Protons, electrons and neutrons
· So an atom of one element must have a different mass from another element, we call this the Mass Number.
· The number of protons determines which element an atom is and the bottom number tell us this, we call this the Atomic number.
Examples:
1) Lithium
7  Li 
3 
The top number is the Mass number. This means that the total number of protons and neutrons are 7.
The bottom number is the Atomic number. This is the number of protons.
Because an atom is neutral, this means that this is also the number of
electrons. This atom has 3 protons and 3 electrons.
If we take the Atomic number (Z) from the Mass number (A) we get the number of
neutrons. 73=4 neutrons.
2) Nitrogen
14  N 
7 
An atom of nitrogen is twice as heavy as an atom of Lithium.
The top number is the mass number. This means that the total number of protons and neutrons are 14.
The bottom number is the Atomic number. This is the number of protons.
Because an atom is neutral, this means that this is also the number of
electrons. This atom has 7 protons and 7 electrons.
If we take the Atomic number (Z) from the Mass number (A) we get the number of
neutrons. 147=7 neutrons.
Isotopes
An atom of the same element that has the same number of protons and electrons but a different number of neutrons.
Measurement of relative masses
· Because Chemistry is about reacting ratios we have to measure the amount of reacting particles of each reactant.
· We now know that each atom (and therefore molecules) have different masses (because they have different numbers of protons and neutrons), we have to be able to weigh out a number of particles of reactants to react with each other.
· Since atoms are so small we give them a mass scale of their own.
· This scale is called:
Unified atomic mass unit, u: 1u = 1.66 x 10^{24}g
· This is basically the masses of a proton / neutron.
· The mass of a carbon  12 atom = 12u
· The mass of 1/12th of a carbon  12 atom = 1u
· We have to state the Atomic mass number of the atom as elements usually have isotopes:
there is a carbon  13 atom containing an extra neutron, this would have an mass of 13u!!
· This is not the only thing we have to be careful of, there are 4 different types of masses and we have to use the correct one depending what we are referring to:
1) Relative isotopic mass:
· We use this one when we are only referring to one isotope of an element
^{16}O has a mass of 16u
2) Relative atomic mass:
· We use this one when we are referring to the mixture of naturally occurring isotopes in of an element (the average by %  later)
3) Relative molecular mass, (Mr):
· We use this one when we are referring to simple molecules. Basically covalently bonded molecules, Cl_{2}, H_{2}O
water, H_{2}O has a mass of 18u
H 1 x 2 = 2
O 16 x 1 = 16
18
4) Relative formula mass:
· We use this one when we are referring to ionic compounds and giant covalent compounds. Such as CaBr_{2}, SiO_{2}
Calcium bromide, CaBr_{2} has a mass of 199.9u
Ca 40.1 x 1 = 40.1
Br 79.9 x 2 = 159.8
199.9u
· However for convenience we often use relative molecular mass instead.
· Use the formula:
RAM = (% x Ar) + (% x Ar) + ....
100
· Mass spectroscopy is the analytical method we use to obtain the % abundances for the isotopes.
Worked example:
RAM = (90.9 x 20) + (0.2 x 21) + (8.9 x 22)
100
RAM = 20.18u
Have a go for Si:
Questions 13 p9 / 2,3 p35
Amount of substance and the mole
· Since atoms are so small and therefore have such a small mass we have to measure them in large numbers.
· These large numbers are called a mole.
· The simplest way to understand the mole is to treat it as a word to describe a number:
Dozen 12
Ton 100
Pony 20
Grand 1000
Mole 6.02x10^{23} (Avogadro's constant, N_{A})
· The mole is such a large number as it takes that many atoms to be able to measure a mass in g.
· It does appear to be quite an unusual number but it has been thought out:
Basically in 12g of carbon12 you would find 6x10^{23} atoms of carbon.
· This number has been chosen to make it fit with the Atomic Masses from the Periodic Table:
1g of ^{1}H atoms would have 6 x 10^{23} atoms of H
16g of ^{16}O atoms would have 6 x 10^{23} atoms of O (atom is 16 x heavier than H)
32g of ^{32}S atoms would have 6 x 10^{23} atoms of S (atom is 32 x heavier than H)
When you think about it like this it actually makes sense!!!
· In fact if you were to measure out 6x10^{23} (A Mole) atoms of any element you would find that its mass is the same as its RAM:
1 Mole of Sodium ^{23}Na 23g mol^{}1
1Mole of Magnesium ^{24}Mg 24g mol^{}1
1 Mole of Iron ^{56}Fe 56g mol^{}1
· A molecule is made up from more than 1 atom so the mass of 1 mole of that molecule will be the sum of the RAM
1 Mole of water H_{2}O 18g mol^{1}
1Mole of Sodium Chloride NaCl 58.5g mol^{1}
Using Moles
· Because we can’t actually count out molecules or atoms (moles) we convert it to something we can measure i.e. mass.
· If 1 Mole of water is 18g then 2 moles would be 36g. 3 moles would be 54g and 0.5 moles would be 9g.
· The numbers are not always this simple so a formula helps.
Questions 1,2 p11 / 4,5 p35 / 2 p36
Empirical Formula is the simplest ratio of atoms in a molecule. Molecular formulae is the actual ratio of atoms in a molecule
· This can be calculated using moles from percentage composition:
Example 1
A sample of iron oxide was found to have 11.2g of iron and 4.8g of oxygen. Calculate the formula of this compound
Fe O
11.2 4.8
Divide by Ar 11.2/55.8 4.8/16
Moles 0.2 0.3
Divide by smallest 0.2/0.2 0.3/0.2
Ratio 1 : 1.5
Ratio 2 : 3
Fe_{2}O_{3}
Example 2
A sample of hydrocarbon was found to have 1.20g of carbon and 0.25g of hydrogen. Calculate the Empirical formula of this compound. Then find out the molecular formula if the Mr = 58
C H
1.20 0.25
Divide by Ar 1.2/12 0.25/1
Moles 0.1 0.25
Divide by smallest 0.1/0.1 0.25/0.1
Ratio 1 : 2.5
Ratio 2 : 5
Empirical formula C_{2}H_{5}
Molecular formula C_{4}H_{10}
Questions 1,2 p13 / 6,7 p35
Avogadro's hypothesis
Equal volumes of gases will have the same number of atoms / molecules
· This makes gases particularly easy to calculate 1 mole of any gas, no matter what it is, occupies the same volume (at RTP)
1 mole of any gas occupies 24dm^{3} (24000cm^{3}) at room temperature and pressure. (1dm^{3} = 1000cm^{3})
· Again the numbers are not always simple so a general formula will help:
Questions 13 p15 / 8 p35 / 3 p36
Concentration:
· So far when we have looked at reacting quantities we have dealt with the mole.
· Many reactions in chemistry involve solutions.
· A solution is expressed as a number of moles in 1dm^{3 } (1 litre or 1000cm^{3}) of solvent (usually water). · Concentration is the number of moles of specified entities in 1 dm^{3} of solution

