Chemistry Unit 1.2 - Formulae, Equations and Moles

Notes on formulae, equations and moles

HideShow resource information
  • Created by: mar-rii
  • Created on: 01-01-10 16:51
Preview of Chemistry Unit 1.2 - Formulae, Equations and Moles

First 242 words of the document:

Chemistry Unit 1.2: Formulae, Equations and Moles
Equations and Moles
Empirical and Molecular Formulae
The empirical formula of a compound tells us the simplest ratio of atoms in that compound. It is
found from experimental data.
A hydrocarbon was found to contain 75% carbon and 25% hydrogen by mass. Determine its empirical
Mass(g) 75 25
Divide by the relative atomic mass 75/12 25/1
Ratio of atoms 6.25 25
Divide by smallest 6.25/6.25 25/6.25
. 1 4
So, the empirical formula of this hydrocarbon is CH4.
The molecular formula of a compound tells us the actual number of atoms in a compound of one
mole of that compound.
It was found that a hydrocarbon had an empirical formula of CH3. It was found that its relative formula mass
(r.f.m.) is 30. Determine its molecular formula.
1. Determine r.f.m. of CH3 = (12x1) + (1x3) = 15
2. Determine the ratio of mass = 30:15 = 2:1
3. Multiply the number of atoms by the ratio to give molecular formula = 2 * CH3 =C2H6
Compound Formulae
Find the Compound Formula for AB where
Step 1:
Step 2: If , they combine like this:
Step 3: If not...: cancel down this fraction into its simplest form, like:
Step 4: Call this 'cancelled down' version of x and y as
Step 5: The Compound can be written as
2. 3 is not equal to 2

Other pages in this set

Page 2

Preview of page 2

Here's a taster:

Formula =
Full and Ionic Equations
A fully balanced equation or stoichiometric equation is one that shows the formulae of reactants and
products and the relative number of particles reacting. e.g.
An ionic equation is one where the ions are represented separately. Only ionic compounds can be
represented as ions. Covalent substances and elements cannot be represented as ions. Also, ionic solids
cannot be represented as free ions because the ions are not free to move.…read more

Page 3

Preview of page 3

Here's a taster:

Moles of CaCO3 = Mass/r.f.m = 10/100 = 0.1moles
3) Work out the amount in moles of CO2:
Moles of CO2 = Moles of CaCO3 = 0.1moles
4) Work out the volume of CO2:
Volume of CO2 = Moles of CO2 x 24 = 0.1 x 24 = 2.4dm3
Gaseous Reactions
The ratio of the volume of a gas produced in a reaction to the number of moles of this gas in the reaction is
constant.…read more


No comments have yet been made

Similar Chemistry resources:

See all Chemistry resources »See all resources »