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1. Stoichiometric relations
By
Dr. Rania S. Seoudi
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2. Part1- Introduction to the particulate nature
of matter and chemical change
2

3. a- The particulate nature of matter
Particles vibrate about
mean position
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Melting
Freezing
Boiling
Condensing
Sublimation
Deposition
The three states of matter
Particles moving
around each other
Particles moving at high
speeds in all directions

4. Comparison between three states of matter
4

5. Providing system with energy
5
Melting: increasing temp.
of solid provide particles
energy to overcome
attraction force. During
melting both solid &
liquid exist together
Freezing: decrease
the energy of the
system by removing
energy
Boiling: increasing temp.
of liquid provide particles
of energy to break
attraction force between
liquid particles. During
boiling both liquid & gas
exist.
Condensing

6. Working principle of refrigerator
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7. Matter
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Elements: one type of atoms Compounds: more than one type of atoms

8. b- chemical change
Elements and compounds:
An element: is a pure substance that contains only one type of atom.
An atom: is the smallest part of an element that can still be recognized as that
element.
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9. Physical and chemical properties
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Chemical combination: Chemical
change
Chemical properties dictate how
something reacts in a chemical
reaction.
Mixture: Physical combination
(physical change)
Physical properties are basically all
the other properties of a substance –
such as melting point, density,
hardness, electrical conductivity etc.

10. The meaning of chemical change
Combination of elements at fixed ratios
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11. Balancing equations
- Chemical reaction: a process in which a substance (or substances ) is changed
into one or more new substances
- Chemist present this changes into chemical equation using symbols
- Reactant → product
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12. Balancing chemical equation
1. Identify all reactants and products and write their
correct formulas on the left side and right side of the equation, respectively.
2. Begin balancing the equation by trying different coefficients to make the number
of atoms of each element the same on both sides of the equation.
3. First, look for elements that appear only once on each side of the equation with
the same number of atoms on each side. Next, look for elements that appear only
once on each side of the equation but in unequal numbers of atoms. Balance these
elements. Finally, balance elements that appear in two or more formulas on the
same side of the equation.
4. Check your balanced equation to be sure that you have the same total number of
each type of atoms on both sides of the equation arrow.
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13. Homogeneous: mixture has the same
(uniform) composition throughout the
mixture and consists of only one
phase. Example sea water.
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Heterogeneous mixture does not have uniform
composition and consists of separate phases.
Heterogeneous mixtures can be separated by
mechanical means.
Mixture

14. Part2- The mole concept
1- Relative masses:
Materials react with each other in certain ratios.
a- Relative atomic masses (Ar) in grams: the average of the masses of the
isotopes in a naturally occurring sample of the element relative to the mass of
1/12 of an atom of carbon-12.
E.g:
silver atomic mass is 107.87 that is equal to relative atomic mass of 107Ag and
109Ag
14

15. b- Relative molecular mass (Mr) in grams: is the sum of the relative atomic
masses of the individual atoms making up a molecule.
Example 1: relative molecular mass of methane (CH4)=
12.01 (Ar of C) + (4x1.01 (Ar of H)) = 16.05g
Example 2:
Relative atomic mass of ethanoic acid or acetic acid (CH3COOH)=
12.01 + (3x1.01)+ 12.01 +(2x16) +1.01= 60.06g
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16. C- Relative formula mass in grams: mass of one formula unit relative to the
mass of 1/12 of an atom of carbon-12 (ions or molecules).
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17. Moles
Mole: is the amount of substance that contains the same number of particles
(atoms, ions, molecules, etc.) as there are carbon atoms in 12 g of carbon-12.
This number is called Avogadro’s constant “L” (or NA) .
Avogadro’s no. has the value 6.02 × 1023 mol−1. So, 12.00 g of carbon-12
contains 6.02 × 1023 carbon atoms.
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18. Number of moles
- The meaning of average atomic mass:
The Ar of oxygen is 16.00, which means that, on average, each oxygen atom is 16
12 times as heavy as a carbon-12 atom. Therefore 16 g of oxygen atoms must
contain the same number of atoms as 12 g of carbon-12, i.e. one mole, or 6.02 ×
1023 atoms.
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Number of moles (n)=
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒖𝒃𝒔𝒕𝒂𝒄𝒆
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔

19. The mass of a molecule
Mass of one mole = the sum of grams of all atoms of the molecule.
Example: the mass of one mole of water H2O=
(2x1.01) + 16 = 18.02grams = 6.02x1023molecules of water.
This means; mass of one molecule of water =
18.02
6.02𝑥 1023 = 2.99x10-23g
Remember: mass of molecule is very small compared to mass of one mole which
is greater than 1.
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mass of one molecule =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′ 𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

