I will upload the rest later :) This is everything you NEED to know, ( ill add in preparing soluble salts when I do it) xxx

1. Quantitative
analysisC3.2
Love , Heba 
HEBA SAEY

2. 1 Mole = 6.023 x 1023
One mole of atoms or a molecule of
any substance will have the same
mass in Grams as the Relative Mass
( Mr and Ar) for that substance
HEBA SAEY

3. MEASURING AMOUNTS
( MASS AND MOLES )
HEBA SAEY

4. Examples :
Substance Relative Mass One molar mass
Carbon ( C ) 12 12 g
Nitrogen ( 𝑁2) Mr = ( 14 x 2 )
28
28 g
Carbon Dioxide (𝐶𝑂2) Mr = ( 12 x 2 )
44
44g
Water (𝐻2 𝑂) Mr = (1 x 2) + 16
Mr = 18
18g
HEBA SAEY

5. Moles & Mass Formula
MASS
Moles x Mr
In GRAMS
Number of Moles given
Example 1) What is the mass in grams
of 3.5 moles of magnesium (mg)
Step 1) Mr of Mg = 24
Step 2) Identify which formula to use
Step 3) Mass = Moles x Mr
Step 4 ) Mass = 3.5 x 24
Step 5) Mass= 84 g
Example 2) How many moles are there
in 66g of Carbon dioxide?
Step 1) Mr of 𝐶𝑂2 = 12+ ( 16 x 2 ) = 44
Step 2) Identify which formula to use
Step 3) Moles = Mass ÷ Mr
Step 4 ) Moles = 66 ÷ 44 = 1.5
Step 5) Moles = 1.5 g
HEBA SAEY

6. HEBA SAEY
SOLUTIONS AND
CONCENTRATIONS

7. Remember :
HEBA SAEY
Mass Concentration = 𝒈/𝒅𝒎 𝟑 𝒐𝒓 𝒈 𝒅𝒎−𝟑
Volume = 𝒅𝒎 𝟑 ( if given in 𝒄𝒎 𝟑 divide by 1000)
Mole concentration = 𝒎𝒐𝒍/𝒅𝒎 𝟑 or 𝒎𝒐𝒍 𝒅𝒎−𝟑
Mass = g ( grams)

8. How much has dissolved?
HEBA SAEY
You can find this out by evaporating the water…
If you have say, 500g of solution , you don’t need to use the whole lot,
you can just use for example 10g…
1) Weigh a clean, dry evaporating basin
2) Weigh 10 g of the solution and put it in the basin
3) Gently heat the basin to evaporate the water from the solution
4) Check if the water seems to have evaporated
5) Weigh the dry basin and remaining solid
6) Reheat and reweigh until there's no further change in mass
STEP 6 IS TO ENSURE ALL WATER HAS EVAPORATED
Calculations on next page

9. Example:
HEBA SAEY
Mass of clean Basin = 54.6g
Mass of Basin with Solid = 56.9g
0.3 g difference..
We used 10g of 500g
0.3g is how much that has dissolved in 10 g
500 g = 10 x 50
0.3 x 50 = 15 g
15 g of substance has dissolved in 500g of
Water

10. Concentration
HEBA SAEY
IMPORTANT FORMULAS
MASS CONCENTRATION
𝑔/𝑑𝑚3
Mass Conc Volume
Mass
EXAMPLE…

11. EXAMPLE WORKING OUT MASS
CONCENTRATION
HEBA SAEY
3.75g of Sodium chloride are dissolved in 250 𝑐𝑚3 of water. What is
the concentration in 𝑔/𝑑𝑚3 of the final solution ?
1) Know which formula your using
2) Mass concentration = 𝑚𝑎𝑠𝑠 ÷ 𝑣𝑜𝑙𝑢𝑚𝑒
3) Mass = 3.27 and Volume = 250/1000= 𝟎. 𝟎𝟐𝟓 𝒅𝒎 𝟑
4) 3.75 ÷ 𝟎. 𝟐𝟓 = 𝟏𝟓
5) Mass concentration = 𝟏𝟓 𝒈/𝒅𝒎 𝟑

