Transcript File

Using the idea of moles to find formulae
The empirical formula of a chemical compound is the simplest
whole number ratio of atoms of each element present in a
compound
• 3.2g of copper reacted with 0.8g of oxygen. What
is the formula of the oxide of copper that was
formed? (At. Mass Cu=64: O=16)
Substance
Copper oxide
1. Elements
Cu
2. Mass of each element
(g)
3.2
3. Mass / Atomic Mass
4. Ratio (divide by lowest)
5. Formula
O
0.8
0.8/16 =0.05
3.2/64 =0.05
1:1
CuO
• We found 0.50g of magnesium reacted with 0.28g
of oxygen. What is the formula of the magnesium
oxide that was formed? (At. Mass Mg=24: O=16)
Substance
Magnesium oxide
1. Elements
Mg
2. Mass of each
element (g)
3. Mass / Atomic
Mass
4. Ratio (divide by lowest)
0.50
5. Formula
O
0.28
0.50/24 = 0.02
1:1
MgO
0.28/16 =0.02
An oxide of hydrogen
Calculate the empirical formula of a compound with the percentage
composition:
C 39.13%; O 52.17%; H 8.700%.
It is clear at this stage that dividing by the smallest has not resulted in
a simple ratio. You must not round up or down at this stage.
You must look at the numbers and see if there is some factor that you
could multiply each by to get each one to a whole number.
If you multiply each by 3 you will get:
C
O
H
3
3
8
So C3H8O3 is the
empirical formulae
You need to be careful about this; the factors will generally be clear
and will be 2 or 3. You must not round 1.33 to 1 or 1.5 to 2.
Exercise 5: Calculation of formulae from experimental data
pp52-