score of pencils (20), a ream of paper (500) so why not a mole of particles (6.02214129(27)×1023 ) or more briefly 6.022×1023
The number is called the Avogadro number. (Symbol: L)
You can see how its calculated if you divide the standard RAM of Carbon 12 by the mass of an atom of carbon 12 (1.99252×10−23 g) that gives you the number of atoms in the 12g of carbon.
Try it for yourself and see that it approximates to the Avogadro number.
Actually, the Relative Atomic Mass of any substance measured in grams contains a Mole of particles.
This mass per mole of particles is called the Molar Mass (symbol M) with units g/mol or g.mol−1
You can probably figure out that therefore the molar mass of carbon-12 is 12.0g.mol−1
And if technetium has a RAM of 99.0 its molar mass is 99.0g.mol−1
And if molar mass applies to all substances then lets see how it works out for ethanol C2H5OH molecules.
C2H5OH
There are two carbon atoms (2*12.0)
There are six hydrogen atoms (6*1.0)
and one oxygen atom (1*16.0)
Total molar mass is 46.0g.mol−1
It is important you watch for or specify the kind of particle that's being referred to
So for example:
1.0 mol of Br i.e. 1 mole of bromine atoms RAM 80 Molar mass 80g.mol−1
1.0 mol of Br2 i.e. 1 mole of bromine molecules RAM 80*2 =160 Molar mass 160g.mol−1
So what would happen if we had say 1.75g of carbon and wanted to know how many moles of atoms of carbon-12 there were in that sample?
Using our analogy with the coins experiment we said we would find the number of coins as follows:
number of coins = mass of all coins /mass of one coin
so to find the number of moles of a substance (its Amount) we divide the mass of that substance by the mass of one mole
In words: number of moles (n) = mass of substance (m)
mass of one mole of the substance (M)
You are probably familiar with this simple little equation n = m/M
It comes in other shapes and sizes which you can find on the web:
These I picked because they happen to be correct unlike several others which use the RAM for the mass of one mole.
RAM has the same integer value but it doesn't carry a unit and so doesn't belong in this equation.
Thing to do now is to get to using this relationship in real problems from books and exam papers so that you are thoroughly saturated in this mathematical chemistry.
There are no short cuts just damned hard work required to really get this.
Some of us will need to do much more than others to train our brains.
Here is a collection of typical exam problems:
Typical problem and its solution:
A sample of magnesium weighs 128g What amount of magnesium is this?
Substitute in to n=m/M
m=128g and M = 24g.mol−1
thus n = 128g/24g.mol−1 = 5.33mol
Some further problems:
Calculate the mass of a) 2.5mol H2 b) 0.75mol copper sulfate CuSO4 c) 0.66 mol ethene C2H4
Calculate the amount of particles in a) 21.0g Chlorine molecules Cl2 b) 156g Helium atoms He
Calculate the amount of carbon dioxide in 1.5g and what mass of ethanol contains the same number of particles?
Calculate the number of particles in a) 15g of iron, b) 12g of zinc, c) 14.4g of phosphorus P4
How many ions are present in a) 0.3mol of potassium hydroxide, b) 20g of copper sulphate
If you need answers you will have to add a comment (below) and email for me to get back to you.
Some related fun mole stuff:
The Mole (1) Relative Atomic Mass and the Mass Spectrometer
The Mole (3) Using the mole to determine the simplest (empirical) formulae
The Mole (4) Using the mole to determine equation stoichiometry
Let's suppose you have an opaque bottle which your friend has weighed for you and put in the bottle a number of 2p coins.
How can you find out how many coins are in the jar without opening it?
What would you need to know about the jar and its contents?
You'd need the mass of a typical 2p coin and the mass of the empty jar.
Here's what you'd do:
First you'd weigh the jar and coins and then subtract the mass of the jar to find the mass of the coins themselves.