12. Concentration
HEBA SAEY
IMPORTANT FORMULAS
MOLE CONCENTRATION
𝑚𝑜𝑙/𝑑𝑚3
Mass Conc Mr
Mole Conc
EXAMPLE…

13. EXAMPLE 1 – converting Mass
concentration to Mole conc.
HEBA SAEY
Find the mole concentration of 196 𝑔/𝑑𝑚3 solution of 𝐻2 𝑆𝑂4
1) Know which formula your using
2) Mole concentration = 𝑚𝑎𝑠𝑠 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 ÷ 𝑀𝑟
3) Mass conc. = 196 𝑔/𝑑𝑚3
4) Mr = ( 1 x 2) + 32 + ( 16 x 4) = 98
5) 196 ÷ 98
6) Mole concentration = 𝟐 𝒎𝒐𝒍/𝒅𝒎 𝟑

14. EXAMPLE 2 – converting Mole
concentration to Mass conc.
HEBA SAEY
Find the mass concentration of 2.5 𝑚𝑜𝑙/𝑑𝑚3 solution of HCL
1) Know which formula your using
2) Mass concentration = 𝑀𝑜𝑙𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 × 𝑀𝑟
3) Mole conc. = 2.5 𝑚𝑜𝑙/𝑑𝑚3
4) Mr = 1 + 35.5 = 36.5
5) 36.5 x 2.5 = 91.25
6) Mass concentration = 𝟗𝟏. 𝟐𝟓 𝒈/𝒅𝒎 𝟑

15. Hard Water
HEBA SAEY

16. Hard water makes SCUM ( a nasty ppt ). This is
because it doesn’t lather with soap.
HEBA SAEY
Hard water contains calcium ions ( 𝑪𝒂 𝟐+)
or / and
magnesium ions (𝑴𝒈 𝟐+)
In some areas water flows over rocks and through soils containing Magnesium/ Calcium ions.
Magnesium Sulfate, 𝑀𝑔𝑆𝑂4 dissolves in water, and so does
Calcium Sulfate ,𝐶𝑎𝑆𝑂4
Calcium Carbonate usually exists as chalk, limestone or marble.
It can react with acid rain to create calcium hydrogencarbonate.
This is soluble and dissolves in water , releasing Calcium ions.
Calcium hydrogencarbonate = 𝐶𝑎(𝐻𝐶𝑂3)2

17. Removing Temporary Hardness –
BOILING.
HEBA SAEY
Temporary hardness is caused by Calcium Hydrogencarbonate
𝑪𝒂(𝑯𝑪𝑶 𝟑) 𝟐
𝐶𝑎(𝐻𝐶𝑂3)2 aq → 𝐶𝑎𝐶𝑂3 𝑠 + 𝐻2 𝑂 𝑙 + 𝐶𝑂2 𝑔
Calcium hydrogencarbonate → 𝑐𝑎𝑙𝑐𝑖𝑢𝑚 𝑐𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒 + 𝑤𝑎𝑡𝑒𝑟 + 𝑐𝑎𝑟𝑏𝑜𝑛 𝑑𝑖𝑜𝑥𝑖𝑑𝑒
.
Calcium hydrogencarbonate decomposes.
The lime scale in your kettle is Calcium Carbonate – its insoluble

18. Removing Permanent and Temporary
Hardness – ion exchange resin
HEBA SAEY
Permanent hardness is caused by 𝑀𝑔𝑆𝑂4 and
𝐶𝑎𝑆𝑂4
EXAMPLE: 𝑁𝑎2 𝑅𝑒𝑠𝑖𝑛 𝑠 + 𝐶𝑎2+ → 𝐶𝑎𝑅𝑒𝑠𝑖𝑛 𝑠 + 2𝑁𝑎+(aq).
Resin is an insoluble solid polymer. Water is supplied trough an ion
exchange resin. The resin contains a lot of Sodium ions ( or hydrogen)
and exchanges them for the C𝑎2+ and M𝑔2+ ions when the water runs
through them .

