Hi, I'm a year 12 student and this May/June I'm going to be taking the AS chemistry CIE. I have just finished physical chemistry and inorganic chemistry and this second term I'm going to start the organic chemistry.

I need any tips on how to study for inorganic chemistry, I'm struggling with inorganic chemistry because it's very difficult to memorise all the equations....

I need any tips on how to study for inorganic chemistry, I'm struggling with inorganic chemistry because it's very difficult to memorise all the equations....

Original post by Sarah123454

Hi, I'm a year 12 student and this May/June I'm going to be taking the AS chemistry CIE. I have just finished physical chemistry and inorganic chemistry and this second term I'm going to start the organic chemistry.

I need any tips on how to study for inorganic chemistry, I'm struggling with inorganic chemistry because it's very difficult to memorise all the equations....

I need any tips on how to study for inorganic chemistry, I'm struggling with inorganic chemistry because it's very difficult to memorise all the equations....

hi, what I find useful to do is to write down all the equations on a flashcard and then revisit them every so often so it jogs your memory, or you could talk to some friends about them too and ask them to test you on them - just keeping them in the back of your mind is something I found helpful 🙂

Original post by Sarah123454

Hi, I'm a year 12 student and this May/June I'm going to be taking the AS chemistry CIE. I have just finished physical chemistry and inorganic chemistry and this second term I'm going to start the organic chemistry.

I need any tips on how to study for inorganic chemistry, I'm struggling with inorganic chemistry because it's very difficult to memorise all the equations....

I need any tips on how to study for inorganic chemistry, I'm struggling with inorganic chemistry because it's very difficult to memorise all the equations....

Don’t bother memorising equations.

The approach I found easier was to learn the formulae of various compounds (or how to work out their formulae) and to work out the equation from there.

Let’s say for example a question asked you to write an equation for the reaction of phosphorus(V) oxide with sodium hydroxide, to form sodium phosphate and one other product.

I would first work out that sodium phosphate is Na3PO4 and recall that phosphorus(V) oxide is P4O10 and sodium hydroxide is NaOH.

Setting up an unbalanced equation:

P4O10 + NaOH —> Na3PO4 + …

Since there are 4 P on the LHS, there should also be four on the right-hand side. So the next thing to try would be scaling up the Na3PO4 by 4:

P4O10 + NaOH —> 4Na3PO4 + …

Since there are 4 x 3 = 12 Na on the RHS, there should be 12 on the RHS. Scaling up the NaOH by 12:

P4O10 + 12NaOH —> 4Na3PO4 + …

As we can see, there are 12 hydrogens on the LHS and none on the RHS. There are also 22 oxygens on the left and just 16 on the right. We know there must be one more product formed and from the fact it is made up only of H and O in a 2:1 ratio, it’s safe to assume the other product is water - 6 molecules of it, in fact:

P4O10 + 12NaOH —> 4Na3PO4 + 6H2O

And that is the final answer!

Original post by TypicalNerd

Don’t bother memorising equations.

The approach I found easier was to learn the formulae of various compounds (or how to work out their formulae) and to work out the equation from there.

Let’s say for example a question asked you to write an equation for the reaction of phosphorus(V) oxide with sodium hydroxide, to form sodium phosphate and one other product.

I would first work out that sodium phosphate is Na3PO4 and recall that phosphorus(V) oxide is P4O10 and sodium hydroxide is NaOH.

Setting up an unbalanced equation:

P4O10 + NaOH —> Na3PO4 + …

Since there are 4 P on the LHS, there should also be four on the right-hand side. So the next thing to try would be scaling up the Na3PO4 by 4:

P4O10 + NaOH —> 4Na3PO4 + …

Since there are 4 x 3 = 12 Na on the RHS, there should be 12 on the RHS. Scaling up the NaOH by 12:

P4O10 + 12NaOH —> 4Na3PO4 + …

As we can see, there are 12 hydrogens on the LHS and none on the RHS. There are also 22 oxygens on the left and just 16 on the right. We know there must be one more product formed and from the fact it is made up only of H and O in a 2:1 ratio, it’s safe to assume the other product is water - 6 molecules of it, in fact:

P4O10 + 12NaOH —> 4Na3PO4 + 6H2O

And that is the final answer!

The approach I found easier was to learn the formulae of various compounds (or how to work out their formulae) and to work out the equation from there.

Let’s say for example a question asked you to write an equation for the reaction of phosphorus(V) oxide with sodium hydroxide, to form sodium phosphate and one other product.

I would first work out that sodium phosphate is Na3PO4 and recall that phosphorus(V) oxide is P4O10 and sodium hydroxide is NaOH.

Setting up an unbalanced equation:

P4O10 + NaOH —> Na3PO4 + …

Since there are 4 P on the LHS, there should also be four on the right-hand side. So the next thing to try would be scaling up the Na3PO4 by 4:

P4O10 + NaOH —> 4Na3PO4 + …

Since there are 4 x 3 = 12 Na on the RHS, there should be 12 on the RHS. Scaling up the NaOH by 12:

P4O10 + 12NaOH —> 4Na3PO4 + …

As we can see, there are 12 hydrogens on the LHS and none on the RHS. There are also 22 oxygens on the left and just 16 on the right. We know there must be one more product formed and from the fact it is made up only of H and O in a 2:1 ratio, it’s safe to assume the other product is water - 6 molecules of it, in fact:

P4O10 + 12NaOH —> 4Na3PO4 + 6H2O

And that is the final answer!

Thankss

Original post by bsingh006

hi, what I find useful to do is to write down all the equations on a flashcard and then revisit them every so often so it jogs your memory, or you could talk to some friends about them too and ask them to test you on them - just keeping them in the back of your mind is something I found helpful 🙂

Thankss

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