Dimensional analysis

Dimensional analysis is quite a simple idea. So we'll start with a fairly simple example:

1 + 1 = 2

This may seem reasonable, but is it actually true? The answer is, it depends on the units. This will only be true if we are using the same units on both sides of the equation. If not, it makes no sense. Returning to money (as peoples' maths always improves here), 1p + 1p = £2. Anyone that thinks it is should send their £2 to me and I'll return your two pennies! We can use this technique to help us check the equations we use in Chemistry:

Example 1

Number of moles (in mol) = concentration (in mol dm^-3) × volume (in dm³)

The minus sign in dm^-3 means per, which means divided by. So mol dm^-3 means moles divided by dm³. Look at the right-hand side of the equation. It says moles divided by dm³ multiplied by dm³. The two dm³ cancel to give mol, which is exactly the same unit as on the left-hand side.

Example 2

As long as you remember the significance of the minus sign with the power, this one is easy. g cm^-3 means grams per centimetre cubed, which means grams divided by centimetres cubed. Exactly the same as on the right-hand side.

Example 3

This heat equation is a useful one for calorimetry experiments:

Heat (in J) = S.H.C (in J g^-1 K^-1) × mass (in g) × temperature change (in K)

Where S.H.C. = specific heat capacity

On the right-hand side we have g and g^-1 (grams divided by grams) - these cancel. We also have K and K^-1 (Kelvin divided by Kelvin) - these also cancel. This only leaves J, which is the same as the left-hand side.

Example 4

A final example from entropy, which can save you some exam marks!

Entropy change, DS, is measured in J mol^-1 K^-1. DH is measured in kJ mol^-1 and temperature, T, is given in Kelvin. Looking at the right-hand side of the equation, we see that we have divided kJ mol^-1 by K. This gives units of kJ mol^-1 K^-1. These are not the normal units of entropy. We should have first converted our DH value in kJ mol^-1 to a value in J mol^-1 (by multiplying by 1000). Dividing by Kelvin gives J mol^-1 K^-1, the usual units. It is possible to give the units of entropy in kJ mol^-1 K^-1. Technically this is not wrong, just a little strange. It's a bit like giving your height in kilometres - not incorrect, but definitely odd!. The metre is a more sensible unit. Also, we cannot add one entropy in kJ mol^-1 K^-1 to another in J mol^-1 K^-1.

Check out your own equations. If the units don't match, your equation is wrong!