1. Quantitative Chemistry
Topic 1
SL + HL

2. 1.1 Mole concept and Avogadro’s Constant
NA =
Number of atoms in 12 gram 12C (def.) =
6.02*1023 units
Mole = n = Amount of substance, with the unit mol

3. 1 mol equals: 6.02*1023 Hydrogen atoms, H
6.02*1023 Hydrogen molecules, H2
6.02*1023 Water molecules, H20
6.02*1023 formula units of Sodium Chloride, NaCl
If you have 1 mol of something, then you have 6.02*1023 units of that.

4. In 1 mol of H2-molecules there is 2 mol Hydrogen atoms 1 = 2
In 1 mol of H2O molecules there is 3 mol of atoms; 2 mol of H-atoms and 1 mol of O-atoms = 3 =2 =1

5. 1.2 Formulas A 12C-atom have by definition the mass =12
All other atoms or molecules masses relates to that mass. How much heavier or lighter they are.
Masses of single atoms and single molecules, single formula units are therefore called:
Relative atomic mass, Ar
Relative molecular mass, Mr
Relative formula mass, Mr
Relative masses have no unit in IB (in other system the unit can be u or amu)

6. Relative mass => Mass of 1 mol
Mass of 1 H-atom: Ar = 1.01
Mass of 1 mol H-atoms = 1.01g Atomic mass
Mass of 1 H2-molecule: Ar = 2.016
Mass of 1 mol H2-molecule = g Molecular
mass
Mass of NaCl = Ar = 58.44
Mass of 1 mol NaCl = g Formula mass
Different masses, same figures

7. Calculate the mass of one mole of a species from its formula.
The term molar mass, M, will be used. Unit: g/mol If you know the formula of a compound, then you know the molar mass of the compound.

8. You find the masses of the atoms in the Periodic table
The Molar mass of water , H2O M = 2*1+16 = 18g/mol The Molar mass of (NH4)2SO4 M= (14+4*1) *4 = 132 g/mol The Molar mass of CuSO4*5H2O M= *4 + 5(2*1 + 16) = g/mol

9. Relationship between the amount of substance in moles, mass and molar mass => The table
Quantity Unit
Mass, m g
Molar mass, M g/mol (usually known)
Mole, n mol
Downwards: divide
Upwards: multiply
You can always do a unit analysis, to check your calculation

10. Examples You have 34 g of Ammonia. How many mole is that?
Write the Formula of the compound.
Write the table
Fill in what you know and ?
Calculate

11. 1. Write the Formula of the compound.
NH3

12. 2. Write the table
NH3 m g M g/mol n mol

13. 3. Fill in what you know and ?
NH3 m 34 g M 17 g/mol (from periodic table) n ? mol

14. 4. Calculate
NH3 Downwards m 34 g => divide M 17 g/mol n 34/17= 2 mol Answer: 2 mol

15. New example
You have 0.50 mol of NaCl. How many gram is that?

16. 1. Write formula 2. Write table
NaCl m g M g/mol n mol

17. 3. Fill in know and ? 4. Calculate
NaCl m ? g M 58.5 g/mol (from periodic table) n 0.50 mol Upwards: multiply 0.5 * 58.5 = g Answer: 29 g (significant figures)

18. Percent Composition of Compounds
Percent composition of a compound is usually the Mass percent of the elements in a compound, e.g. H2O M = 18 g/mol
Mass% H = 2/18 = = 11.1%
Mass% O = 16/18 = = 88.8%
(Very seldom is the % composition referred to number of atoms; e.g. H2O have 3 atoms
% H= 2/3 = 67% and %O = 1/3 = 33%)

19. Empirical and Molecular formula
Molecular formula: Shows the actual number of each atom/element in a compound, e.g.
Ethane C2H6
Glucose C6H12O6
Empirical formula: Shows only the ratio of the elements in a compound, e.g.
Ethane CH3
Glucose CH2O
(Formulas of salts is empirical formulas)

20. Determining the Empirical formula
You can determine the empirical formula from the percentage composition
Use the table
Assume that you have 100g of the compound
Calculate number of moles
Compare Mole-ration. The ratio give the formula

21. Example A compound is found to consist of:
70.58 % C, 5.93 % H, % O (=100%)
What’s the formula for this compound?
Assume 100 g of the compound.
Calculate number of moles

22. (divide with the lowest)
C H O
Mass% (=100%)
m g (=100g)
M g/mol
n mol
Check the ratio
(divide with the lowest)

