The ability to lay out calculations in a comprehensible manner depends on

- knowing what you are doing and NOT being reliant on rote formulae;
- realising that calculations need linking words and phrases to make them readable;
- a correct use of units
*throughout the calculation*.

Teachers I have spoken to often like my list
less the further they go down it. The last item in particular raises comments such as 'my
students have enough problems without bringing in this idea.' My response is that the
problems can arise because they *don't* use this idea! Like most worthwhile things it
involves an initial effort that makes succeeding work much easier, so overall needs less
work to achieve a given end.

The advantages of using units throughout include:

- an awareness that equations are not merely symbols, but express relationships between physical quantities;
- an check on whether the equations used are in fact right, because of the in-built check on the units of the answer;
- a gradual awareness of what sort of magnitude of answer is reasonable in a given set of circumstances.

Here is a simple titration example to show what I mean, easily extended to other types of calculation.

25.0 cm^{3} of a solution of sodium carbonate was titrated with 0.108 mol dm^{-3}
hydrochloric acid solution using methyl orange indicator. The volume needed was 27.2 cm^{3}.

Find the concentration of the sodium carbonate solution in mol dm^{-3}, and in
g dm^{-3} of the anhydrous salt.

The reaction is: Na_{2}CO_{3}
+ 2HCl à 2NaCl + H_{2}O + CO_{2}

Amount of hydrochloric acid used = 0.0272 dm^{3}
× 0.108 mol dm^{-3}

= 0.00294 mol

Thus amount of sodium carbonate = ½ x
0.00294 mol = 0.00147 mol in 25.0 cm^{3}

Thus concentration of sodium carbonate solution = 0.00147 mol /0.025
dm^{3 }

= 0.0588 mol dm^{-3}.

The molar mass of anhydrous sodium carbonate is

{(2×23) + 12 + (3×16)}g mol^{-1} = 106 g mol^{-1};

The concentration of the sodium carbonate is therefore

0.0588 mol dm^{-3} × 106 g mol^{-1} = **6.233 g dm ^{-3}**.

Note the following, which I promise you will increase your understanding and your marks if you do them too:

- I have used the units throughout;
- I have used the word
*amount*in its technical chemical sense, that is a quantity of moles; - I have calculated the molar mass of sodium carbonate by writing out each atomic mass explicitly - you'd be amazed how many marks disappear because molar masses are wrong but the examiner can't tell whether it's due to arithmetic or chemistry (you could, of course, in your everyday work look the value up in a Data Book, in which case say you have done so);
- I have not used the dimensionless quantity
*relative*molecular mass. I cannot think of any calculation in the A level course that does so!

You might think that this is quite a performance for a relatively easy calculation. Part of your apprenticeship is, however, learning to solve problems using in-built checks whilst also realising what you're doing. This method achieves both things.