Section 1 e) Summary

This topic is about chemical formulae and equations, how to write and balance them correctly, as well as how to calculate reacting masses.

In a chemical reaction, the number and types of particles should be the same on both sides. But as different molecules require different numbers of each element, the ratio of each molecule created is relative. We must balance the equation to ensure the number of particles, for example:

H2 + O2 --> H2O
The above equation is unbalanced. There are two particles of hydrogen, and two particles of oxygen on the reactants side, but on the products side there are two particles of hydrogen and only one particle of oxygen.
We can remedy this by first changing the particle with the deficit:
H2 + O2 --> 2H2O
Now there are two oxygen particles on either side of the equation. But there are also double the number of hydrogen particles in the products than in the reactants, but this is easily fixed:
2H2 +O2 --> 2H2O
Now there are 4 hydrogen and 2 oxygen on both sides of the equation: it is balanced.

We can use balanced chemical equations to figure out the reacting masses, as it gives us a ratio of the moles, which when multiplied by Mr gives the mass, as shown in this formula triangle:

For example, if we have this balanced equation:

Fe2O3 + 3CO --> 2Fe + 3CO2

And we are told that there is 75 grams of iron oxide, and that we have to calculate what mass of iron can be derived from this with excess carbon monoxide, it can be calculated as follows:

Mr of Iron Oxide: 160

Mass of Iron Oxide: 75 g

75 / 160 = 15/32 mol

Ratio of Fe2O3 : Fe is 1:2

15/32 x 2 = 15/16 mol

Mr of Fe: 56 

56 x 15/16 = 52.5g

52.5 grams of iron can be derived from this reaction 

However, the actual iron we get from this reaction in reality is lower. This is because some may be lost in separation etc., this is called the yield. 

Percentage yield is how much is actually collected of the amount that is expected, and can be calculated with the following equation:

Yield = (mass of product collected x 100) / mass of product calculated

This tells us what percentage of the product is lost in this method, and can be used to compare different methods, as well as give a more accurate prediction of what the actual yield will be. 

For this topic, it is important to know the specifics of two different reactions, which simply demonstrate that you know how to calculate reacting masses and percentage yield:

Metal Oxide
Weigh a crucible with its lid, then place magnesium inside and weigh again, recording the measurements.
Heat it strongly over a roaring bunsen flame, and gently open the lid with tongs momentarily, to allow oxygen to get in without letting the magnesium oxide escape.
Weigh it every so often so you can see how the mass is changing. When the mass stops changing, the reaction is complete.
Record the final mass of the magnesium oxide and crucible, and use the numbers you recorded earlier to determine the mass of magnesium and the mass of oxygen.
Using each element's relative atomic mass, the empirical formula can be determined.

Salt (crystallisation)
Weigh an evaporating basin, then add hydrated copper sulfate and weigh again.
Heat the solution over a bunsen burner, gently stirring to ensure even heating.
Stir until the solution loses colour, indicating all water has been lost.
Weigh again to find the mass of water lost, and also the mass of anhydrous copper sulfate.
Divide mass by Mr of the copper sulfate, and do the same with water.
Simplify this to find the ratio of water to copper sulfate, and round to the nearest whole number,
e.g. CuSO4 ● 5H2O