Fundamental Chemistry: Theory and Practice Carnegie College
Topic 2 – The Mole
Acknowledgements
No extract from any source held under copyright by any individual or organisation has been included in this booklet.
The author would like to thank the following people for their assistance in the development of these materials:
Alan McDowall
Contents
Topic 2 - The Mole 1
Using the Mole 3
Avogadro’s Constant 3
Gram Formula Mass (gfm) 3
Putting it All Together 7
Concentration 8
Putting it All Together 9
Molar Gas Volume 12
The Mole Progress Checklist 15
Answer to SAQs 16
© Carnegie College DH2K 34
Fundamental Chemistry: Theory and Practice Carnegie College
Topic 2 – The Mole
Topic 2 - The Mole
By the end of this section you should be able to:
· Outline the principal of the mole
· Understand Avogadro’s constant
· Use Avogadro’s constant in calculations
· Calculate gram formula masses
· Calculate concentration, number of mole and volumes from given information
· Understand and use the concept of molar gas volume.
Science is built on measurements; we measure length in metres; we measure time in seconds and so on. Chemists concern themselves with the study of atoms, molecules and ions; these are very small, and come in very large numbers. To help deal with the scale of things in chemistry a special unit is used, just as a baker would talk of having three dozen rolls, when they have thirty-six rolls, chemistry has its very own unit for measuring a specific quantity. This unit is called the mole (symbol: mol). The word mole is derived from the Latin for ‘heavy mass’. The mole is the central unit of chemistry, and will be used through this course.
We often use units to indicate a specific number of items, such as:
Dozen = 12 objects
Score = 20 objects
Gross = 144 objects
As chemistry involves very large number of things, atoms, molecule ions etc, the mole corresponds to a very large number:
Mole = 602000000000000000000000
To simplify this large number we can re-write this as:
6∙02 x 1000000000000000000000000
and just as we can write 100 as 10 x 10 or 102, we can write
100000000000000000000000 as 1023
So 1 mole 6∙02 x 1023 objects.
Question
If I have 1 mole (1 mol) of copper (Cu) atoms. How many Cu atoms will I have?
Answer: 6∙02 x 1023 Cu atoms.
Question
If I have one mol of water molecules, (H2O), how many water molecules will I have?
Answer: 6∙02 x 1023 H2O molecules
If I have one mole of sodium ions (Na+), how many Na+ ions will I have?
Answer: ______
The value of the mole, 6∙02 x 1023 may seem an odd number, but is based on factual observation. Carbon is the central component of many areas in chemistry, and so the mole is defined as the number of 12C atoms there are in 12g.
If we measure out 12 atomic mass units (amu) of Carbon C, we would have one atom of carbon. The six proton and six neutrons would give a mass of 12. If we weighed out 16 amus of Oxygen (O), we would have one atom of oxygen.
So if we weigh out 12 amus of C, 16 amus of O and 56 amus of Fe, we would have one of each type of atom. One O, one C and 1 Fe.
Weighing things out in amu’s would be pretty tricky!
By keeping the proportions the same we can use units that we can work with.
If we weigh out 12 grams of C, and 16 grams of O and 56 grams of Fe, we would have the same number of atoms in each. This is because we have just amplified the numbers from amu to grams.
So if we express a formula mass in grams we will have the same number of atoms.
So 63·5 grams of Copper (Cu) and 200·6 grams of Mercury (Hg), both contain the same number of atoms.
If we express any element by its formula mass in grams we will always have the same number of atoms.
So if I weigh out 12∙0g of carbon (graphite, diamond or fullerenes,) it would contain
6∙02 x 1023 atoms of carbon.
Using the Mole
Just as kilograms are used to measure mass of a substance, the mole is used to measure the amount of substance. When it is used in this form it is given the symbol, n. So the amount of a substance in a sample could contain 1∙5 mol of Iron, (Fe) atoms. This would be written as n(Fe) = 1∙5 mol, just the same as if its mass was 1∙5kg, it would be written as m = 1∙5kg.
In chemistry the amount of material is measured in moles.
Avogadro’s Constant
Avogadro’s number is a fundamental constant in chemistry. It is given the symbol L.