· We use square brackets to denote concentration [X]. (also known as molarity, M).
· For example – a solution of sodium hydroxide has a concentration of 1.0 Mol dm^{3} (1.0M)
[NaOH_{(aq)}] = 1.0M This means there is 1 mole of sodium hydroxide dissolved in 1dm^{3} of water.
[NaCl_{(aq)}] = 2.0M This means there is 2 moles of sodium chloride dissolved in 1dm^{3} of water.
[KOH_{(aq)}] = 0.5M This means there is 0.5 moles of potassium hydroxide dissolved in 1dm^{3} of water.
· This is fine so long as we keep making up 1dm^{3} solutions.
· Most lab experiments only use a 100cm^{3} or so and making up 1000cm^{3} would be waste so we need to scale down:
· For example:
500cm^{3} [NaOH_{(aq)}] = 1.0M
0.5 moles of sodium hydroxide dissolved in 0.5 dm^{3}
2000cm^{3} [NaCl_{(aq)}] = 2.0M
4 moles of sodium chloride dissolved in 2 dm^{3} of water.
100cm^{3} [KOH_{(aq)}] = 0.5M
0.05 moles of sodium hydroxide dissolved in 0.1dm^{3}

· The values are not usually as nice as this so we can use the following formula:
Concentration = No. of moles C = Moles
(mol dm^{3}) Volume(dm^{3}) V (dm^{3})
· Basically the number of moles per volume of solution.
· We usually like our formula to have n, number of moles at the start:
No. of moles = Concentration (Mol dm^{3}) x Volume (dm^{3}) Moles = C x V (dm^{3})
· However this formula assumes we are working in dm^{3} and we usually work in cm^{3}.
· dm^{3} and cm^{3 } are related by a factor of 1000  you must convert into dm^{3} by dividing cm^{3} by 1000
Standard solutions
· We can make solutions of known concentration using volumetric flasks. The easiest way of learning this is to try an example.
· We need 250cm^{3} of 0.1 mol dm^{3 }solution of sodium hydroxide.
· Use the formula to calculate the No. of moles of sodium hydroxide –
No. of moles = Concentration (Mol dm^{3}) x Volume (dm^{3}) [250 / 1000 = 0.25]
No. of moles = 0.1 Mol dm^{3} x 0.25 dm^{3}
No. of moles = 0.025 Mol
· Now we know how many moles of sodium carbonate we need in 250 cm^{3} to make a 0.800 Molar solution.
· We now need to convert moles into a mass –
Mass = No. Moles x Mr Mr (NaOH) = 1x23 = 23
1x16 = 16
Mass = 0.025 x 40 1x 1 = 1
40 gMol^{1}
Mass = 1g
· Weigh this out in a beaker.
· Dissolve in distilled water and pour into the graduated flask.
· Add more distilled water to the beaker to wash the solute into the graduated flask.
· Repeat the last step several times to ensure all the solute is in the graduated flask.
· Fill the graduated flask with distilled water so the meniscus sits on the line.
· Stopper the flask and invert several times to ensure mixing.
Mass Concentrations:
· This is the mass in g dissolved in 1 dm^{3} of solution
· This means that the concentrations are measured in g dm^{3} instead
· The concentration for the solution made above would be 1 g dm^{3} as a mass concentration.
Concentrated vs diluted:
· This is down to the amount of solute dissolved in solution:
· Concentrated : Lots of solute per dm^{3}
· Diluted : A small amount of solute per dm^{3}
A16P: Part 1: Making a standard solution of ethanedoic acid
Questions 13 p17 / 9,10 p35 / 4 p36
· Mole calculations can now be used to calculate reacting amounts / product amounts.
· This is done by using the balanced chemical equation and moles calculations using masses, gas volumes and concentrations.
· ALL of these require the use of the mole:
Mass / mole calculations:
· Example: 2.4g Magnesium reacts in air to form magnesium oxide. Calculate the mass of magnesium oxide made:
STEP1: Write a balanced chemical equation and add the amounts given and question mark what you are asked to work out:
STEP2: Check the state symbol of your starting mass to decide which moles equation you will use  (s)  means you use Moles = mass / Ar
STEP3: Use the reacting ratios to work out how many moles you have made (or need):
