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Summary AQA Chemistry - Amount of Substance

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I completed the A-Level in one year teaching most of the course to myself, earning an A overall. This is a summary of the textbook chapter with all details you will need for the exam, I found this most useful as it streamlines the information and keeps everything I needed in one place. Comparing ma...

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  • November 8, 2023
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Amount of Substance
2.1 Avogadro’s constant and the mole
2.5 Balanced equations and calculations:
Relative formula mass: the relative molecular mass of
an ionic compound as it doesn’t exist as a molecule Stoichiometry: the ratio in which reactants and products
Avogadro’s constant: the number of atoms in 12g of are produced in simple whole numbers
C12
the mole: the amount of substance that contains
How to balance-
avogadro’s constant (6.022x10^23)
1. Write out the symbol equation for the reaction
2. Start by balancing oxygen, make sure there’s the
2.2 Moles in solutions:
same number of oxygen atoms on each side
3. Repeat for the rest of the elements
Concentration: tells us the amount of solute dissolved in a 4. Count the atoms on each side to make sure it’s
known volume balanced
5. Make sure the charges are also balanced
concentration = the number of moles 6. Add state symbols
volume

For ionic equations-
2.3 The ideal gas equation: 1. Write the balanced symbol equations
2. Then write each aqueous compound as it’s ions eg.
The volume of a gas will change under the effects of NaOH = Na+, OH-
pressure and temperature. Therefore the ideal gas 3. Strike out ions that appear on both sides, the spectator
equation combines all the theories around constant ions
relationships into one equation. 4. The remaining ions/ compounds that aren’t aqueous
form the ionic equations
Pressure x volume = moles x temperature x gas constant
(Pa) (M3) (K) (8.31) Spectator ions: ions in the equation that don’t take part in
the ionic equation

You can rearrange this formula to find any of the
values or the molecular mass of a gas. Once you have 2.6 Atom economy and percentage yield
found moles, you can apply this to a different
formula to find the molecular mass Atom economy: a theoretical idea on how much of a
desired product you should be able to make
% atom economy = mass of desired product. x100
2.4 Empirical and molecular formulae Total mass of reactants

Empirical formula: the simplest whole number ratio of Percentage yield: the practical efficiency of a reaction
atoms of each element in a compound. shows you where reactants may be lost (loss of mass,
Eg. C2H4 = CH2 incomplete reactions)
How to find: Yield= the number of moles of a product X 100
1. Work out the number of moles of each element Theoretical number of moles of a product
2. Divide the other atoms by the smallest number of
moles to obtain a ratio Common exam questions:
3. Multiple by a 2 if the ratio leaves you with decimals/
fractions “Define avogadro’s constant”

Molecular formula: the actual whole number ratio of “Evaluate the reaction if the atom economy is… and the
atoms of each element in a compound. percentage yield is…”
It’s not always the same as the empirical formula so you
must calculate the multiplier to find the molecular “Calculate the empirical formula, how does this differ to the
formula. molecular formula”

Multiplier= Relative molecular mass of compound • ideal gas equation calculations
Relative mass of empirical formula of compound • balanced equation calculations
• titration

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