Notesale: Turn your study into money

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A
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F
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Like the word "dozen" represents the number 12, so
"mole" represents the number 6 x 1023
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It is a very very big number (6 followed by 23 zeros)
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Needless to say, chemists are concerned with atoms, ions, molecules and compounds
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The number is called Avogadro's number
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So, one mole of carbon atoms has a mass of 12 grams
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A
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For example, look up the relative atomic mass of sodium (Na), (the larger number above it in the
periodic table)
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One mole of helium has a mass of 4 g,
One mole of neon has a mass of 20 g,
One mole of magnesium has a mass of 24 g,
One mole of calcium has a mass of 40 g
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As discussed ealier, the number of protons added to the number of neutrons is known as the
relative atomic mass
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00 g
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The very accurate atomic mass scale is based on a specific isotope of carbon, carbon - 12,

12

C=

12
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The strict definition of relative atomic mass (Ar) is that it is equals average mass of all the isotopic
atoms present in the element compared to 1/12th the mass of a carbon-12 atom
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Calculate its relative atomic
mass
Solution: Think of the data based on 100 atoms, so 75 have a mass of 35 and 25 atoms have a
mass of 37
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5
So the relative atomic mass of chlorine is 35
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5
What about elements which exist as molecules or compounds?

Relative Formula Mass of a Compound (R
...
M)
To calculate the mass of one mole of a compound, the number of each type of atom in the
compound is multiplied by that atoms relative atomic mass and all those numbers added together
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f
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m
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The molar mass is
obtained from r
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m, r
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m, or Mr by simply adding g (grams)
If all the individual atomic masses of all the atoms in a formula are added together you have
calculated the relative formula mass (for ionic compounds) or molecular mass (for covalent
elements or compounds)
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can be used for any element or compound
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Anything with at least two atoms requires the term relative formula/molecular mass
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Relative atomic masses, Ar: H = 1, Cl = 35
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5 = 71 respectively
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Relative atomic masses are Ca=40, H=1 and O=16
Solution: Mr = 40 + 2 x (16+1) = 74
Mass from amount:
The key mathematical equation needed here is:
Mass (g) = relative formula mass (g mol-1) x amount (mol)
Using the triangular relationship from above if the mass section is covered over then the amount
multiplied by the relative formula mass gives the mass
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Examples
i
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25 mol of NaCl?
Solution: 0
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5 g mol-1 × 0
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63 g
ii
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026 kg
Amount from Mass
The key mathematical equation needed here is
Amount (mol) = mass (g)/relative formula mass (g mol-1)
Using the triangular relationship from above if the amount section is covered over then the mass
divided by the relative formula mass gives the amount Example calculations
i
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5 g mol-1
= 2 mol
ii
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4 g of Al2(SO4)3?
Solution: 68
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4 g/342 g mol-1
= 0
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Example calculations
What is the molar mass of a compound for which 0
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2 mol
= 210 g mol-1
Every mole of any substance contains the same number of the defined species
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023 x
1023 'defined species' per mole
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in 18g (H2O = 1+1+16 = 18, H = 1; O = 16)
Note
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Molar mass is a term used to describe the mass of one mole i
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the relative
atomic/formula/molecular mass in grams (g)
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Or you could say 2 moles of ammonia is formed from 1 mole of nitrogen
molecules (N2) and 3 moles of hydrogen molecules (H2)
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5
moles of O2 molecules)
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i
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or g of Z = mole of Z x atomic or formula mass of Z
iii
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Example 1: How many moles of potassium ions and bromide ions in 0
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So there will be 0
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Example 2: How many moles of calcium ions and chloride ions in 2
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So there will be 2
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5 moles of calcium ions and 2
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Using the Avogadro Constant, you can actually calculate the number of particles in known quantity
of material
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Moles water = 1/18 = 0
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0556 x 6 x 1023 = 3
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Calculating % Composition (from masses of each element)
Divide the mass of each element by the total mass of the compound and multiply by 100
2
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What is the percentage of oxygen in carbon dioxide gas?
Solution: mass of oxygen in one mole of carbon dioxide gas = 2×16 g = 32 g
mass of one mole of carbon dioxide = 12 g + (2×16 g) = 44 g
percentage of oxygen = (32 g/44 g) × 100 = 72
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Example 2
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A typical question here involves the calculation of the lowest possible whole number ratio of atoms
in an organic molecule, given the percentage elemental composition for that molecule
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3, 3/4 instead of whole
numbers, convert to the fraction and multiply all ratios by the denominator or the fraction