Avogadro’s constant (L) = 6∙02 x 1023
Avogadro’s constant is used to convert an amount expressed in moles into an actual number, n of objects.
Avogadro’s constant is the number of atoms, ions or molecules in 1 mole.
Gram Formula Mass (gfm)
1 mole of atoms/molecules/ions will equal the formula mass expressed in grams. The mass of Aluminium (Al) is 27∙0 amu, therefore if I weighed out 27∙0 grams of Al, I would have one mole of Al atoms, which would contain 6∙02 x 1023 Al atoms.
If I weighed out 4∙0g of Helium (He) I would have one mole of He atoms, 6∙02 x 1023 He atoms. If I weighed out 2∙0g, I would have 0∙5 moles of He, or
6∙02 x 1023 ÷ 2 = 3∙01 x 1023 He atoms.
The atomic mass expressed in grams is the gram formula mass.
Example:
If I had 18g of water (H2O), how many moles of H2O would I have?
First calculate the gram formula mass(gfm) of H2O:
Mass of Atom / Number of Atoms / Total Mass(amu)Hydrogen = 1 amu / 2 / 2
Oxygen = 16 amu / 1 / 16
Total mass / 18
So the gfm of H20 is 18g. So if you weigh out 18∙0g of water it will contain one mole of H2O water molecules. That is, it will contain 6∙02 x 1023 molecules of H2O.
If I had 8g of methane (CH4), how many moles of CH4 would I have?
First calculate the gfm of CH4
Mass of Atom / Number of Atoms / Total Mass (amu)Hydrogen = 1 / 4 / 4
Carbon = 12 / 1 / 12
Total mass / 16
Then use the formula no of moles = mass ÷ gram formula mass
n = m/gfm = 8/16
Answer = _____0∙5 moles_____
1
Calculate the gram formula mass of the following:
Example: Water (H2O)
Mass of atom / Number of atoms / Total MassHydrogen = 1 / X 2 / 2
Oxygen = 16 / X 1 / 16
Total / 18 amu
(a) Hydrochloric acid (HCl)
Mass of Atom / Number of Atoms / Total Mass(b) Sulphuric Acid (H2SO4)
Mass of Atom / Number of Atoms / Total MassSAQ 1 continues over the page
SAQ 1 continued
(c) Sodium Chloride (NaCl)
Mass of Atom / Number of Atoms / Total Mass(d) Ethanoic Acid (C2H4O2)
Mass of Atom / Number of Atoms / Total Mass(e) Sodium Sulphate (Na2SO4)
Mass of Atom / Number of Atoms / Total Mass(f) Calcium Hydroxide (Ca(OH)2)
Mass of Atom / Number of Atoms / Total MassCalculate the formula mass of the following:
(g) Silver nitrate (AgNO3)
Mass of Atom / Number of Atoms / Total MassSAQ 1 continues over the page
SAQ 1 continued
(h) Calcium Carbonate Formula (CaCO3)
Mass of Atom / Number of Atoms / Total Mass(i) Zinc Chloride (Zn Cl2)
Mass of Atom / Number of Atoms / Total Mass(j) Zinc Sulphate Formula (Zn SO4)
Mass of Atom / Number of Atoms / Total Mass(k) Copper (II) Sulphate Formula (Cu SO4)
Mass of Atom / Number of Atoms / Total Mass(l) Sodium Bromide Formula (NaBr)
Mass of Atom / Number of Atoms / Total MassCheck your answers with those give at the end of this booklet.
Putting it All Together
We now know that the formula mass, the number of moles and the gram formula mass are all interconnected. So we can express them as an equation:
Mass = Number of moles x gram formula mass
m = n x gfm
We can rearrange this to:
Number of moles (n) = Mass(m) ÷ gram formula mass(gfm). We can use a knowledge triangle to express this relationship.
Concentration
So far we have looked at the amount or mass of molecules we have, and at how we can determine how many atoms or molecules we have.
However, it is unlikely that a patient would be given a specific mass of Sodium or Potassium, or drug. They are more likely to be given a drug at a specific concentration. Concentrations can be derived in many ways, but they all refer to a particular mass within a set volume, such as grams per litre or parts per million.
One of the most common types of concentration used involves the mole. When someone has their cholesterol levels measured it is generally reported as moles per litre. This is how we are going to learn to express concentrations.