23. C H O
4 : : 1
Empirical formula
C4H4O

24. Molecular formula
If you know the Empirical formula, C4H4O, and you know the Molar mass of the compound, 136 g/mol then you can calculate the Molecular formula. C4H4O M = 68 g/mol To Low C8H8O2 M = 136 g/mol Correct C12H12O3 M = 204 g/mol To High

25. 1.3 Chemical Equations
A chemical reaction describes what is going on in a chemical reaction.
Shows the Reactant(s)
Shows the Product(s)
Show the Ratio between the compounds (if correct balanced)
Can show in which state the compounds are:
(s) solid, (l) liquid,
(g) gas, (aq) water solution (the most common)

26. Examples Reactants Products Propane + Oxygen  Carbon dioxide + water
Word equation gives only the compounds.
Equation must be balanced
C3H O2  CO H2O
Now we have a balanced equation = same number of all elements on both sides of the reaction arrow.
The green numbers = subscripts; cannot be changed (a certain compound have a certain formula, if you change the formula you change the compound)
The red numbers = coefficients; changes so that the reaction will be balanced (coefficients is valid only for a specific reaction)

27. What does an equation tell us?
C3H8(g) + 5 O2(g)  3 CO2(g) H2O(l)
The state of the compounds
That you need 5 Oxygen molecules/Propane
1 Propane molecule will produce 3 Carbon dioxide molecules and 4 Water molecules
Etc
Or multiples thereof

28. Calculate mass etc in a reaction Examples
If you start with 15 g of propane: Which mass of oxygen do you need and which mass of carbon dioxide and water do you get?
Write the balanced reaction equation
Write the table
Fill in known and ?

29. C3H O  3 CO H2O
m g ? ? ?
M 44g/mol g/mol g/mol g/mol n
Calculate mol of propane (down =divide)
Go to next compound on the “mol level”. Check the Ration between the compounds.

30. C3H O  3 CO H2O
m g ? ? ?
M 44g/mol g/mol g/mol g/mol
n mol 1:5 mol 5:3 1.02mol 3:4 1.36mol
1. Go upwards = multiply to get the masses

31. C3H O  3 CO H2O
m g g g g
M 44g/mol g/mol g/mol g/mol
n mol 1:5 mol 5:3 1.02mol 3:4 1.36mol

32. More examples
Sodium reacts with chlorine and the product is sodium chloride. If you want to make 10 g of sodium chloride how many grams of sodium and chlorine do you need?
Write the balanced equation
Write the table
Fill in known and ?

33. 2 Na Cl2  2 NaCl
m ? ? g
M g/mol n
Mol of NaCl? Down = divide
Go on “mol level” to other compounds. Check Ratio

34. 2 Na + Cl2  2 NaCl m ? ? 10g M 23g/mol 71g/mol 58.5g/mol
n 0.17mol 2:1 mol 1:2 0.17mol
1. Up = multiply
m g g g

35. More examples Ammonia, NH3, can be made from Nitrogen and Hydrogen.
What mass of ammonia can you get from 168 g of Nitrogen and 24 g of Hydrogen?
Write balanced equation
Write table
Fill in known and ?
Calculate mol

36. N2 + 3 H2  2 NH3 m 168 g 24 g ? M 28 g/mol 2 g/mol 17 g/mol
n mol mol
Theoretical ratio between N2 and H2 => 1:3
Actual ratio between N2 and H => 1:2
We need 18mol of H2 if all N2 should react : Nitrogen is in excess. Hydrogen is limiting reactant, and decides amount of product that can be formed.
m g g g
M g/mol g/mol g/mol
n mol 3:2 8mol

37. 1.4 Theoretical and Percent yield
The theoretical yield in the example was 136g. In real life nothing is perfect. Some of the reactants may not react or they react and become something else etc. Lets say that you only end up with 105g product. Percent yield = 105/136 = 77%

38. A gas have under standard condition the mol volume: 22.4 dm3/mol

39. Reacting gas volumes
For a gas at a constant temperature and pressure, the volume is directly proportional to the number of moles of gas. The mol ration  The volume ratio

40. If you have 2 dm3 of Hydrogen gas, which volume of Oxygen do you need for complete reaction and which volume of water vapour will you get? (Constant: Pressure and Volume)
2 H2(g) O2(g)  2 H2O(g) dm ? ?
Mol Ratio : : Volume Ratio
=>
2 H2(g) + O2(g)  2 H2O(g)
2dm dm dm3

41. Under other conditions
The ideal gas equation
pV=nRT
p = Pressure; Pa
V = Volume; m3 OBS! Usually in chemistry dm3
n = Mole; mol
T = Temperatur; K OBS! K = oC
R = Gas constant; J/mol*K