Calculating Empirical Formula (from experimentally determined masses)
multiply the mass of each element (in grams) by 1/molar mass of that element
Continue with steps 3 & 4 from above
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An organic compound contains carbon and hydrogen only in the ratio of 85
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3 % hydrogen
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The empirical formula for the compound is CH2
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Knowing the molar mass of the compound enables the molecular formula of it to be found
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The actual molar mass of the compound in the previous example is 42 g mol-1
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The value of n = 42 g mol-1 / 14 g mol-1 = 3
The molecular formula is (CH2)3 or more properly C3H6
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The relative atomic
masses of the elements (Ar) are given in the tabular format method of solving the problem
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35g of aluminium was heated in oxygen until there was no further gain in weight
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55g
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Note: to get the mass of oxygen reacting, all you have to do is to subtract the mass of metal from
the mass of the oxide formed
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Ratios

Aluminium, Al (Ar

Oxygen, O (Ar = 16)

Comments and tips

1
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55 − 1
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2 g

not the real atom ratio

1
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2

=27)
Reacting mass
mass in g

moles (

/Ar)

/27 = 0
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05

Simplest whole

/0
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05 x 40 = 2

/16 = 0
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075
/0
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5 (then

can now divide by
x 2 smallest ratio number or

= 3) or 0
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24% carbon,
4
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72% chlorine
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Deduce the empirical and molecular formula
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24 g

Chlorine (Ar =
35
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04 g

71
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24

/12 = 2
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02

/2
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04

/1 = 4
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04

/2
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72

/35
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02 mol can now divide by

2
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02 = 1

smallest ratio
number

trial and error
therefore the simplest ratio = empirical formula for the chlorinated hydrocarbon = CH2Cl BUT the
molecular mass is 99, and the empirical formula mass is 49
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5) AND

99

/49
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Underneath an equation you can add the appropriate atomic or formula masses
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It also allows you to predict what mass of products are formed (or to predict what is needed to
make so much of a particular product)
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g
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g
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NOTE: The symbol equation must be correctly balanced to get the right answer!
Example 1:
a
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tonne of copper?
Solution: 2CuO(s) + C(s)→ 2Cu(s) + CO2(g) (atomic masses Cu=64, O=16, C=12)Formula
Mass ratio is 2 x (64+16) + (12) → 2 x (64) + (12 + 2x16)
= Reacting mass ratio 160 + 12 → 128 + 44
(In the calculation, impurities are ignored)
12 of C makes 128 of Cu
Scaling down numerically: mass of carbon needed
= 12 x 16/128 = 1
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How many tones of copper can be made from 640 tones of copper oxide ore?
Solution: 160 of CuO make 128 of Cu (or direct from formula 80 CuO → 64 Cu)
Scaling up numerically: mass copper formed
= 128 x 640/160 = 512 tones
Example 2:
What mass of carbon is required to reduce 20 tonne of iron(II) oxide ore if carbon monoxide
is formed in the process as well as iron? (Atomic masses: Fe = 56, O = 16)
Solution: Reaction equation: Fe2O3 + 3C → 2Fe + 3CO
Formula mass Fe2O3 = (2x56) + (3x16) = 160
160 mass units of iron oxide reacts with 3 x 12 = 36 mass units of carbon
So the reacting mass ratio is 160: 36
So the ratio to solve is 20: x, scaling down,
x = 36 x 20/160 = 4
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Note: Fe2O3 + 3CO → 2Fe + 3CO2 is the other most likely reaction that reduces the iron ore
to iron
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So the volumes have equal moles of separate particles in them
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This is 24dm3 (24 litres) or 24000 cm3, at room temperature and pressure (r
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p
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In this table, N =6
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t
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t
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r
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p = 250C and 1atmosphere pressure
s
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p = 00C and 1 atmosphere pressure
RMM