20. The number of particles
One mole of O2 contains 6.02 × 1023 O2 molecules.
O2 contains two atoms so no of atoms = 2x 6.02 × 1023 = 1.204 × 1024 atoms
One mole of O2 contains two atoms
one mole of O2 contains 6.02 × 1023 O2 molecules
one mole of O2 contains 1.204 × 1024 O atoms
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21. Example:
0.1 mol of H2O contains 0.1x 6.02x 1023 H2O molecules
H2O contains two hydrogen atoms so 0.1mole H2O= 0.1x 2x 6.02x 1023 =
1.204x1023 Hydrogen atoms.
H2O contains one oxygen atom so 0.1mol H2O= 0.1 x 1x 6.02x 1023 = 6.02x1022
oxygen atoms.
0.1 H2O contains 3atoms= 0.1x 3x 6.02x 1023 = 1.0806x 1023 particles
0.3mol atoms
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22. 22

23. 2- Empirical and molecular formulas
a- Percentage composition of a compound:
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% by mas of an element =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑥 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠
𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠

24. b- Empirical and molecular formulas
Empirical formula: the simplest whole number ratio of the elements present in a
compound.
Molecular formula: the total number of atoms of each element present in a
molecule of the compound. (The molecular formula is a multiple of the empirical
formula.)
A molecular formula is a whole number multiple of the empirical formula.
Meaning: the molecular formula C2H4
has empirical formula (CH2)n where n=2
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25. Part3- Reacting masses and volumes
1- Calculations involving moles and masses
Mass is conserved for any reaction:
For 55.85g Fe reacts with 32.06g S to produce 87.91g FeS
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- Calculated yield predicts the experimental yield or the % of the yield compared
to the expected (efficiency of a reaction).
% yield =
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑡ℎ𝑒𝑜𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
x100

26. 26
Formula Formula for solving moles questions involving masses
An alternative way of doing these questions is to use a formula.
𝒎 𝟏
𝒏 𝟏 𝑴 𝟏
=
𝒎 𝟐
𝒏 𝟐 𝑴 𝟐
Where:
m1 = mass of first substance
n1 = coefficient of first substance (number in front in the chemical
equation)
M1 = molar mass of first substance

27. Calculating the actual yield and % reaction yield
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% yield=
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑥100

28. Limiting reactant
Limiting reactant: the reactant that is used up before other reagent in a reaction.
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29. 2- calculations involving volumes of gases
I- Real gases and ideal gases
- Ideal gas: model for the behaviour of real gases.
- Real gas: a- molecules have no volumes.
b- no force exit between molecules (except when they collide).
- Real gases deviate from ideal gases because of attractions between gases
molecules. Real gas deviate at very high pressure or very low temperature.
- Volume of gas depend on the number of molecules (Vα no. of moles of gas) not
nature of gas.
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30. II- Using volumes of gases
Avogadro’s law: equal volumes of ideal gases measured at the same temperature and
pressure contain the same number of molecules.
Means:
Number of molecules of (100cm3 H2 at 25oC and 100kPa) =
Number of molecules of (100cm3 NH3 at 25oC and 100kPa).
This means:
Volume can be used instead of moles
One mole of H2 reacts with one mole of Cl2 or one volume of H2 reacts with one
volume of Cl2
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31. III- Converting volumes of gases to
number of moles
STP = standard temperature and pressure = 273K and 100kPa (1bar)
100kPa = 1.00 x 105Pa
- Volume of 1Mole of gas under certain conditions is molar volume
- Molar volume: of an ideal gas at
- STP= 22.7dm3mol-1 or 2.27x10-2m3mol-1
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Number of moles=
𝑣𝑜𝑙𝑢𝑚𝑒
𝑚𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒
𝑥100

32. Kelvin absolute zero
Kelvin absolute zero is the lowest possible temperature at which everything has
the lowest energy state.
Absolute zero = -273oC
-273OC (celsius) = 0K
- Converting Kelvin to Celsius: Add 273
i.e: 25oC = 273 +25= 298K
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33. IV- Conversions between volume units:
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34. V- formula for solving moles questions
involving volumes of gases
where:
m1 = mass of first substance (in g)
n1 = coefficient of first substance
M1 = molar mass of first substance
V1 = volume of first substance if it is a gas
V2 = volume (in dm3) of second substance if it is a gas
n2 = coefficient of second substance
Mv = molar volume of a gas = 22.7 dm3 at STP
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𝑉1
𝑛1
=
𝑉2
𝑛2
𝑚
𝑀
= V
𝑚1
𝑛1 𝑀1
=
𝑉2
𝑛2 𝑀 𝑣

