Formulae, Equations and Amount of Substance
Equations, Yield and Atom Economy
Pearson Edexcel International A Level Chemistry
Balancing equations and state symbols
- Balance one element at a time, changing only the numbers in front of the formulae and never the formulae themselves.
- Every equation carries a state symbol on every formula.
| Substance | State symbol |
|---|---|
| solid, including a precipitate | (s) |
| liquid, including water | (l) |
| gas | (g) |
| dissolved in water | (aq) |
Ionic equations
Method: an ionic equation
- Write the full balanced equation with state symbols.
- Split every aqueous ionic compound into its ions.
- Cancel the spectator ions, the ions that appear unchanged on both sides.
| Reaction | Ionic equation |
|---|---|
| displacement of a metal | Mg(s) + Fe2+(aq) → Fe(s) + Mg2+(aq) |
| displacement of a halogen | 2I−(aq) + Br2(aq) → I2(aq) + 2Br−(aq) |
| precipitation | Ba2+(aq) + SO42−(aq) → BaSO4(s) |
| acid with a base | MgO(s) + 2H+(aq) → Mg2+(aq) + H2O(l) |
| acid with a carbonate | CO32−(aq) + 2H+(aq) → CO2(g) + H2O(l) |
Never write an ionic equation with uncancelled spectator ions.
Observations
| Reaction | Observations |
|---|---|
| metal with an acid | effervescence; the magnesium dissolves |
| copper in silver nitrate solution | colourless to blue solution; silver solid |
| precipitation | the colour of the precipitate, such as white |
- Never write a white precipitate when a metal reacts with an acid.
- Name the colour. Never write just the solution changes colour.
Reacting masses
Method: the mass of one substance from the mass of another
- Amount in moles of the substance you are given, from its mass and its Mr.
- Multiply by the ratio of the balancing numbers, to reach the amount in moles of the substance you want.
- Mass = amount × Mr, using the Mr of the substance you want.
| Quantity | Symbol | Unit |
|---|---|---|
| mass | m | g |
| amount | n | mol |
| molar mass | M | g mol−1 |
The limiting reagent
Method: finding the limiting reagent
- Amount in moles of every reactant.
- Divide each amount by the balancing number in front of that reactant in the equation.
- The smallest answer is the limiting reagent. Every other reactant is in excess.
- Every step after this one uses the amount of the limiting reagent only.
One reactant is added in excess to ensure all the other reactant is used up.
Percentage yield
percentage yield = mass of product obtained ÷ theoretical mass
× 100
- The theoretical mass is the mass of product the balanced equation gives from the starting material at 100 per cent yield.
- A percentage yield is always below 100 per cent: write reaction incomplete, transfer losses or side reactions. Never write spillages as the reason for a low yield.
Method: percentage yield
- Amount in moles of the starting material, from its mass and its Mr.
- Multiply by the ratio in the equation, to reach the expected amount of the product.
- Theoretical mass = expected amount × the Mr of the product.
- Divide the mass of product obtained by the theoretical mass, then multiply by 100.
Masses of two different substances
Never write the mass of product divided by the mass of the starting material. One gram of the starting material does not make one gram of product, so the two masses cannot be compared directly: convert through moles to the theoretical mass of the product, and divide by that.
Percentage atom economy
percentage atom economy = molar mass of the desired product ÷
sum of the molar masses of all products × 100
Method: percentage atom economy
- Balance the equation, and keep every balancing number.
- On the top, the molar mass of the desired product, multiplied by its balancing number.
- On the bottom, add up the molar mass of every product, each one multiplied by its balancing number.
- Divide, and multiply by 100.
- Give the atom economy as a percentage. Never write the atom economy left as a decimal fraction.
- A route with fewer waste products has a lower total molar mass of products, so a higher atom economy.
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