Molecules

SO2
CO2

1 mole
1 mole

64g
44g

N
N

V dm3
V dm3

NH3

1 mole

17g

N

V dm3

H2

1 mole

2g

N

V dm3

1 mole of any gas always contains the same number of molecules (6
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When gases combine, they do so in small volumes which bear a simple ratio to one another and to
the volume of product if gaseous
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Some handy relationships for substance Z below:
moles Z = mass of Z gas (g) / atomic or formula mass of gas Z (g/mol)

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gas volume of Z = moles of Z x molar volume
moles of Z = gas volume of Z / molar volume
Example 1: What is the volume of 3
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5/2 = 1
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75 x 24 = 42 dm3 (or 42000 cm3)
Example 2: Given the equation MgCO3(s) + H2SO4(aq) → MgSO4(aq) + H2O(l) +CO2(g)
What mass of magnesium carbonate is needed to make 6 dm3 of carbon dioxide at r
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p?
[Ar's: Mg = 24, C = 12, O = 16, H =1 and S = 32]
Answer;
a
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25 mol of gas
From the equation, 1 mole of MgCO3 produces 1 mole of CO2, which occupies a
volume of 24 dm3
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25 moles of MgCO3 is need to make 0
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25 x 84 = 21g
b
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Formula masses: MgCO3 = 84 (from above), CO2 = 12 + 2x16 = 44
MgCO3: CO2 equation ratio is 1 : 1
so 84g of MgCO3 will form 44g of CO2
44g of CO2 will occupy 24dm3
so scaling down, 6 dm3 of CO2 will have a mass of 44 x 8/24 = 11g
if 84g MgCO3 → 44g of CO2, then
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This means the molecule ratio of the equation automatically gives us the gas volumes ratio of
reactants and products, if all the gas volumes are measured at the same temperature and
pressure
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Example 1: HCl(g) + NH3(g) → NH4Cl(s)
Answer: 1 volume of hydrogen chloride will react with 1 volume of ammonia to form solid
ammonium chloride
e
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25cm3 + 25cm3 → products or 400dm3 + 400 dm3 →products (no gas formed)
Example 2: N2(g) + 3H2(g) → 2NH3(g)
Answer: 1 volume of nitrogen reacts with 3 volumes of hydrogen to produce 2 volumes of
ammonia

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g
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What volume of oxygen is required to burn 25cm3 of propane, C3H8
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so actual ratio is 25 : 5x25, so 125cm3 oxygen is needed
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What volume of carbon dioxide is formed if 5dm3 of propane is burned?
heoretical reactant-product volume ratio is C3H8 : CO2 is 1 : 3
so actual ratio is 5 : 3x5, so 15dm3 carbon dioxide is formed
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What volume of air (1/5th oxygen) is required to burn propane at the rate of 2dm3 per
minute in a gas fire?
Theoretical reactant volume ratio is C3H8 : O2 is 1 : 5
so actual ratio is 2 : 5x2, so 10dm3 oxygen per minute is needed,
Therefore, since air is only 1/5th O2, 5 x 10 = 50dm3 of air per minute is required
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So any chemical equation shows in some way the overall chemical change of
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The arrow → means the direction of change from reactants to products
No symbols or numbers are used in word equations
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The word equation is presented to summarise the change of reactants to products
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The equation can be written as;
ZnS(s) + O2(g) → ZnO(s) + SO2(g)
This is an example of a stoichiometric, or normal chemical (or symbol) equation
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Rules on Balancing Symbol equations
1
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write a word equation with appropriate reactants on the left and products on the right
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Writing the correct symbol or formula for each equation component
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g
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a part of the formula
e
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Ca(OH)2 means 1 Ca and 2 OH's (or 2 O's and 2 H's in total)
Numbers before a formula double or treble it etc
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g
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NOTE: If the number is 1 itself, by convention, no number is shown in a formula or before a
formula
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Using numbers if necessary to balance the equation
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If all is correct, then the sum of atoms for each element should be the same on both side of the
equation arrow
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in other words: atoms of products = atoms of reactants
This is a chemical conservation law of atoms and later it may be described as the 'law of
conservation of mass
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the equations are first presented in 'picture' style and then written out fully with state
symbols
c
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Ionic Equations
In many reactions only certain ions change their 'chemical state' but other ions remain in exactly
the same original physical and chemical state
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The ionic equation represents the 'actual' chemical change and omits the spectator ions
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A base can be defined as a proton
acceptor
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Insoluble salt formation: An insoluble salt is made by mixing two solutions of soluble
compounds to form the insoluble compound in a process called 'precipitation'
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a
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silver nitrate + sodium chloride → silver chloride + sodium nitrate