STEP4: Check the question/ state symbol to decide whether to convert it to mass / concentration / volume  (s) = mass
· Example: 2.3g sodium reacts with excess sulphuric acid to form sodium sulphate and hydrogen gas. Calculate the volume of hydrogen made:
STEP1: Write a balanced chemical equation and add the amounts given and question mark what you are asked to work out:
STEP2: Check the state symbol of your starting mass to decide which moles equation you will use  (s)  means you use Moles = mass / Ar
STEP3: Use the reacting ratios to work out how many moles you have made (or need):
STEP4: Check the question/ state symbol to decide whether to convert it to mass / concentration / volume  (g) = volume
Concentration / mole calculations:
· Example: 2.3g sodium reacts with 250cm^{3} sulphuric acid to form sodium sulphate and hydrogen gas. Calculate the concentration of sodium sulphate solution made:
STEP1: Write a balanced chemical equation and add the amounts given and question mark what you are asked to work out:
STEP2: Check the state symbol of your starting mass to decide which moles equation you will use  (s)  means you use Moles = mass / Ar
STEP3: Use the reacting ratios to work out how many moles you have made (or need):
STEP4: Check the question/ state symbol to decide whether to convert it to mass / concentration / volume  (aq) = concentration or volume
A16PPart 2: Titrating your standard solution of ethanedoic acid
Questions 13 p21 / 4 p36
Use the fill in sheet
Interpretation:
Hydrated  Crystals that contain water
Anhydrous  Crystals that do not contain water
CuSO_{4}.H_{2}O
CuSO_{4}.5H_{2}O
CuSO_{4}.5H_{2}O
CoCl_{2}.6H_{2}O
Na_{2}SO_{4}.10H_{2}O
From empirical formula
MgCl_{2}H_{10}O_{5 } à MgCl_{2}.5H_{2}O
Na_{2}CH_{20}O_{13 } à Na_{2}CO_{3}.10H_{2}O
CaN_{2}H_{8}O_{10 } à Ca(NO_{3})_{2}._{4}H_{2}O_{ }
From mole calculations (example)
Mass of hydrated MgSO_{4}.xH_{2}O = 4.312g
Mass of anhydrous MgSO_{4} = 2.107g
Crystal, MgSO_{4}  Water, H_{2}O  
Masses of each  2.107g  (4.312  2.107) 
2.107g  2.205g  
Moles of each  2.107 / Mr  2.205 / Mr 
2.107 / 120.4  2.205 / 18  
0.0175  0.1225  
Divide by the smallest  0.0175 / 0.0175  0.1225 / 0.0175 
1  7 
So the formula of hydrated MgSO_{4}.xH_{2}O = MgSO_{4}.7H_{2}O
Now calculate the Formula from the practical you completed earlier.
Questions 12 p27 / 15 p35
Titrations  Volumetric analysis
· This technique can be used to find:
Concentration
Mr
Formula
Water of crystalisation
· To do this you react a certain volume of a solution with an unknown concentration with a solution of known concentration.
· Using moles and reacting ratios, you can calculate the concentration of this solution. This is known as titration.
· The only requirement is that you can tell when one solution has completely reacted with the other.
· Between acids and alkalis, we use indicators to let us know when the resulting solution is neutral.
· An indicator will change colour at the ‘end point’ (neutral).
· Common indicators are:
Indicator  Acidic colour  Base colour  End point colour 
Methyl orange  Red  Yellow  Orange 
Bromothymol Blue  Yellow  Blue  Green 
Phenylphthalein  colourless  Pink  Pale pink 
· Rinse the burette with distilled water then acid. · Fill the burette to the graduation mark ensuring the air bubble is removed from the tap. · Rinse a pipette with alkali, fill and transfer a known volume to a clean, dry conical flask. · Add some indicator. · Run the acid into the alkali and stop when the colour changes. This is your ‘range finder’. · Wash the conical flask with pure water. · Repeat the previous steps but stop a few cm^{3} before the colour change. · Add drop wise until the first drop changes the colour of the indicator. · Repeat until you get 2 results that agree within 0.1cm^{3}.

