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# How I Think About Ratio Problems

This is a slight diversion from the normal blogs about programming to some simple mathematics. The hope is that by describing the way I think about the type of problem below may help other people who are struggling and think in a similar way to me.

This blog was inspired by a friend of mine who is working on her GCSE Chemistry, and asked for some explanation of a problem.

The problem was really two problems, and I have split them a little to make life easier.

Problem part A: I have $50g$ of solution $20\%$ of the weight of this solution is Iron ions How many grams of iron is in the solution?

Problem part B: The iron ions in the solution are all from iron chloride ($FeCl_2$) what is the weight of Iron Chloride in the solution?

My approach to both of these problems is the same (and I think fairly unusual.

#### Part A:

$20\%$ iron means that if I have $100$ of the thing (the unit doesn’t matter) then $20$ of it is iron and $80$ is other stuff. A way of writing this is:

$$ 20_{Fe} + 80_{stuff} = 100_{solution}$$

However, I don’t always have 100 of the solution so I need to be able to change the equation to deal with this, to do this I’ll multiply both sides by a number $a$ giving:

$$a \times (20_{Fe} + 80_{stuff}) = a \times 100_{solution}$$

Now I can choose $a$ to turn $100$ into $50$ (in this case it’s easy $a$ is a half). This gives the formula:

$$0.5 \times (20_{Fe} + 80_{stuff}) = 50_{solution}

Multiplying out gives:

$$10_{Fe} + 40_{stuff} = 50_{solution}$$

So if we have $50g$ of solution we have $10g$ of Iron.

#### Part B:

To do this section we need to know how much iron weighs compared to iron chloride. This is exactly what the atomic weight numbers on the periodic table are for. Iron has an atomic weight of $56$, and chlorine has an atomic weight of $35.5$. This means that Iron Chloride has atomic weight described by:

$$ 56_{Fe} + (2 * 35.5)_{Cl} = 127_{FeCl_2}$$

This works just like the formula in part a e.g., if we have $127g$ of $FeCl_2$ then there are $56g$ of iron and $71g$ of chlorine. To make this adjustable we need a multiplier again:

$$ a \times (56_{Fe} + 71_{Cl}) = a \times 127_{FeCl_2} $$

This time we know how much iron we want (we want $10$) so we need to choose $a$ to turn $56$ into $10$. To do this we can go through $1$, if we divide $56$ by $56$ then we get $1$, we can then multiply $1$ by $10$ to get ten. e.g. $a = \frac{10}{56}$.

Applying this to the above formula gives:

$$\frac{10}{56} \times (56_{Fe} + 71_{Cl}) = \frac{10}{56} \times 127_{FeCl_2}$$

Multiplying out gives:

$$ 10_{Fe} + 12.7_{Cl} = 22.7_{FeCl_2}$$

So given we know we have $10g$ of iron we must have 22.7g of iron chloride.

## Comments on this problem

Whilst I’m quite happy with the mathematics of this problem I’m less convinced about how often a problem like this will really occur. It seems much more likely in an experiment that the amount of iron chloride in a solution is know, and you may want to calculate the amount of iron in the solution. The again what would I know. I’m not a chemist!