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Lead (II) iodide can be made by mixing lead (II) nitrate solution with potassium iodide
solution
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Calcium carbonate forms on e
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mixing calcium chloride and sodium carbonate solutions
Calcium chloride + sodium carbonate → calcium carbonate + sodium chloride
CaCl2(aq) + Na2CO3 (aq) → CaCO3(s) + 2NaCl(aq)

Ionically : Ca2+(aq) + CO32-(aq) → CaCO3(s)
The spectator ions are Cl- and Na+
Redox reaction analysis:
a
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The sulphate ion SO42-(aq) is called a spectator ion, because it doesn't change in
the reaction and can be omitted from the ionic equation
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b
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Ionic Equation is: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
Zn oxidised by electron loss, H+ reduced by electron gain
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So we need a standard way of comparing the concentrations of solutions
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Note: 1dm3 = 1 litre = 1000ml = 1000 cm3, so dividing cm3/1000 gives dm3, which is handy to
know since most volumetric laboratory apparatus is calibrated in cm3 (or ml), but solution
concentrations are usually quoted in molarity, that is mol/dm3 (mol/litre)
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e
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You need to be able to calculate:
The number of moles or mass of substance in an aqueous solution of given volume and
concentration
The concentration of an aqueous solution given the amount of substance and volume of
water
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molarity of Z = moles of Z / volume in dm3
molarity x formula mass of solute = solute concentration in g/dm3, dividing this by 1000
gives the concentration in g/cm3
(concentration in g/dm3) / formula mass = molarity in mol/dm3,
moles Z = mass Z / formula mass of Z
1 mole = formula mass in grams

Example 1
What mass of sodium hydroxide (NaOH) is needed to make up 500 cm3 (0
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5M
solution? [Ar's: Na = 23, O = 16, H = 1]
1 mole of NaOH = 23 + 16 + 1 = 40g
for 1000 cm3 (1 dm3) of 0
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5 moles NaOH
which is 0
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Example 2
How many moles of H2SO4 are there in 250cm3 of a 0
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8 moles H2SO4
but there is only 250cm3 of solution, so scaling down
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8 x (250/1000) = 0
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2 x 98 = 19
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95g of potassium bromide was dissolved in 400cm3 of water
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[Ar's:
K = 39, Br = 80]
moles = mass / formula mass, (KBr = 39 + 80 = 119)
mol KBr = 5
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05 mol
400 cm3 = 400/1000 = 0
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05/0
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125M
Example 4
What is the concentration of sodium chloride (NaCl) in g/dm3 and g/cm3 in a 1
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Masses: Na = 23, Cl = 35
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5 = 58
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5 x 58
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8 g/dm3, and concentration = 87
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0878 g/cm3
Example 5
A solution of calcium sulphate (CaSO4) contained 0
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Calculate
the concentration in (a) g/dm3, (b) g/cm3 and (c) mol/dm3
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concentration = 0
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25 g/dm3, then since 1dm3 = 1000 cm3
b
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25/1000 = 0
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At
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mass CaSO4 = 40 + 32 + (4 x 16) = 136
moles CaSO4 = 0
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00368 mol
concentration CaSO4 = 0
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00184 mol/dm3

Titration: Acid and Alkali
Titrations can be used to find the concentration of an acid or alkali from the relative volumes used
and the concentration of one of the two reactants
...

1
...

