1.3 CALCULATING REACTING MASSES & EMPIRICAL AND MOLECULAR FORMULA

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Hi everyone! How are you today? This is our second post on chemical calculations. How are you feeling after the first one? Did you practice some more questions with my youtube channel afterwards?

Ok, so this unit will mostly be examples. I will go through different scenarios and explain everything on the way. I hope this way, you will see what steps you need to take in different types of questions.

We are going to start with calculations invloving reacting masses:

In reactions, for products to be formed, we need reactants. And from a given amount of reactants, we can find the mass of products at the end. As long as we know the number of moles (mole ratio) and balanced equation. (It can also work the other way. If you know the mass of products formed you can find what mass of reactants would be needed). And when we talk about a molar ratio:

Stoichiometry - molar relationship between reactants and products in a chemical equation.

This is basically the ratio of moles between reactants and products. As shown in the example: calcium carbonate has a 1:1 ratio with calcium oxide and in the second one, oxygen has a 1:2 ratio with magnesium oxide. This is due to the numbers at the front of the molecule (it shows you the number of moles of this substance).

These are the steps for doing such calculations:

1. Find the substance for which we know both the mass and relative molecular mass (Mr). Then find the number of moles for this substance using n=m/Mr;

2. Using the big numbers at the front of substances, figure out a ratio for the substance you just found the 'n' for and the substance you want to find the mass for. This is then used to find the number of moles for the substance with unknown mass.

3. Lastly, use the 'n' found and the Mr of the substance with unknown mass in order to find the mass (using n=m/Mr rearranged to m=nxMr).

And now examples:

e.g (1)

1. We want to find the number of moles for the molecule with known mass which is calcium carbonate. In order to do this, find the Mr and the mass you already have. Put it into n=m/Mr and your done with this step. (This shouldn't be hard because the last post was all about it).

2. Figure out the ratio between calcium carbonate and calcium oxide. This is 1:1 because there are no numbers in front of the molecules. This means the number of moles for calcium carbonate (the one you found in the previous step) is the same as the one in calcium oxide.

3. Lasly, find the Mr of calcium oxide (one with unknown mass) and use the number of moles just found to calculate the mass of calcium oxide produced.

e.g.(2)

1. Again, find the Mr of hydrochloric acid and using this find the number of moles of hydrochloric acid.

2. Figure out a ratio. In this example this is 2:1. We are trying to find the number of moles for magnesium chloride, so what you need to do is divide 'n' by 2.

3. Now that we know the 'n' of the 'unknown mass' substance, rearrange the equation to find mass (of course you need to calculate the Mr for magnesium chloride first).

When they say something is in excess, there is enough of it for the reaction to be complete. For us this means you don't have to take it into consideration for this type of question.

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Empirical formulae:

Empirical formulae - this is the simplest formla. It shwows the simplest ratio of elements present in a compound. e.g. for benzene (C6H6), the empirical formulae is CH. ( simplest ratio of 6:6 is 1:1).

Again there are couple of steps you need to follow to complete this type of question ( you can find a different route for yourself that you like but this is the one I found helpful and easiest to work with).

e.g.(1)

1. The first step is to find the Mr of each atom. You do this by looking at the periodic table and finding the relative atomic mass.

2. The next step is to write yourself the mass of each of these atoms. This you will be given in the question. If it is not given as a mass, it will be a percentage and you use it exactly the same as if it was a mass.

3. Now, you want to find the number of moles of these atoms using your Mr and mass and the equation n=m/Mr.

4. Whatever your number o moles will be you need to divide both of the them by the smaller number. So, in this example we have the number of moles of 0.1 (this is the smallest number ) and what you do is divide the number of moles of the other atom by this 0.1.

5. The number you got in the previous step is your ratio of atoms. In this example you had 1 and 1 so the ratio is 1:1, MgO.

e.g.(2)

1. Again, find the relative atomic mass of all the atoms.

2. Then, write yourself down the percentage of each of these atoms. (This is what I told you before, you might get a percenatge instead of mass but it works exactly the same).

3. Now, the number of moles for each atom using n=m/Mr.

4. It is time to find the ratio, so what you do is divide each atom by the smallest number of moles. In this example this is 0.9. Therefore, you divide each number of moles by 0.9 and this will give you the ratio.

5. The ratio is 2:5:1, so, C2H5Br.

What we are going to learn about now is the molecular formula:

As we said before empirical formula is the simplest ratio of atoms in a molecule and:

Molecular formula - it shows the actual number of atoms of each element in a molecule. (It is the multiple of empirical formula).

If the empirical formula of benzene is CH (Mr=13) and benzene has a Mr of 78 this means the actual ratio must be different than the CH. 78/13=6. Therefore, it is actually C6H6 (ratio 6:6).

I have examples again:

e.g.(1)

Every time you are trying to find the molecular formula, you first will probably be asked to find the empirical formula (or you will be give that).

a) As before, follow the steps and you will find the empirical formula. If you are confused looking at the picture and you have no idea where I got these numbers from, look back at the examples of calculating empirical formula and follow the steps. You are left with C4H8O.

b) Now, it is finding the molecular formula. In order to do this you need to have two Mr numbers. One for the empirical formula we just found, which is 72. And one for the actual formula (this one you need to be given in the question). You then divide the empirical formula Mr over the actual. If it equals to 1 then it means the empirical formula is the same as the molecular formula. If the number is bigger than one then you need to multiply each element by that number in the empirical formula.

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Finding 'x' in a formula:

Now, the last part: finding how many H2O's is present in a compound. As in the picture above, you have a formula but the number of waters is missing. This is the number you need to find. You can do this with only having the mass of the compound.

e.g.(1)

1. Calculate the number of moles of the other part than water. This is called the anhydrous part (it is called anhydrous because it doesn't evaporate). In order to do this, you need to calculate the Mr (not counting the water) and mass you already have (mass of the anhydrous part, 0.774g).

2. Now, calculate the number of moles of water that has evaporated. You can find the Mr of water with no problem. And your mass will be the overall mass - mass of the anhydrous part.

3. All you do now, is the number of moles of water/ number of moles of anhydrous part. This is because you have more water than the anhydrous part and you need to find out how many times more.

e.g.(2)

You usually will have mass in grams and you would need to find the molar ratio. However, this time you already have the mass in g/mole, therefore the ratio is already established.

As before, you have the molar mass of the whole compound and then the molar mass of the anhydrous part only, so to find the molar mass of just water you need to take away one from the other.

The next step is to use this molar mass for water and divide by its Mr. This will give you the missing 'x'.

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!