Unit Analysis

A convenient method for many calculations is unit analysis, which relies on cancellation of units and conversion ratios to perform calculations. You’ve already done unit analysis in calculations such as:
$$ 2\: \text{m}\, \times \frac{100\, \text{cm}}{1\, \text{m}} = 200\, \text{cm}$$

The key considerations when performing calculations with unit analysis are:

1. Units cancel across ratios

The central premise of unit analysis is that units will cancel across a ratio (i.e. on the top and bottom of a fraction). When setting up your calculation or choosing your ratios, make sure to check that your intermediate units will cancel out, leaving you with your desired unit at the end. (Remember that for values that aren’t already fractions, you can imagine them as $\frac{\text{value}}{1}$ if you need!)

$$ \require{cancel} 3.25\, \cancel{\text{g}\, H_2 O} \times \frac{1\, \text{mol}\, H_2 O}{18.02\,\cancel{\text{g}\, H_2 O}} = 0.180\, \text{mol}\, H_2 O$$


$$3.25\, {\text{g}\, H_2 O} \times \frac{18.02\,{\text{g}\, H_2 O}} {1\, \text{mol}\, H_2 O}= 58.6\, \frac{g^2}{\text{mol}}\, H_2 O\qquad (?!?!?)$$

2. Show ALL your units (including the “of what”)

In longer calculations, it’s easy to get mixed around and cancel units that look the same but actually aren’t. You can avoid this by making sure to label all your units all the time – including the “of what” part – not just “g” but “g H2O“.

For example, if you wanted to calculate the mass of Al that could react with 25 mL of 0.10 M HCl according to the reaction $$2\, Al\, \text{(s)}\, + \, 6\, HCl\, \text{(aq)}\, \rightarrow \, 2\, AlCl_3 \, \text{(aq)}\, +\, 3\, H_2 \, \text{(g)}\, $$ the following calculation appears to be correct: $$ \require{cancel} 0.025\,\text{L} \times \frac{0.1\, \text{mol}}{\text{L}} \times \frac{6\,\text{mol}}{2\,\text{mol}} \times \frac{36.46\, \text{g}}{1\,\text{mol}} \\ 0.025\,\cancel{\text{L}} \times \frac{0.1\, \bcancel{\text{mol}}}{\cancel{\text{L}}} \times \frac{6\,\cancel{\text{mol}}}{2\,\bcancel{\text{mol}}} \times \frac{36.46\, \text{g}}{1\,\cancel{\text{mol}}} =0.27\,\text{g} $$

However, this is not the correct answer! Filling in the “of what” in the units reveals: $$\require{cancel} 0.025\,\text{L}\, HCl \times \frac{0.1\, \text{mol}\, HCl}{\text{L}\, HCl} \times \frac{6\,\text{mol}\, HCl}{2\,\text{mol}\, Al} \times \frac{36.46\, \text{g} \,HCl}{1\,\text{mol}\, HCl} \\ 0.025\,\cancel{\text{L}\, HCl} \times \frac{0.1\, \text{mol}\, HCl}{\cancel{\text{L}\, HCl}} \times \frac{6\,\cancel{\text{mol}\, HCl}}{2\,\text{mol}\, Al} \times \frac{36.46\, \text{g} \,HCl}{1\,\cancel{\text{mol}\, HCl}} = 0.27\, \frac{(\text{mol}\, HCl)(\text{g}\,HCl)}{(\text{mol}\, Al)}$$

Hopefully it is obvious that the “units” of the final answer in this calculation don’t make sense … that is our clue that something went wrong in the calculation. In this case there’s two errors: the mole ratio is inverted and the wrong molecular mass was used (HCl instead of Al). When we fix the errors, the answer looks much better: $$\require{cancel} 0.025\,\text{L}\, HCl \times \frac{0.1\, \text{mol}\, HCl}{\text{L}\, HCl} \times \frac{2\,\text{mol}\, Al}{6\,\text{mol}\, HCl} \times \frac{26.98\, \text{g} \,Al}{1\,\text{mol}\, Al} \\0.025\,\cancel{\text{L}\, HCl} \times \frac{0.1\, \bcancel{\text{mol}\, HCl}}{\cancel{\text{L}\, HCl}} \times \frac{2\,\cancel{\text{mol}\, Al}}{6\,\bcancel{\text{mol}\, HCl}} \times \frac{26.98\, \text{g} \,Al}{1\,\cancel{\text{mol}\, Al}} = 0.022\, \text{g}\,Al$$

Writing out calculations with the full units can take up a little more space, but the built in error checking is guaranteed to save you at least once on an exam or lab report!