1. Quantitative & Analytical
Cover Sheet
Vocab List

2. QA1: What gives fireworks their colour?
The colour in fireworks comes from the heating of metal salts
Ion- Atom or group of atoms that is permanently charged
Cation- Atom (metal) or group of atoms with a permanent positive charge (Li+, Na+, K+, Ca2+)
A colour is produced when the salts of these cations are heated
Practical- flame tests
The Ammonium (NH4+) cation can be tested by reacting ammonium salts with NaOH to produce NH3 ammonia gas
NH4Cl(s) + NaOH(aq) NaCl(aq) + H2O(l) + NH3(g)

3. QA2: What ion is present?
Other cations (Cu2+, Fe2+, Fe3+) can be tested for by reacting salts of these cations with sodium hydroxide and observing the colour change
Anion- Atom (non-metal) or group of atoms with a permanent negative charge (Cl-, Br-, I-, SO42-, CO32-)
Salts of these anions can be tested for using various solutions
Practical- Ions Testing
Read t/b 11.5; ans qus 1-3
Read t/b 11.6; ans qus 1-3

4. QA3: How can you tell which gas you have?
Practical- Gas tests
Chemical test for water
Water can attach itself to a compound
The compound is then said to be hydrated, e.g. hydrated copper (II) sulphate, and has a distinct colour (blue)
Upon heating the compound will change colour as the water is removed
The resulting compound is said to be anhydrous (without water), e.g. anhydrous copper (II) sulphate (white)
CuSO4.5H2O(s)  CuSO4(s) + 5H2O(g)
blue white
If water is added to the anhydrous compound it will become hydrated and change colour (white to blue)
Physical test for water
Pure water will boil at 100oC (however, impurities will increase the boiling point)
Read t/b 11.4; ans qus 1-3

5. QA4: How can Cl have 35.5 for its relative atomic mass?
Atomic Number
Here is some data about elements taken from the Periodic Table.
What is the atomic number (proton number) of the elements?
Mass Number
The mass of each atom is the number of protons + neutrons that are present. (Remember that electrons have a relatively tiny mass).
Protons + neutrons = Mass number (nucleon number). 23 Na 11 56 Fe 26
119 Sn 50 19 F 9 27 Al 13 11 26 50 9 13
Atom
Protons
Neutrons
Mass Number
Hydrogen 1
Lithium 3 7
Aluminium 13 14

6. How many neutron? Calculate the number of neutrons in these atoms.
Number of Neutrons = mass number - atomic number
Atom
Mass Number
Atomic Number
Number of Neutrons
Helium 4 2
Fluorine 19 9
Strontium 88 38
Zirconium 91 40
Uranium
238 92 2 10 50 51
146

7. Isotopes
Atoms of a given element always have the same number of protons, but they may have a different numbers of neutrons.
Atoms that differ in this way are called isotopes.
Isotopes have the same number of protons and the same number of electrons.
So the chemical properties are identical
However, because the mass number is different (different number of neutrons) the physical properties such as melting point and density are different.

8. Isotopes: Carbon
Natural samples of elements are often a mixture of isotopes. About 1% of natural carbon is carbon-13. C 13 6 1% C 12 6
99% 6 6 7
Protons
Electrons
Neutrons

9. H H H Isotopes: Hydrogen
Hydrogen exists as 3 isotopes although Hydrogen-1 makes up the vast majority of the naturally occurring element. H 1 H 2 1 H 3 1
Protons
Electrons
Neutrons
Hydrogen
Protons
Electrons
Neutrons
(Deuterium)
Protons
Electrons
Neutrons
(Tritium)

10. Isotopes: Chlorine
About 75% of natural chlorine is 35Cl the rest is 37Cl.
The Relative Atomic Mass (RAM) of chlorine can be calculated:
RAM of Cl = [(75 / 100) x 35] + [(25 /100) x 37]
= 35.5g
Neon is made of 90% 20Ne and 10% 22Ne, calculate its RAM
Uses of Isotopes
Medical – cancer treatment (chemical tracers)
Industrial – check for leaks in oil and gas pipelines; radiocarbon dating using 14C
Isotopes Worksheet
Read t/b 2.2; ans qus 1-3

11. Relative Atomic Mass (Ar)
The atoms of each element have a different mass.
Carbon is given a relative atomic mass (RAM) of 12g.
The RAM of other atoms compares them with carbon.
Eg. Hydrogen has a mass of only one twelfth that of carbon and so has a RAM of 1.
Below are the RAMs of some other elements.
Element
Symbol
Times as heavy as carbon
R.A.M
Helium He
one third
Beryllium Be
three quarters
Molybdenum Mo
Eight
Krypton Kr
Seven
Oxygen O
One and one third
Silver Ag
Nine
Calcium Ca
Three and one third 4 9 96 84 16
108 40

