Table of Contents
Quantitative Chemistry.................................................. 7
Atomic Structure............................................................. 29
Periodicity........................................................................ 39
Bonding............................................................................ 51
Energetics......................................................................... 67
Kinetics............................................................................. 81
Equilibrium...................................................................... 87
Acids & Bases................................................................... 97
Oxidation & Reduction................................................... 107
Organic Chemistry.......................................................... 119
Measurement & Data Processing ................................ 143
Answers............................................................................. 153
Chapter 1
IB SL Chemistry
Yield: Theoretical, Experimental and Percentage
The theoretical yield is simply the answer to a stoichiometry question – it’s how much product
you are supposed to get at the end of your experiment
The experimental yield is what you actually do get. In most cases the experimental yield is less
than the theoretical yield, but not always.
Reasons for low yield
Reasons for high yield
loss during transfer
insufficient drying
equilibrium / reaction did not go to completion
gain of oxygen (oxidation)
side reactions
side reactions
impure reactants (already reacted somewhat)
Definition
Percentage yield is the ratio of experimental to theoretical expressed as a percent.
Percent Yield =
1.11 Learning
Check
Experimental Yield
× 100%
Theoretica
al Yield
1. A student determines that the theoretical yield of her preparation of aspirin should produce 4.3 g of
product. After drying and weighing her product, she obtained 3.8 g. What was her percent yield?
2. Sodium thiosulphate may be produced by boiling solid sulphur, S8(s)in a solution of sodium nitrite,
Na2SO3(aq), according to the reaction
S (s) + 8Na SO (aq) → 8Na S O (aq)
8
2
3
2 2
3
If a student starts with 15.50 g of sulphur and an excess of Na2SO3, determine the theoretical yield
of product. If only 62.5 g of product is collected, determine the percent yield.
Empirical & Molecular Formulae
A formula is simply the molar ratio of elements in a compound. Water has 2 moles of hydrogen
for every mole of oxygen. This ratio is fixed for a given compound. If you have a compound of 2
moles of hydrogen for 2 moles of oxygen, you don’t have water; you have hydrogen peroxide – a
compound with properties different from water
As long as we know the ratio of the elements in any units, we can convert to moles and we can
determine the formula.
Lots of time these questions show up on Paper 1, and that’s a good thing. Because you can’t have
a calculator, you must be given information in simple multiples of the relative mass.
Example
What is the empirical formula of a compound that contains 46 g of sodium, 64 g of sulphur
and 48 g of oxygen?
If you look at the periodic table, you notice that the molar masses are 23, 32, and 16
respectively. Therefore you must have 2 moles of sodium, 2 moles of sulphur and 3
moles of oxygen. Na2 S2 O3!
Example
Which of the following compounds has the greatest empirical mass?
a) C6H6 b) C20H40
c) C4H10
d) CH3
Solution: (c)-the empirical formulae are CH, CH2, C2 H5 and CH3 respectively
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C.Lumsden
IB SL Chemistry
Quantitative Chemistry
Calculations from percentage information
1.
Assume 100 g of compound – therefore %’s become grams
2.
Convert grams to moles by dividing by the molar mass
3.
Divide by the smallest value to get whole number ratios
Compound Q is analysed and found to contain 85.6% carbon and 14.4% hydrogen. Determine
the empirical formula of Q.
Process
Information
Carbon
Problem solving
steps
Problem solving
Example
steps
Hydrogen
85.6% carbon
14.4%
Assume 100g
85.6 g
14.4 g
Divide by Mr
85.6 g ÷ 12.01= 7.13 mol
14.4 g ÷ 1.01 = 14.26 mol
Divide by smallest
7.13 ÷ 7.13 = 1
14.26 ÷ 7.13 = 1.999
Ratio
1
2
The formula of compound Q is therefore CH2.
Sometimes formulae do not have a ratio against one, for example, Fe2O3. If you do the previous
calculations, and you find that at the end, you have a formula of 1:1.5, then you will have to double all subscripts to attain a whole number ratio of 2:3. Try the following.
1. Compound X is analysed and found to contain 82.63% carbon and 17.37% hydrogen. Determine the
empirical formula of X.
Be Careful
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2. If compound Y contains 89.92% carbon and 10.08% hydrogen. What is the empirical formula of
compound Y?
3. What is the empirical formula of a compound containing 92.24% carbon and 7.76% hydrogen?
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