35. VI- Macroscopic properties of ideal gases
1- Relationship between pressure and volume
(Boyle’ law)
- At a constant temperature, the volume of a fixed mass of an ideal
gas is inversely proportional to its pressure.
- This means if pressure is doubled at constant
Temp. then the volume will be halved.
-
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The relationship between pressure and
volume of a fixed mass of an ideal gas at
constant temperature.
Pα
𝑘
𝑉
P=
𝑘
𝑉
PV=k

36. 36

37. 2- The relationship between volume and
temperature (Charles’ law)
- The volume of a fixed mass of an ideal gas at constant pressure is
directly proportional to its kelvin temperature.
- This means that if temp. in kelvin is doubled at
constant pressure then the volume will be doubled.
- Remember: An ideal gas can never liquefy
because there are no forces between the molecules.
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Vα𝑇

38. 38

39. 3- The relationship between pressure and
temperature
Gay-Lussac’s Law
- For a fixed mass of an ideal gas at constant volume, the pressure is
directly proportional to its absolute temperature:
- This means: If the temperature (in Kelvin) is
doubled at constant volume then the pressure
will be doubled.
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Pα𝑇

40. 4- The overall gas law equation
- An ideal gas is one that obeys the three gas laws exactly.
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𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
P and V can be in any units
T must be in Kelvin

41. 5- Ideal gas equation
- R is gas constant = 8.31JK-1mol-1 when P in Nm-2 or Pa,
V in m3 and T in K
- Remember: dm3= (1/1000) m3
- (Gay-Lussac’s Law: at constant volume the pressure is directly proportional to
absolute temperature (means temp. in Kelvin).
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PV= nRT

42. Gas Laws
Bolyle’s law: Vα 1/P (at constant n and T)
Charles’ law: V α T ( at constant n and P)
Avogadro’s law: V α n ( at constant P and T).
Gay- Lussac’s law: Pα T (at constant V and n)

43. 3- Calculations involving solutions
Solution
Definitions involving solutions:
Solute: a substance that is dissolved in another substance.
Solvent: a substance that dissolves another substance (the solute).
The solvent should be present in excess of the solute.
Solution: the substance that is formed when a solute dissolves in
a solvent.
e.g. sodium chloride solution: sodium chloride is solid (solute), water is the
solvent, and the mixture is the solution
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44. Note: the volume of solution depend on the attraction forces between solute and
solvent not the addition of the two volumes
44

45. Concentration
Concentration: the amount of solute dissolved in a unit volume of solution.
Volume of solvent expressed in : dm3
solute unit: g or mol
Concentration unit: gdm-3 or mole dm-3
Concentration expressed as M
Concentration (mol dm-3) =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 (𝑚𝑜𝑙)
𝑣𝑜𝑙𝑢𝑚𝑒 (𝑑𝑚3
)
Or concentration (gdm-3)=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 (𝑚𝑜𝑙)
𝑣𝑜𝑙𝑢𝑚𝑒 (𝑑𝑚3
)
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46. Concentration of very dilute solutions
Ppm= 1g of solute in 1million grams of solution.
Concentration =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑥 106
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Ppb= 1g of solute in 1 billion grams of solution (109)
Ppt= 1g of solute in 1 billion grams of solution (1012)
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47. Titrations
Titration is a technique for finding the volumes of solutions that react exactly
with each other. One solution is added from a burette to the other solution in a
conical flask
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48. Equation for solving moles questions
involving solutions
𝑐1 𝑣1
𝑛1
=
𝑐2 𝑣2
𝑛2
where:
c1 = concentration of first substance
v1 = volume of first substance
n1 = coefficient of first substance
c2 = concentration of second substance
v2 = volume of second substance
n2 = coefficient of second substance
48

49. Water of crystallisation
Water of crystallisation: substances crystallise with water as an integral part of
the crystal lattice.
e.g. hydrated copper sulfate (CuSO4·5H2O) and hydrated magnesium chloride
(MgCl2·6H2O).
49

50. Back titration
50

51. Linked reactions
Linked reaction is a compound reaction:
Product of first reaction further react:
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52. Scientific law
Scientific law is a general statement (mathematical form) involves the relation
between various quantities.
Theory
Theory: way of explaining the scientific law.
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Further information