12. Relative Molecular (Formula) Mass (Mr)
RAMs can be used to calculate relative molecular mass Mr (also known as the relative formula mass when considering ionic compounds)
To calculate this we simply add together the relative atomic masses of all the atoms shown in the formula. (N=14; H=1; Na=23; O=16; Mg=24; Ca=40)
Relative Molecular Mass Worksheet
Substance
Formula
Formula Mass
Ammonia
NH3
Sodium oxide
Na2O
Magnesium hydroxide
Mg(OH)2
Calcium nitrate
Ca(NO3)2
14 + (3x1)=17
(2x23) + 16 =62
24+ 2(16+1)=58
40+ 2(14+(3x16))=164

13. Relative Atomic Mass (Ar)
Element
Symbol
Times as heavy as carbon
R.A.M
Helium He
one third
Beryllium Be
three quarters
Molybdenum Mo
Eight
Krypton Kr
Seven
Oxygen O
One and one third
Silver Ag
Nine
Calcium Ca
Three and one third

14. Relative Molecular (Formula) Mass (Mr)
Substance
Formula
Formula Mass
Ammonia
NH3
Sodium oxide
Na2O
Magnesium hydroxide
Mg(OH)2
Calcium nitrate
Ca(NO3)2

15. QA5: How can the same number of any gas molecules have the same volume?
The Mole (Triple)
The mole is a measure of quantity in chemistry.
One mole of a substance is 6.02 x 1023 particles (atoms, molecules, formulae, ions or electrons) of that substance.
This number (6.02 x 1023) is called Avagadro’s number or the Avagadro constant.
The mass of one mole is equal to its relative atomic OR molecular(formula) mass, in grams.
Eg: 1 mole of helium has a mass of 4g, 1 mole of carbon has a mass of 12g etc. This is sometimes called the molar mass.

16. QA5: What is the Mole? The Mole (Double)
The mole is a measure of quantity in chemistry.
The mass of one mole is equal to its relative atomic OR relative molecular (formula) mass, in grams.
Eg: 1 mole of helium has a mass of 4g, 1 mole of carbon has a mass of 12g etc. This is sometimes called the molar mass.

17. The Mole and mass. m n Mr m= n x Mr n =
Moles of a substance = Mass of substance
Molar Mass of substance
This can be summarized using the following triangle to help to remember-
n = moles, m = mass (g), Mr = molar mass (g mol-1)
E.g. 1: How many moles are present in 640g of copper atoms?
Moles = Mass / Molar mass = 640/63.5 = 10.1 moles.
E.g. 2: What is the mass of 2 moles of sodium fluoride?
2 = Mass / 42 So Mass = 2 x 42 = 84g
Moles and mass worksheet m n Mr m=
n x Mr
n =

18. The Mole and gas (Triple)
Moles of a substance = Volume of gas Molar volume of gases
The ‘Molar Volume of gases’ is a constant of 24 dm3 (24,000 cm3) at room temperature and pressure
This can be summarized using the following triangle to help to remember-
n = moles, v = volume (dm3),MV = Molar volume (dm3)
E.g. 1: How many moles are present in 720 cm3 of oxygen?
Moles = Vol. / Molar vol. = (720/1000) / 24 = 0.03 moles.
NB: 1 dm3 = 1000 cm3 so you must convert the units
E.g. 2: What is the volume of 3.5 moles of Krypton atoms?
3.5 = vol. / 24 So vol. = 3.5 x 24 = 84 dm3
Moles and gas worksheet v n MV v=
n x MV
n =

19. QA6&7: How can you use ‘CSI’ data to work out the formula for a compound?
Empirical Formula
This is the simplest ratio of elements in a compound
A Molecular formula is the actual ratio of elements in a compound
E.g. the molecular formula of glucose is C6H12O6 the empirical formula would be CH2O
If the mass or % by mass of each element in a compound is known, this can be used to calculate an empirical formula
The empirical formula can then be developed into a molecular formula using the Mr

20. Empirical Formula (worked example)
0.7g nitrogen reacts with oxygen to form 2.3g of an oxide with an Mr of 92g. Work out the empirical and molecular formula?
N O (elements in the compound)
= (mass or % by mass of each element)
0.7/ / (divide by the atomic mass)
0.05/ / (divide by the smaller number)
1 2
NO (this is the empirical formula)
14 + (2x16) = 46g (this is the empirical mass)
92/46 = (this tells you by how much to x the ratios)
N2O (this is the molecular formula)
Empirical and Molecular Formula worksheet

21. Working out formula using Experimental Techniques
The formula for many types of compounds can be worked out experimentally:
Metal oxide- metals can be heated with steam OR in a crucible to determine the mass of oxide produced
Water- can be electrolysed to produce twice the volume of hydrogen compared to oxygen (H2O)
Salts with water of crystallisation (hydrated salts)- can be heated in a crucible to determine the mass of the anhydrous salt as water evaporates
Finding the Formula worksheet

