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- Practice Balancing Eqns
- Types of Chemical Reactions
- Solution Concentrations
- Calculate RMM
- Uses of transition metals
- Acids and Bases
- Strong + Weak Acids
- Properties of transition metals
- Le Chatelier's Principle
- Main science page
Solutions and Concentration Calculations
This is a common sort of chemistry exam or homework question to come across,
and you are usually required to calculate something like how many grams
or moles of a chemical are in a given volume of solution or what volume
of solution will you need to have a certain amount of some chemical. They
are an important type of chemical calculation because when carrying out
reactions, you need to know exactly how much of each chemical is being used
if you are working with solutions. Many other types of calculations in chemistry
involving solutions depend on getting these ones right.
I hope you find this chemistry tutorial very useful, and try some of those in the menu as well.
How to do these chemical calculations
The method to solve these chemistry problems is sometimes presented in a magical maths formula. But some students have problems using this magical formula as they are not sure what each bit of it means. So here's a straightforward way to do the sums. There are several versions of these questions, but once you learn how each type is done, it gets easier.
Remember that concentration is the number of moles of a chemical dissolved in 1 dm3,
and as you are not always given that, it is moles divided by volume measured in dm3
Type One - given a number of grams of a chemical present in a certain volume of solution, what volume of solution will contain a certain number of moles of that chemical?
Ex 1 If a solution contains 4.9 grams of H2SO4 in 100 ml (or 100cc or 0.1 litre or 0.1 dm3), what volume of solution will contain 2 moles of H2SO4?
Step 1 - convert grams of H2SO4 into moles
Calculate the relative molecular mass of molar mass of H2SO4 by adding together all the atomic masses for each atom present - H = 1, S = 32, O = 16, so the RMM of H2SO4 is 98 (if you need help with calculating rmm, use our tutorial and practice examples pages in the menu).
We have 4.9 g of H2SO4, so that gives us 4.9/98 = 0.05 moles, which are present in the 0.1 dm3
Step 2 - calculate the moles per dm3 (also written as mol dm-3) i.e. the concentration of the solution.
We have 0.05 moles present in 0.1 dm3, so dividing moles by volume will give us the concentration. We MUST use dm3 (also called litres) as the volume in this calculation. (a common trick is to give the volume in ml or cc or cm3, in which case you divide by 1000 to get the dm3).
So it's 0.05/0.1 = 0.5 mol dm-3
Step 3 - calculate the volume that contains the desired number of moles.
To do this we divide the required number of moles by the concentration
In our example it becomes 2/0.5 = 4dm3 (4 litres), the final answer.
It's important to get this basic calculation correct, as the other variations on solution calculations are just this set of steps with some extra steps added on.
Side note: dm3 means the volume of the solution, while dm-3 means "per dm3", for example in mol dm-3 = moles per dm3
Type Two - given the number of moles of some chemical in a given volume of a solution, what volume will contain a certain number of grams.
These solution calculations start at Step 2 above (the basic solution calculation backbone) and just need extra steps to complete them. Let's see how it's done.
Ex 2 If a solution contains 0.025 moles of H2SO4 in 0.5dm3, what volume of this solution will contain 9.8 g of H2SO4?
Step 1 - calculate the concentration of the solution, the moles per dm3
Divide the number of moles by the volume in dm3, in this case 0.025/0.5 = 0.05 mol dm-3
Step 2 - convert the required 9.8 g H2SO4 into moles, by dividing by the relative molecular mass of H2SO4
Well, earlier we calculated the rmm of H2SO4 as 98, so I will not repeat that bit here, the last part of this step is the 9.8/98 = 0.1 moles.
Step 3 - So we simply have to find what volume of solution of concentration 0.05 mol dm-3 is needed to get a total of 0.1 moles of the acid.
We do that by dividing the required moles by the
In this case it's 0.1/0.05 = 2 dm3 and note that here the answer is always a volume measured in dm3.
Type Three - give the concentration of a solution, which volume would contain a certain number of grams of the chemical given
This is the reverse of the above Type One, isn't it, and quite similar to Type Two (you'll start to notice these similarities and that makes things easier as we go on).
Ex 3 What volume of a solution of 0.2 mol dm-3 NaOH would contain 1 g of NaOH?
Step 1 - calculate the rmm (relative molecular mass) of NaOH
We do this by adding up the ram (relative atomic mass) of each atom in NaOH - Na = 23, O = 16 and H = 1, which gives us 40
Step 2 - convert the required number of grams into moles
We do this by diving the required number of grams by the rmm of the NaOH
so it's 1/40 = 0.025 moles required
Step 3 - find the volume of the solution that contains the required number of moles
To get the answer we divide the required number of moles by the solution's concentration
so it's 0.025/0.2 = 0.125 dm3 - the answer is always a volume in dm3
Type 4 - given the concentration of a solution, how much would be needed to react completely with a given volume of another solution whose concentration is also given.
Ex 4 What volume of NaOH solution of concentration 0.025 mol dm-3 will be required to react completely with 200 cm3 of H2SO4 of concentration 0.010 mol dm-3
For solution calculations like these, there is a mathematical equation which you can juggle about to get th eanswer, but converting everything into moles makes things very much clearer.
Step 1 - use the balanced equation for the reaction to find the ratio of the moles reacting
In this case it is 2NaOH + H2SO4 ---> Na2SO4 + H20
i.e. two moles of NaOH reacts with every one mole of H2SO4
Step 2 - find how many moles of H2SO4 are being used in the reaction.
Multiply the volume in dm3 by the concentration in mol dm-3
In this case that's 200/1000 x 0.010 = 0.2 x 0.010 = 0.0020 moles of H2SO4
Step 3 - find the number of moles of NaOH needed to react with this amount of H2SO4
So that's 2 x 0.0020 = 0.0040 moles of NaOH.
Step 4 - find the volume of NaOH that contains that number of moles
Here we divide the required number of moles of NaOH by the NaOH solution's concentration
So it's 0.0040/0.025 = 0.16 dm3 (or 160cm3)
The above question could have been made more complicated by giving you the number of grams of NaOH dissolved in the given volume.
That would mean you'd start by doing the first two steps of question Type 1, to get the concentration.
Or you could have be given the number of grams of H2SO4 dissolved in a given volume, instead of its concentration.
So again you'd have to use those first two steps of Type 1 to get the concentration.
Given grams, convert it to moles by dividing by the relative molecular mass.
Ensure any volume in the calculation is in dm3
Compare any question to the above types and check whether all you need to do is combine part of one type with another type.
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