GCSE AQA Science higher formulas

This has all the formulas needed for GCSE AQA additional chemistry, and may be useful for any other GCSE syllabus. It also has examples of how to use each formula layed out in easy steps.

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  • Created by: lalala
  • Created on: 29-04-12 15:16
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The Relative Atomic Mass (Ar)
The relative atomic mass of an element (Ar)
compares the mass of atoms of the element
with 12C isotope.
It is an average value for the isotopes of the
element.
Ar of an element= look at element on periodic
table and it is the biggest number of the two
(the mass number)…read more

Slide 3

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The relative formula Mass (Mr)
The relative formula mass (Mr) of a compound is the
sum of the relative atomic masses (sum of Ar) of the
atoms.
Mr of an element= add all of the Ar results together.
*Remember to multiply the Ar of each atom by how ever
many atoms of that element there are in the compound
E.g. KNO3
Mr= Ar of K + Ar of N + 3(Ar of O)
39 + 14 + 3(16)
Mr= 39 + 14 + 48
Mr of KNO3 = 101…read more

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Percentage Mass
Percentage mass=
Ar of element you are finding X number of atoms of that element
Total Mr of the compound
*Then multiply your answer by 100 to get a percentage*
E.g. Calculate the percentage mass of Mg in, MgO.
Mg= Magnesium Percentage mass=
Ar = 24 24 x 1
Number of atoms = 1 40
= 0.6
Total Mr of compound=
0.6 x 100
Ar of Mg + Ar of O = 60%
= 24 + 16 Percentage mass of Mg in MgO = 60%
=40…read more

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Percentage Composition
(Empirical Formula)
Only for Higher Tier
There are 7 steps:
1) Separate the compound into it's different elements and write
each one down
2) Write the experimental mass given to you in the question and
write this underneath
3) Write down the Ar of each element
4) Divide the experimental mass of each element by the Ar of each
5) You now have a ratio, but the ratio must be a whole number
(times by 10, 100 or 1000 depending on the decimals to get a
whole number).
6) Cancel the ratio down to it's simplest form (by dividing by a
number they all have in common
7) Write it out in it's formula
Examples on next slide...…read more

Slide 6

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E.g. Find the simplest formula of an oxide of iron produced by reacting 1.12g of
iron with 0.48g of oxygen.
Element
Element Fe (iron) O (Oxygen)
Experimental mass ­ 1.12 0.48
provided by the question
Ar of each element ­ from 56 16
the periodic table (mass
number)
Experimental mass divided 1.12 0.48
by the Ar of each element 56 16
(moles). = 0.02 = 0.03
Ratio ­ the number the 0.02 X 100= 2 0.03 X 100= 3
moles all have in common to
make them a whole number
usually x100
Formula: Fe2O3…read more

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