If you then divide the mass of the coins on their own by the mass of one 2p coin you will have the number of coins in the jar.
And you are saying: why isn't the mole like that?!!
Well, if you were a bank cashier you would use the coin method all the time to count the coins in a bag.
And that's the point I think, the Mole is really about using the numbers in simple problem calculations so that we get savvy at solving them.
Let's get to the mole and see what can do with it.
First the mole is a number of particles.
We have names for numbers of things: a dozen eggs (12), a score of pencils (20), a ream of paper (500) so why not a mole of particles (6.02214129(27)×1023 ) or more briefly 6.022×1023
The number is called the Avogadro number. (Symbol: L)
You can see how its calculated if you divide the standard RAM of Carbon 12 by the mass of an atom of carbon 12 (1.99252×10−23 g) that gives you the number of atoms in the 12g of carbon.
Try it for yourself and see that it approximates to the Avogadro number.
Actually, the Relative Atomic Mass of any substance measured in grams contains a Mole of particles.
This mass per mole of particles is called the Molar Mass (symbol M) with units g/mol or g.mol−1
You can probably figure out that therefore the molar mass of carbon-12 is 12.0g.mol−1
And if technetium has a RAM of 99.0 its molar mass is 99.0g.mol−1
And if molar mass applies to all substances then lets see how it works out for ethanol C2H5OH molecules.
C2H5OH
There are two carbon atoms (2*12.0)
There are six hydrogen atoms (6*1.0)
and one oxygen atom (1*16.0)
Total molar mass is 46.0g.mol−1
It is important you watch for or specify the kind of particle that's being referred to
So for example:
1.0 mol of Br i.e. 1 mole of bromine atoms RAM 80 Molar mass 80g.mol−1
1.0 mol of Br2 i.e. 1 mole of bromine molecules RAM 80*2 =160 Molar mass 160g.mol−1
So what would happen if we had say 1.75g of carbon and wanted to know how many moles of atoms of carbon-12 there were in that sample?
Using our analogy with the coins experiment we said we would find the number of coins as follows:
number of coins = mass of all coins /mass of one coin
so to find the number of moles of a substance (its Amount) we divide the mass of that substance by the mass of one mole
In words: number of moles (n) = mass of substance (m)
mass of one mole of the substance (M)
You are probably familiar with this simple little equation n = m/M
It comes in other shapes and sizes which you can find on the web:
These I picked because they happen to be correct unlike several others which use the RAM for the mass of one mole.
RAM has the same integer value but it doesn't carry a unit and so doesn't belong in this equation.
Thing to do now is to get to using this relationship in real problems from books and exam papers so that you are thoroughly saturated in this mathematical chemistry.
There are no short cuts just damned hard work required to really get this.
Some of us will need to do much more than others to train our brains.
Here is a collection of typical exam problems:
Typical problem and its solution:
A sample of magnesium weighs 128g What amount of magnesium is this?
Substitute in to n=m/M
m=128g and M = 24g.mol−1
thus n = 128g/24g.mol−1 = 5.33mol
Some further problems:
Calculate the mass of a) 2.5mol H2 b) 0.75mol copper sulfate CuSO4 c) 0.66 mol ethene C2H4
Calculate the amount of particles in a) 21.0g Chlorine molecules Cl2 b) 156g Helium atoms He
Calculate the amount of carbon dioxide in 1.5g and what mass of ethanol contains the same number of particles?
Calculate the number of particles in a) 15g of iron, b) 12g of zinc, c) 14.4g of phosphorus P4
How many ions are present in a) 0.3mol of potassium hydroxide, b) 20g of copper sulphate
If you need answers you will have to add a comment (below) and email for me to get back to you.
Some related fun mole stuff:
The Mole (1) Relative Atomic Mass and the Mass Spectrometer
The Mole (3) Using the mole to determine the simplest (empirical) formulae
The Mole (4) Using the mole to determine equation stoichiometry
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