If we dissolve one mole of material in one litre of water we will have formed a one mole per litre solution. This would be written as 1 mol l-1.
So if we dissolve the gram formula mass (58∙5g) of Sodium Chloride in one litre of water we will have a 1 mol l-1 solution.
If we dissolve half of the gram formula mass (29∙25g) of sodium chloride in one litre we would have a 0∙5mol l-1 solution.
Question:
If we dissolve 5.85g of sodium chloride in one litre of water, what will the concentration of the solution be? That’s right O.1Imoll-1
If we know any two of the following three we can calculate the third:
Number of moles (n)
Volume (l)
Concentration (mol l-1)
Number of moles (n) = Volume (l) x Concentration (mol l-1)
Volume (l) = Number of moles (n) ÷ Concentration (mol l-1)
Concentration (mol l-1) = Number of moles (n) ÷ Volume
Putting it All Together
Remember there are 1000 cm3 in 1 litre and 1 cm3 = 1 ml. Volumes should be expressed in litres.
You may also see concentration expressed as moles per decimetre (molldm3) this is the same as moll-1.
Example: How many moles?
How many moles of Sodium Hydroxide must be dissolved to make up the following solution?
200 cm3 of 1 mol l-1
From the above equations we know:
Number of moles (n) = Concentration (mol l-1) x Volume (l)
Therefore n = 1 x 0∙2. Remember as concentration is expressed as moll-1 volume must always be converted to litres.
Number of moles is 0∙2 moles
250 cm3 of 0∙25 mol l-1
Number of moles (n) = Concentration (mol l-1) x Volume (l)
Therefore n = 0∙25 x 0∙25 = 0·0625 moles
Since concentration is expressed as mol l-1 volume must always be in litres.
Calculate the number of moles in each of the following.
2
1 (i) 50 cm3 of 0∙25 mol l-1
(ii) 2000 cm3 of 1 mol l-1
(iii) 300 cm3 of 0∙25 mol l-1
What is the concentration?
What is the concentration of the following solutions of Hydrochloric acid?
1 mole of acid dissolved in 500 cm3 of solution
Concentration (mol l-1) = Number of moles (n) ÷ Volume
= 1 / 0∙5
Concentration = 2 mol l-1
0∙5 mols of acid dissolved in 250 cm3 of solution
Concentration (mol l-1) = Number of moles (n) ÷ Volume
= 0∙5 ÷ 0∙250
Concentration = 2 mol l-1
Try the following:
2 i) 2∙5 mole of acid dissolved in 350 cm3 of solution
ii) 1∙25 mole of acid dissolved in 500 cm3 of solution
iii) 0∙01 mole of acid dissolved in 100 cm3 of solution
SAQ 2 continues over the page
SAQ 2 continued
What is the volume?
Calculate the volume of the following solutions:
Example:
1∙5 mol l-1 solution containing two moles of solute?
Volume (l) = Number of moles (n) ÷ Concentration (mol l-1)
= 2 / 1.5
Volume = 1∙33 l
0∙5 mol l-1 solution containing five moles of solute?
Volume (l) = Number of moles (n) ÷ Concentration (mol l-1)
= 5 ÷ 0∙5
Volume = 10∙0 l
Calculate the volume of each of the following solutions:
3) i) 2∙5 mol l-1 solution containing 0∙5 moles of solute?
ii) 5 mol l-1 solution containing 1∙75 moles of solute?
iii) 0∙75 mol l-1 solution containing 2∙8 moles of solute?
Check your answer with those given at the end of this booklet.
Molar Gas Volume
Amedeo Avogadro observed that one mole of H2(g) occupies the same volume as a one mole sample of O2(g) (when at 20ºC and one atmosphere of pressure).
This volume is 24 litres, although this volume actually depends on temperature and pressure.
One mole of any gas contains the same number of molecules, and hence will occupy the same volume. Under standard conditions of temperature and pressure this is the molar gas volume, and for an ideal gas is 24l. (This is a general observation and in reality the actual value varies.)
So the volume that a gas occupies (V) will be equal to the number of moles of gas (n) times the molar volume (Vmol).
V = n x Vmol
Once again this can be shown in a knowledge triangle.