1.3 ATOM ECONOMY AND % YIELD, CONCENTRATION OF SOLUTIONS & ACID-BASE TITRATION CALCULATIONS
- Chemistry at glance
- Feb 10, 2020
- 5 min read
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Hi!! In this post, you are going to learn about atom economy, % yield, concentration of solutions and titration calculations. But let's start with atom economy and percentage yield first. I also have examples for you prepared. However, if you want to practice some more questions, visit my youtube channel or email me and I will send you a copy of these questions.
Definition at the start:
Atom economy= mass of required product (or the Mr)/total mass of reactants (or the Mr) x100
This shows you how efficient the reaction is. The higher the atom economy the more efficient the reaction is. High atom economy is good for environment beasue this means less waste is created.
e.g.
We don't have any mass in the question, therefore, you know that you need to use the relative molecular mass instead of mass. You are looking to find the atom economy for calcium oxide, so find the Mr of calcium carbonate -100.1 and the Mr of calcium oxide -56.1.
Now, divide the Mr of calcium oxide over the Mr of all the reactants and multiply by 100. And this is how you calculate your atolm economy.
Another definition:
Percentage yield = mass (or moles) of product obtained/ theoretical mass of product x100
This time, you have the actual mass of a product that has been produced in an experiment and you need to find the theoretical mass (which is the mass that should be formed in this type of reaction according to the moles of each atom) and multiply by 100.
e.g.
It is easiest for me to write the equation down and then all the information you already know around. So, in this example 0.5g were formed from 5g. This is the actual mass produced from an experiment. Now, you need to find the theoretical mass. This is the mass that would be formed if there were no losses to the environment.
Find the number of moles of the reactant and use this to find the number of moles of the product (you do this by establishing the ratio, in this example it is 1:1). Then using this number of moles, find the mass of the product that should be produced.
This should be a bigger mass than the actual mass produced. This is because during experiment, some product may be lost.
The last step is to put the smaller mass over the bigger mass and multiply by 100. This is your percentage yield.
Concentration of solutions
number of moles (n)= concentration (c)/ volume (v)
This is another equation that you will not be given in a exam, you need to remember it off by heart. You can also rearrange this equation depending on what you are trying to find.
When using this equation, you need to make sure the volume is in decimeters cubed NOT centimeters.
e.g.(1)
Looking at this question, you should straight away realise that you need to find the number of moles and establish a ratio between that reactant and product.
Let's start with the number of moles of copper (II) sulphate solution, and for this we use our new equation n=cv.
Then, we find the ratio, which is 1:1 and calculate the mass of our product using n=m/Mr rearranged. This will give you a mass of barium sulphate produced.
e.g.(2)
In order to change from mol/dm^3 to g/dm^3 you need to multiply the number by the Mr of the compound. This is exactly what you do in this example. Multiply the concentration by the Mr of sodium carbonate.
e.g.(3)
This is a question involving solubility. You usually calculate solubility in grams per 100g. You can transfer it to grams per cm cubed.
Here are the steps:
1. Find the Mr of the compound for which you are trying to find the concentration.
2. Find the number of moles for this compound, using the grams that are written in the solubility which is 31.6g.
3. You can transfer cm cubed to dm cubed by dividing it by 1000. This is again using the solubility, 100cm^3 divided by 1000 is 0.1dm^3.
4. And now you can find the concentration as you already know the volume (0.1dm^3) and the number of moles (0.313mol) using n=cv.
You're done!
Our last thing to do now, for this topic is to learn how to do acid-base titration calculations:
Acid-base titration - it is a type of volumetric analysis (which is finding the concentration).
In order to do this type of calculation you need to know the concentration and volume of one solution (standard solution) and find the concentration of another solution (standardise it) with known volume. But this will be much easier to explain with an example.
e.g.(1)
1. We know the volume and concentration of sodium hydrogencarbonate solution, so we can find the number of moles of this solution.
2. The next thing, as always, is to find the molar ratio. In this example it is 2:1, so you need to divide the numer of moles by 2.
3. Next, calculate the concentration of the second solution for which we know the volume (and now the number of moles).
4. The question asks us to find the concentration in grams/dm cubed. Therefore, the next step is to find the Mr of solution.
5. And lastly, multiply the Mr by the concentration.
That's all!
e.g.(2) This a slightly more difficult example and you'll see why. But you can do it!
1. This is differet from the previous one, because we have a mass provided and we can find the Mr and this is how you are going to calcuate the number of moles. (Not using concentration and volume this time).
2. We are told that only 25cm^3 were used from 250cm^3 and we calculated the number of moles for 250cm^ therefore you need to divide this number by 10. This gives you the number of moles of sodium hydrogencarbonate in 25cm^3.
3. Now, that you have the number of moles, you need to establish the molar ratio. This is 1:1 in this example. We are provided with the volume of HCl and we just found the number of moles. Therefore, we can calculate the concetration using n=cv.
4. Again, we are asked to find it in g/dm^3, so multiply by the Mr of HCl. And that's all.
ALWAYS CONVERT UNITS IN QUESTIONS! REMEMBER THAT FOR N=CV YOU NEED DM^3, NOT CM^3.
PS. Please remember, I am only a student, and as anyone, I can make mistakes. If you think you can see one, don't hesitate and comment (either here on on my youtube channel) Thank you!
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