19. Titrations
HEBA SAEY

20. Basic’s about Titrations
HEBA SAEY
Acid-base titration is a neutralisation reaction
𝐻+ 𝑖𝑜𝑛𝑠 𝑓𝑟𝑜𝑚 𝑎𝑛𝑑 𝐴𝐶𝐼𝐷 𝑟𝑒𝑎𝑐𝑡 𝑤𝑖𝑡ℎ 𝑂𝐻− 𝑓𝑟𝑜𝑚 𝑎 𝑠𝑜𝑙𝑢𝑏𝑙𝑒 𝐵𝐴𝑆𝐸
IONIC equation:
𝐻+
(aq) + 𝑂𝐻−
→ 𝐻2 𝑂
REMEMBER : ACID + ALKALI → 𝑆𝐴𝐿𝑇 + 𝑊𝐴𝑇𝐸𝑅
Titrations help find out exactly how much of acid is
required to neutralise a quantity of alkali ( or vice versa)

21. How to do a titration - Step 1
HEBA SAEY
Using a pipette and pipette filler, add some
alkali ( About 25𝑐𝑚2
) to a conical flask ,
along with two or three drops of indicator.
( use of indicator depends on strength)
Type of Indicator Strength of Acid Strength of Alkali
Phenolphthalein Weak Strong
Methyl orange Strong Weak
Any Strong Strong

22. How to do a titration - Step 2
HEBA SAEY
-Fill a burette with acid.
-Using the burette add the acid to the
alkali at a time giving it a swirl

23. HEBA SAEY
How to do a titration - Step 3
The indicator changes colour when
all the alkali has been neutralised.
( stop adding in alkali). Repeat this
whole experiment 2/3 times
E.G Phenolphthalein turns colourless in acids and
pink in Alkalis

24. HEBA SAEY
Moles
VolumeConc.
Calculations – Working out number of moles to find Conc.
𝑁𝑎0𝐻 𝐻2 𝑆04
Moles in Balanced Eq. 2 1
Moles in Experiment 0.025 x 0.1 = 0.0025 0.0025/2 = 0.00125
Volume 25 /1000 = 0.025 30/1000 = 0.03
Concentration 0.1 0.00125/0.03= 0.0416
𝐸𝑥𝑎𝑚𝑝𝑙𝑒: 25𝑐𝑚3 𝑜𝑓 𝑆𝑜𝑑𝑖𝑢𝑚 ℎ𝑦𝑑𝑟𝑜𝑥𝑖𝑑𝑒 ℎ𝑎𝑠 𝑎 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 0.1 𝑚𝑜𝑙/𝑑𝑚3 it
𝑡𝑎𝑘𝑒𝑠 30𝑐𝑚3 𝑜𝑓 𝑆𝑢𝑙𝑓𝑢𝑟𝑖𝑐 𝑎𝑐𝑖𝑑 𝑡𝑜 𝑛𝑒𝑢𝑡𝑟𝑎𝑙𝑖𝑠𝑒 𝑖𝑡 .
𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑐𝑖𝑑?
Start off by writing the balanced equation then adding in the information given to you.
2𝑁𝑎0𝐻 + 𝐻2 𝑆04 → 𝑁𝑎2S𝑂4 + 2𝐻2 𝑂
ANSWER : 0.0416 𝑚𝑜𝑙/𝑑𝑚3

25. HEBA SAEY
Moles
VolumeConc.
Calculations – Working out number of moles to find Conc.
𝐶𝑎(𝑂𝐻)2 2𝐻𝐶𝑙
Moles in Balanced Eq. 1 2
Moles in Experiment 0.025 x 0.5 = 0.0125 0.0025/2 = 0.00125
Volume 25/1000 = 0.025 32.5/1000 = 0.0325
Concentration 0.5 0.00125 / 0.0325 = 0.77
𝐸𝑥𝑎𝑚𝑝𝑙𝑒: 32.5 𝑐𝑚3
𝑜𝑓 𝐻𝑦𝑑𝑟𝑜𝑐ℎ𝑙𝑜𝑟𝑖𝑐 𝑎𝑐𝑖𝑑 𝑖𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑛𝑒𝑢𝑡𝑟𝑎𝑙𝑖𝑠𝑒 25𝑐𝑚3
𝑜𝑓
0.5𝑚𝑜𝑙/𝑑𝑚3 𝑜𝑓 𝐶𝑎𝑙𝑐𝑖𝑢𝑚 ℎ𝑦𝑑𝑟𝑜𝑥𝑖𝑑𝑒. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑐𝑖𝑑.
Start off by writing the balanced equation then adding in the information given to you.
𝐶𝑎(𝑂𝐻)2 + 2𝐻𝐶𝑙 → 𝐶𝑎𝐶𝑙2 + 2𝐻2 𝑂
ANSWER : 0.77 𝑚𝑜𝑙/𝑑𝑚3