22. Formula Equations Working out formula worksheet
To balance a formula equation the number of elements on both sides of the equation must be the same
A balanced formula equation shows the ratio of moles of each species (element or compound) reacting and being produced in a reaction.
The numbers before each species are called the ‘coefficient’.
State symbols can be included in subscript brackets after each species: (s) for solid, (l) for liquid, (g) for gas and (aq) for aqueous solution
When ionic compounds are reacting together they are most likely to be in an aqueous solution state (aq)
The state of other substance can be worked out using melting points and boiling points
Chemical Equations worksheets
Read t/b 4.2; ans qus 1-3
Read t/b 4.3; ans qus 1-3

23. QA8: What mass to react (how much will I get)?
Conservation of Masses
New substances are made during chemical reactions
However, the same atoms are present before and after reaction. They have just joined up in different ways.
Because of this the total mass of reactants is always equal to the total mass of products.
This idea is known as the Law of Conservation of Mass.
Reaction
but no
mass change

24. There are examples where the mass may seem to change during a reaction.
Eg. In reactions where a gas is given off the mass of the chemicals in the flask will decrease because gas atoms will leave the flask. If we carry the same reaction in a strong sealed container the mass is unchanged.
Gas given off.
Mass of chemicals in flask decreases Mg
HCl
Same reaction in sealed container:
No change in mass
11.71

25. Reacting Masses and Equations
By using an equation the stoichiometric mass(es) of product(s) can be calculated by information given about the reactants
Example 1- Calculate the mass of carbon dioxide produced when 24 g of carbon reacts with excess oxygen?
carbon + oxygen carbon dioxide
C + O2  CO2

26. Example 2- What mass of aluminium chloride is produced when 100 g of chorine gas reacts with an excess of aluminium?
aluminium chlorine aluminium chloride
Al + Cl2 AlCl3
2Al + 3Cl2 2AlCl3

27. Reacting Masses in Industrial Processes
Industrial processes use tonnes of reactants not grams.
Equations are still used to calculate masses of products.
We simply swap grams for tonnes.
Example 3- What mass of CaO does 200 tonnes of CaCO3 give?
CaCO3  CaO CO2

28. Iron is extracted from iron oxide Fe2O3 in a blast furnace
Example 4- What mass of Fe does 100 tonnes of Fe2O3 give?
Fe2O CO  2Fe CO2

29. Ammonia is made from nitrogen and hydrogen
Example 5- What mass of NH3 is formed when 50 tonnes of N2 is completely converted to ammonia?
Reacting Masses Worksheet
N H2  NH3

30. Percentage Yield (Triple)
For various reasons, not all reactions produce a 100% yield
However, it is important to try and produce as high a yield as possible
It the pharmaceutical industry, some drugs are synthesised in 10 steps, each step needing to have a yield of at least 90%
The percentage yield can be calculated as:
% yield = experimental yeild x 100
theorectical yield

31. Percentage Yield (worked example)
The electrolysis of water forms H2 and O H2O  2H2 + O2
What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O?
Percentage Yield Worksheet

32. QA9/10: What can you work out with volumes and concentration?
To make a concentrated solution- either dissolve more solute in the same vol of solvent or dissolve the same amount of solute in a smaller volume.
The Mole and solutions
Moles of a substance = Concentration x Volume
This can be summarized using the following triangle to help to remember-
n = moles, v = volume (1 dm3 = 1000 cm3 = 1 L), c = concentration (mol dm-3)
E.g. 1: How many moles are present in 250cm3 of 2 mol dm-3 hydrochloric acid
Moles =conc. x vol. = 2 x (250/1000)=0.5 moles.
E.g. 2: What is the conc. of 2 moles of 50 cm3 potassium iodide solution
c = n / v
c= 2 / (50/1000)= 40 mol dm-3 (M)
M stands for molarity = mol dm-3 n c v n=
c x v
v =
c =

33. Reacting solutions and Equations
By using an equation the stoichiometric solution volumes of a substance can be calculated by information given about another substance
Example 1- What volume of a 0.5 mol dm-3 solution of sulphuric acid is needed to react with 5 g of sodium hydroxide?
sulphuric acid+ sodium hydroxidesodium sulphate+ water
H2SO4 (aq) NaOH (s)  Na2SO4 (aq) H2O (l)

34. Example 2- What concentration of 25 cm3 solution of copper chloride is needed to produce 9 g of silver chloride?
Silver nitrate+ copper (II) chloridesilver chloride+copper nitrate
2AgNO3 (aq) CuCl2 (aq)  2AgCl (s) + Cu(NO3)2 (aq)
Equations and solutions calculations
Titration Key Assessment
Titration Problems