sharepoint 2016 workflow history report


Syllabus > Stoichiometry
> Moles

  syllabus stoichiometry faq software forum contact add
to favorites

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

Mole concept & Avogadro’s constant

1.1.1: Describe the mole concept and apply it to substances.
The mole concept applies to all kinds of particles: atoms, molecules,
ions, formula units etc. The amount of substance is measured in units
of moles. The approximate value of Avogadro’s constant (L), 6.02 x 1023
mol-1, should be known.

The structure of matter

It is now accepted that matter in all its forms is made up of indivisible
particles that themselves have mass. These particles are called atoms,
molecules and ions. The nature of the substance is dictated by the atoms
elements that have bonded together to make the bulk substance. This may
be an ionic structure, a covalent structre or a metallic structure.


Giant ionic structure giant covalent structure simple covalent giant metallic structure

Molecules are made up of two or more atoms chemically bonded together.

Ions are specialised atoms or groups of atoms chemically combined together
that have lost or gained electrons and posess an overall electrical charge.

The fundamental particle that is the building block of matter is therefore
the atom. There are about 90 naturally occuring types of atoms each with
a different arrangement of sub-atomic particles (protons, neutrons and
electrons) and consequently different masses.

The structure of matter is one of the following:

  • atoms —> molecules —> bulk compound or element
  • atoms —> bulk element
  • atoms —> ions —> bulk ionic compound

The masses that are measured in the laboratory are masses corresponding
to vast numbers of tiny atoms or molecules. Logically atoms that are heavier
will register larger masses for equal numbers of atoms.

Relative atomic mass

If one carbon atom has a mass of 12 atomic mass units and one magnesium
atom has a mass of 24 atomic mass units, then as a magnesium atom is twice
as heavy as a carbon atom it follows that this ratio will be maintained
for any number of atoms.

On the atomic mass scale the carbon 12 isotope is designated a value
of 12 atomic mass units and all other masses are measured relative to
this (relative atomic mass)

The mole concept

It is convenient to consider the number of atoms needed to make 12g of
carbon and for this number to be given a name – one mole of carbon atoms.
This allows us to talk about relative quantities of substances in the
macroscopic world and to know the relative number of atoms (or smallest
particles) in each bulk substance.

The actual number of atoms that is needed to give the relative atomic
mass expressed in grams is called Avogadro’s number (symbol L)

Avogadro’s number = 6,02 x 1023

Definition of a mole

There are two useful definitions.

  1. The relative atomic (molecular) mass of a substance expressed in grams
  2. An Avogadro number of particles of any substance

Example 1:

one mole of carbon = 12 g

magnesium atoms are twice as heavy as carbon atoms therefore 1
mole of magnesium = 24g

Example 2.

equal masses of carbon and magnesium contain different numbers
of atoms.

6g of carbon contains 6/12 moles of carbon =0,5 moles

6g of magnesium contains 6/24 moles of magnesium =0,25 moles


Sodium carbonate crystals (27.8230g) were dissolved in water and
made up to 1.00 dm3. 25.0 cm3 of the solution were neutralised by
48.8 cm3 of hydrochloric acid of conc 0.100 mol dm-3. Find n in
the formula Na2CO3.nH2O

48.8 cm3 of 0.1M HCl = 0.00488moles

Na2CO3 + 2HCl –> NaCl + CO2 + H2O

therefore moles of Na2CO3 = 0.00488/2 = 0.00244moles

This is in 25cm3 therefore the moles in 1000cm3 = 0.00244/0.025

If the formula = Na2CO3.nH2O

Then the neutralisation has measured only the Na2CO3

Therefore the mass of Na2CO3 = RMM x no of
moles = 106 x 0.0976 = 10.3456g

The remaining mass must be due to water = 27.823 – 10.3456 = 17.4774g

RMM of water = 18 therefore this is equivalent to 17.4774/18 moles
= 0.971

Thuus the mole ratio of Na2CO3 to water in
the original compound = 0.096 : 0.971

or approximately 1 :10

The formula is therefore Na2CO3.10H2O

Example 3.

How many atoms are ther in 24g carbon

24g of carbon = 24/12 moles = 2 moles

1 mole of atoms = 6,02 x 1023

therefore 2 moles of carbon contains 2 x 6,02 x 1023
atoms = 1,204 x 1024 atoms

1.1.2: Calculate the number of particles and the amount
of substance (in moles). Convert between the amount of substance (in moles)
and the number of atoms, molecules or formula units

1 mole = 6.02 x 1023 formula units of that substance.

We can also talk about the atoms within molecules.

For example 1 mole of water contains 2 moles of hydrogen atoms and 1
mole of oxyten atoms. It is a simple matter of multiplying the moles of
the compound by the atoms or ions that make it up.


Having problems with this topic?

Why not try out the new interactive ebook on Stoichiometry and
the Gas laws?

CS1 Colourful Solutions – download a trial run today.

Useful links



IB Chemistry 1 SL Questions

Uploaded by Camilla Clrt
Paper 1, Paper 2, Markscheme Full description
Copyright: © All Rights Reserved
Download as RTF, PDF, TXT or read online from Scribd
Flag for inappropriate content


You are on page 1of 36

Share this document

Share or Embed Document

Sharing Options

  • Share on Facebook, opens a new window
  • Share on Twitter, opens a new window
  • Share on LinkedIn, opens a new window
  • Share with Email, opens mail client
  • Copy Text

Related Interests

  • Mole (Unit)
  • Gases
  • Acid
  • Ammonia
  • Chemical Compounds

You’re Reading a Free Preview

Pages 4 to 36 are not shown in this preview.
Buy the Full Version