42. Example
6.0 g Carbon burns in Oxygen. Give the volume of formed Carbon dioxide.
Temperature = 400K
Pressure = 1.0 kPa = Pa
Balanced Equation
Table
Fill in known and ?
Calculate

43. C + O2  CO2 m 6. 0g M 12g/mol n 0. 50 mol  1 : 1  0
C + O2  CO2 m 6.0g M 12g/mol n 0.50 mol  1 : 1  0.50 mol Use the gas law: pV = nRT V = nRT/p = 0.50*8.314*400/ = 0.017m3 V= 17 dm3

44. Solute + Solvent  Solution
1.5 Solutions
Solute Solvent  Solution
= Substances = the liquid
You often have to dissolve chemicals to make them react

45. The composition of Solutions can be given in many ways, e.g.
Mass percent = Mass of substance/Mass of solution
Volume percent = volume of solute/ totale Vol
Mol fraction = Xa = na/(na+nb) 
Molality = moles of solute/kg of solvent
Gram/dm (IB sullabus)
Molarity = moles of solute/ dm3 of solution (IB sullabus)
Concentration in mol/dm3 is often represented by square brackets around the substance under consideration, eg [HCl]

46. Examples
You dissolve 59.5 g of Potassium Bromide in water to 750 cm3. What’s the concentration/molarity?
Write compound formula
Write the table; add volume and concentration
Fill in known and ?

47. KBr m 59,5 g M 119 g/mol n V 0.75 dm3 c ? OBS! Don’t mix up the V in solutions and the V in gas equation. Gas equations relation to the table is the number of moles; n

48. KBr Down = divide m 59,5 g M 119 g/mol n 0. 500 mol V 0. 75 dm3 c 0
KBr Down = divide m 59,5 g M 119 g/mol n mol V 0.75 dm3 c mol/dm3 [KBr] = M (Capital M as a unit is often used instead of mol/dm3, don’t mix up with M for Molar mass) Square brackets = concentration of….

49. More Examples
How many grams of NaCl is it in 200 cm3 of a solution with the concentration of 0.50 mol/dm3?
Write compound formula
Write the table; add volume and concentration
Fill in known and ?

50. NaCl m. Up = multiply M 58. 5 g/mol n V 0. 200 dm3 C 0
NaCl m ? Up = multiply M 58.5 g/mol n V dm3 C 0.50 mol/dm3 ___________________________________________________ m 5.9 g n 0.10 mol

51. More examples
The concentration of chloride ions, if you dissolve 10 g AlCl3 in 0.50 dm3 in water?
Write equation
Write the table
Fill in known and ?

52. AlCl3 + aq  Al3+ (aq) + 3 Cl- (aq)
Salts dissociate in water solutions:
AlCl3 + aq  Al3+ (aq) + 3 Cl- (aq)
m g
M g/mol
n  :  V
c ?

53. AlCl3 + aq  Al3+ + 3 Cl- m 10g OBS! [Cl-] = 0.45 mol/dm3
M g/mol
n mol  :  mol
V 0.5 dm dm3
c 0.15 mol/dm mol/dm3
OBS!
[Cl-] = 0.45 mol/dm3
[AlCl3] = [Al3+] = 0.15 mol/dm3

54. More Examples
You have 0.25 dm3 of NaOH solution with a concentration of M. Which volume of a 0.10 M Sulphuric acid do you need to neutralise the NaOH solution?
The mass of the formed Sodium sulphate?
Write balanced equation
Write the table
Fill in known and ?

55. 2 NaOH + H2SO4  Na2SO4 + 2 H2O M ? M 142 g/MOL n 2:1 1:1
V dm ?
C mol/dm mol/dm3

56. up =multiply down =divide 2 NaOH + H2SO4  Na2SO4 + 2 H2O M 8
up =multiply down =divide 2 NaOH + H2SO4  Na2SO4 + 2 H2O M 8.9g M 142 g/mol n 0.125mol 2:1 mol 1:1 mol V 0.25 dm dm3 C 0.50mol/dm mol/dm3

57. Dilution You have 0.75 dm3 of 0.3 M HCl. You add 0.25 dm3 of water.
The new concentration?
Write equation
Write the table
Fill in known and ?

58. HCl(aq) before  HCl(aq) after m M n 0. 225 mol 1:1 0. 225 mol V 0
HCl(aq) before  HCl(aq) after m M n mol 1:1 mol V 0.75 dm dm3 c 0.3 mol/dm mol/dm3 ( New volume = V = = 1.00 dm3)