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Applied Chemistry Fundamental
Physical Chemistry

Chemical Equations

Basic Chemical Equation

Chemical equations are vital in Chemistry for us to represent what is happening in a reaction.

The first, and simplest way of reprenting a chemical reaction is to do it in words the following shows the reaction when magnesium is burned, producing a bright blue light.

magnesium + oxygen magnesium oxide

However, in Chemistry chemical symbols are used to abbreviate the equation, these can be found on a periodic table. There is a small 2, after the O since there are two oxygen molecules bonded.

Mg + O2 MgO

The next step is to balance the equation so there are the same number of molecules on both sides. We need to double the Mg, so that one goes to each of the oxygen and that makes 2 of the product.

2Mg + O2 2MgO

Finally we add on state symbols to show what state each chemical is in when it is reacted. An (s) stands for solid; (l) for liquid; and (g) for gas. The fourth state symbol is (aq), this stands for aqueous and means the substance is dissolved.

2Mg(s) + O2(g) 2MgO(s)


Using the the wealth of information that is packed into a chemical equation you can do some calculations, or even vice versa: do some calcualtions to get a chemical equation. I will start with the former.

Using the relative atomic masses (RAMs) and relative formula masses (RFMs) you can work out what amount of a certain substance you can get. Say you wanted to know how much hydrogen you would get when reacting 28g of lithium with water. The full balanced equation for this reaction is:

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

  • You first find out the RAM of 2Li, which is 14. Therefore each part of the equation must be doubled to get the mass in grams because 28 is double 14.
  • The RAM of H2 is 2.
  • Double this number to get 4.
  • You will get 4 grams of hydrogen when you react 28g of lithium with as much water as is required.
  • Now we will look at finding out the formula of a reaction using the results from an experiment. 4.14g of lead combines with 0.64g of oxygen.

  • The RAM of lead (Pb) is 207, and of Oxygen (O) is 16.
  • You divide the masses by their RAMs giving 0.02 for lead and 0.04 for oxygen.
  • Now you take the smallest number (in this case 0.02) and divide those two again.
  • So 0.02/0.02 =1 (Pb)
  • and 0.04/0.02 = 2 (O)
  • Therefore the formula is PbO2