2
...

In most volumetric calculations of this type, you first calculate the known moles of one reactant
from a volume and molarity
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2M
hydrochloric acid
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Calculate the molarity of the sodium hydroxide and concentration in g/dm3
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2 = 0
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003 mol NaOH in 25 cm3
scaling up to 1000 cm3 (1 dm3), there are
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003 x (1000/25) = 0
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12M or mol dm-3
since mass = moles x formula mass, and Mr(NaOH) = 23 + 16 + 1 = 40
concentration in g/dm3 is 0
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41g/dm3
Example 2:
20 cm3 of a sulphuric acid solution was titrated with 0
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If the acid
required 36 cm3 of the alkali KOH for neutralisation what was the concentration of the acid?
equation 2KOH(aq) + H2SO4(aq) →K2SO4 + 2H2O(l)
mol KOH = 0
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0018 mol
mol H2SO4 = mol KOH / 2 (because of 1 : 2 ratio in equation above)
mol H2SO4 = 0
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0009 (in 20 cm3)
scaling up to 1000 cm3 of solution = 0
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045 mol
mol H2SO4 in 1 dm3 = 0
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045M or mol dm-3
since mass = moles x formula mass, and Mr(H2SO4) = 2 + 32 + (4x16) = 98
Concentration in g/dm3 is 0
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41g/dm3

How to carry out a titration
The diagrams show the typical apparatus (1)-(6) used in manipulating liquids and on the left a brief
three stage description of titrating an acid with an alkali:

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The operation is an example of volumetric (titrimetric) analysis
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Volumetric analysis is a powerful technique, which is used in a variety of ways by chemists in many
different fields
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The description should include
which reagent is placed in the burette, name of indicator (but no reason for choice of indicator),
detection of endpoint and what should be observed, and repetition for accuracy
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The definite amount of material is measured by weighing, and the definite volume of solution
prepared in a volumetric flask
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Tip the solid from a weighing bottle into a large (250 cm3) beaker and add about 50 cm3 of distilled
water from a wash bottle
...

Take great care not to lose any of the solution and remember to wash the solution off the stirring
rod back into the beaker
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Now fill the flask to within about 1 cm of the calibration mark on the neck
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Stopper the flask and invert a number of times to thoroughly mix the contents
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Before use, a pipette must be washed out with the solution it is to measure
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Let the solution down until the bottom of the meniscus just touches the calibration mark
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Allow the pipette to drain for about 20 seconds, then touch the tip to the surface of the liquid in the
conical flask
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Use of a Burette
The burette is designed to deliver definite but variable volumes of liquid
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Clamp the burette vertically in a stand
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Fill the burette carefully using a beaker and a filter funnel
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Read the burette scale by observing
the position of the bottom of the liquid meniscus, making sure your eyes are level with the
graduation mark
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Record the volume reading to the nearest 0
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Indicator
Colour in Acid
Neutral color

Colour in alkali

Litmus

Red

Purple

Blue

Phenophthalein

Colourless

Colourless

Pink

Methyl orange

Red

Orange

Yellow

Universal indicator

Red

Green

Blue/Violet

Bromothymol blue

Yellow

Green

Blue

Titration Technique
When performing a titration, place the conical flask containing the aliquot on a white tile under the
burette so that the tip of the burette is inside the mouth of the conical flask to avoid splashing
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Add a few drops of a suitable indicator to the solution in the conical flask
...

When the end-point is reached, as shown by the indicator changing colour, quickly close the tap
...

Now repeat the titration with a fresh aliquot
...

Record the volume
...


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05 cm3 (approximately 1 drop)
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10 cm3 and, strictly, you should repeat the
titrations until this is achieved
...

With practice, your technique should improve so that you should not need to do more than 4
titrations (1 trial + 3 accurate)
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05 cm3
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CT

1 mark

D

1 mark

AC

1 mark

PA

1 mark

FA

1 mark

Total

5 marks

Specimen titration Table II

Calculating the Concentration of a Solution from Titration Data
When you have finished this section you should be able to:
Calculate the concentration of a solution from titration data and the balanced equation
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