· Record results in a table like the one below:

Range finder / titration / cm3 
Run 1 / cm3 
Run 2 / cm3 
2^{nd} burette reading 



1^{st} burette reading 



Volume added 



25cm^{3} of 0.01M sulphuric acid was added to exactly neutralize 50cm^{3} of sodium hydroxide. Calculate the concentration of the sulphuric acid:
2NaOH_{(aq)} + H_{2}SO_{4(aq)} à Na_{2}SO_{4(aq)} + 2H_{2}0_{(l)}
No. of moles = Concentration (Mol dm^{3}) x Volume (dm^{3})
(H_{2}SO_{4})
No. of moles = 0.01 Mol dm^{3} x 0.25 dm^{3}
(H_{2}SO_{4})
No. of moles = 25 x 10^{3} Moles
(H_{2}SO_{4})
3 Use the ratio to work out the number of moles in the sample of alkali
2 : 1 NaOH : H_{2}SO_{4}
2x the number of moles for H_{2}SO_{4}
_{ }No. of moles = 25 x 10^{3} x 2
(NaOH)
No. of moles = 50 x 10^{3} Mol
(NaOH)
4 Calculate the concentration. (Convert to g dm^{3} if necessary)
Concentration (Mol dm^{3}) = No Moles
Volume (dm^{3})
Concentration (Mol dm^{3}) = 50 x 10^{3}
0.050 (dm^{3})
Concentration (Mol dm^{3}) = 1 Mol dm^{3}
5 Convert to g dm^{3} if required (usually for solubility)
Mass = No. moles x Mr (NaOH)
Mass = 1 x 40
Concentration = 40 g dm^{3}
U1 A20: Finding the Mr of a monobasic acid
Finding the concentration of NaOH
Finding the concentration of limewater
Finding the purity of a sample of sodium carbonate solution
Finding the Mr of washing soda
Questions 12 p29 / 17 p35
Questions 12 p33 / Remaining questions p35  37