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IGCSE Coordinated Science: Stoichiometry

By (Administrator)

4.0 Stoichiometry

1 Use the symbols of the elements to write the formulae of simple compounds.

Using the symbols for elements from the periodic table, we can write the formulae for chemical compounds. Below are the formulae of some chemical compounds that you may have come across before:

Water = H2O

Hydrochloric acid = HCl

Sulphuric acid = H2SO4

2.Deduce the formula of a simple compound from the relative numbers of atoms present.

It is possible to work out the chemical formulae of a compound when given the elements present.

For example, if we are told that a certain compound contains both sodium and chlorine, we can deduce its formulae to be NaCl.

Remember when working out the chemical formulae to take into consideration the valency of the elements present.



3 Deduce the formula of a simple compound from a model or a diagrammatic representation.

You also need to be able to work out the chemical formulae of a compound from a diagram:


A simple way would be to simply count the number of electrons in each atom!

The red dots represent the electrons of the element on the left and the blue crosses represent the electrons of the element of the right.

Since the diagram features an ionic compound, we have to take in account the single electron given from the positive ion to complete its full outer shell.

In total, the element on the left has 11 red dots, or electrons; if we look at the periodic table, the element with 11 electrons is sodium.

Likewise, the element on the right has 17 crosses, or electrons. Using the periodic table, the element with 17 electrons is Chlorine.

4 Construct and use word equations.

The compound contains sodium and chlorine; hence we can deduce its identity to be sodium chloride.

5 Determine the formula of an ionic compound from the charges on the ions present.

To work out the formulae of an ionic compound using the ions present simply remember that the charge on any ionic compound must be neutral, or 0.

For example, we are told that a compound has Fe2+ ions present as well as Cl- ions.

Using that information, for every Fe2+ present in the compound there must be 2 Cl- ions to balance out the charge. Therefore, the formula of the ionic compound must be FeCl2.

7 Deduce the balanced equation for a chemical reaction, given relevant information

Balancing Equations:

To balance an equation, you have to remember that there must be same number of atoms of each element present on both sides of the equation:

1)                  ___ NaNO3 + ___ PbO —->___ Pb(NO3)2 + ___ Na2O

2    NaNO3 + _1_ PbO —->_1_ Pb(NO3)2 + _1_ Na2O

Both sides of the equation has the same numbers of atoms of each element: 2Na; 2NO3; Pb; O

2)            ___ AgI + ___ Fe2(CO3)3 —-> ___ FeI3 + ___ Ag2CO­3

3)            C2H4O2 + ___ O2 —–> ___ CO2 + ___ H2O

4)            ___ ZnSO4 + ___ Li2CO3 —-> ___ ZnCO3 + ___ Li2SO4

5)            ___ V2O5 + ___ CaS —–> ___ CaO + ___

6)            ___ Mn(NO2)2 + ___ BeCl2 —-> ___ Be(NO2)2 + ___ MnCl2

7)            ___ AgBr + ___ GaPO4 —-> ___ Ag3PO4 + ___ GaBr3

8)            ___ H2SO4 + ___ B(OH)3 —–> __ B2(SO4)3 + ___ H2O

9)            ___ S + ___ O2 —–> ___ SO2

10)          ___ Fe + ___ AgNO3 —–> ___ Fe(NO3)2 + ___ Ag


8 Define relative atomic mass, Ar.

Relative Atomic Mass: is the measure of the mass of one atom of an element.

On the periodic table, to work out the relative atomic mass of an element, you simply look at the Mass Number of the element, a sum of the number of the protons and neutrons present.

For example, the isotope Carbon-12 has 6 protons & 6 neutrons, so its relative atomic mass is 12.

However, remember that there are different isotopes of elements as well; the isotope Carbon-13 has 6 protons & 7 neutrons, so its relative atomic mass is 13.

This is the reason why elements like Chlorine has a Mass Number of 35.5. They have different isotopes with different masses. For chlorine, it is the average mass of the two isotopes Chlorine-35 & Chlorine-37.

Define relative molecular mass, Mr, as the sum of the relative atomic masses (relative formula mass or Mr will be used for ionic compounds).

Relative Molecular Mass: is the mass of a single molecule of a compound. It is the sum of the relative atomic mass of the atoms present within the compound.

For example: To work out the relative molecular mass of NaNO3, you would find the relative atomic mass of the elements present:

Na = 23

N = 14

O = 16

The relative molecular mass of NaNO3 would be:

23 + 14 +(3 x 16) = 85


Click here to go to the next lesson: 4.2 The Mole Concept

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8 Responses to IGCSE Coordinated Science: Stoichiometry

  1. -.-' on March 2, 2014 at 5:08 PM

    Can you put up the answers to the balancing equations sums? Thanks a load.

    • James Yang on March 15, 2014 at 11:19 PM

      Hi, the answers have been uploaded. Please select the “Answer” button to reveal the answer.

  2. Revy on April 23, 2013 at 12:17 AM

    “The red dots represent the electrons of the element on the left and the blue crosses represent the electrons of the element of the right.” in 4.1.3 in syllabus should be “The red dots represent the electrons of the element on the RIGHT and the blue crosses represent the electrons of the element of the LEFT.”
    else that would make no sense in my opinion, not entirely sure though, thanks in advance though, if you’re going to change it :)

  3. Zilla on January 14, 2013 at 12:44 AM

    You made a mistake with the chlorine isotopes. There are chlorine-35 and chlorine-37 and the percentages of each type in. The world determine the average number

  4. Nikhil on April 23, 2012 at 6:59 AM

    On the syllabus point 3, the description given does not match the diagram. Please check over.

  5. NLai on April 10, 2012 at 2:49 AM

    Where’s #6 of the syllabus “Construct and use symbolic equations with state symbols, including ionic equations”?


  6. Nikhil on March 26, 2012 at 2:08 PM

    The example given for relative molecular mass is incorrect? 23 + 14 (3 X 16) = 53?

    • Acceleratedstudynotes on March 26, 2012 at 2:27 PM

      Thanks, I’ve made the necessary correction.