Simple chemical formulae
& chemical equations
Syllabus:
At the end of the this section you should:
1.21 use the state symbols (s),(l),(g) and (aq) in chemical equations to represent solids, liquids, gases and aqueous solutions respectively
1.22 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation
Finding the Formula of a Metal Oxide
Non-metal oxides are covalent compounds
And so are generally gases.
We can form an Oxide by burning an element in air.
If the element is a non-metal the gas formed will escape.
But not if it’s a metal element.
Metal Oxides are Ionic compounds.
So they are solids.
Finding the Formula of a Metal Oxide
If we use Magnesium
Magnesium + Oxygen → Magnesium Oxide
2 Mg(s) + O2(g) → 2 MgO(s)
This means
SOLID
This means
Gas
These are called
COEFFICIENTS
Finding the Formula of a Metal Oxide
STEP 1
Weigh the Crucible and the lid.
STEP 2
Weigh the Crucible and the lid with the Magnesium in.
We can now work out how much Magnesium we are using by subtracting
Finding the Formula of a Metal Oxide
STEP 3
Heat the crucible strongly for several minutes.
The lid must be used to prevent the Magnesium Oxide powder escaping.
But you must occasionally open it to let more Oxygen in.
The reaction is over when there are no more white flashes when you lift the lid
Finding the Formula of a Metal Oxide
STEP 4
Re-weigh the Crucible, lid and Magnesium Oxide.
Wait for the crucible to cool down a little.
Don’t touch it with your fingers!
Carry the crucible as shown in the photo.
Maths!
B - A = Mass of Magnesium = 45.48 - 45.00 = 0.48g
C - A = Mass of Magnesium Oxide = 45.80 - 45.00 = 0.80g
Mass of Magnesium Oxide - Mass of Magnesium = Mass of Oxygen gained = 0.80 - 0.48 = 0.32g
What is weighed? | Mass (g) | |
Empty crucible and lid | 45.00 | A |
Crucible, lid & Magnesium | 45.48 | B |
Crucible, lid & Magnesium Oxide | 45.80 | C |
More Maths!
Moles = Mass ÷ Molar Mass I
Moles of Magnesium = 0.48 ÷ 24 = 0.02 moles
Moles of Oxygen = 0.32 ÷ 16 = 0.02 moles
So the ratio of Magnesium: Oxygen is 0.02: 0.02 which is the same as 1:1
So the simplest formula for Magnesium Oxide is Mg1O1
Or just MgO.
Losing Oxygen.
Unreactive metals bond weakly.
Oxygen can be removed from Copper Oxide by passing gas over it.
So weighing the Copper Oxide before and after would tell us how much Oxygen was lost
Further Maths
A - C = Mass of Copper Oxide = 31.50 - 30.50 = 1.00 g
B - C = Mass of Copper = 31.27 - 30.50 = 0.77 g
Mass of Copper Oxide - Mass of Copper = Mass of Oxygen lost = 1.00 - 0.77 = 0.23 g
What is weighed? | Mass (g) | |
Test-tube and Copper Oxide | 31.50 | A |
Test-tube and Copper | 31.27 | B |
Empty Test-tube | 30.50 | C |
Is there no end to all the Maths?
Moles = Mass ÷ Molar Mass
Moles of Copper = 0.77 ÷ 63.5 = 0.0121
Moles of Oxygen = 0.20 ÷ 16 = 0.0125
Ratio Cu:O : 0.0121:0.0125 which is almost exactly 1:1
So, the simplest formula for Copper Oxide is Cu1O1
We would write as CuO
The Formula of water
To find the formula of water we simply split it up into the elements from which it is made.
We can do this by passing a dc electric current through it.
In the lab we use a piece of equipment called the Hoffman Voltameter.
Formula of water
Water doesn’t usually conduct because it has no charged particles.
It is covalent; not Ionic.
Adding some acid helps split it into ions.
H2O(l) --> H+(aq) + OH-(aq)
Ions
Positive ions are called Cations.
They are attracted to the negative electrode - the Cathode.
Negative ions are called Anions.
They are attracted to the positive electrode - the Anode.
Redox
When H+ ions go to the negative terminal (Cathode) they pick up electrons.
2H+(aq) + 2e- → H2(g)
This is Reduction
We can prove the gas made is Hydrogen with the “Pop Test”
Redox
When OH- ions go to the positive terminal (Anode) they lose electrons.
4OH-(aq) - 4e- → O2(g) + 2H2O(l)
This is Oxidation
We prove the gas is Oxygen by relighting a glowing splint
Formula
When we look at the gases made its obvious that twice as much Hydrogen is made as Oxygen.
This suggests a formula of H2O
But it doesn’t really prove it because some gas may dissolve
Formula
But combining the two half-equations shows that twice as many Hydrogen ions are produced when water splits.
2H+(aq) + 2e- → H2(g)
4OH-(aq) - 4e- → O2(g) + 2H2O(l)
We must have the same number of electrons in both half-equations so we can combine them.
4H+(aq) + 4e- → 2H2(g)
4OH-(aq) - 4e- → O2(g) + 2H2O(l)
4OH-(aq)- 4e - +4 H+(aq)+ 4e ---> 2H2O(l) + O2(g) + 2H2(g)
Water of Crystallisation
Many salts contain water as part of their structure.
Copper Sulphate is white when anhydrous (without water)
And blue when hydrated (containing water)
Finding the formula
If we weigh the hydrated salt and then weigh it again after removing the water by heating, we can work out how much water has been removed.
But we must be careful to completely dehydrate the salt without decomposing it.
The method
We would first have to weigh an empty crucible.
Then weigh the crucible and Hydrated Copper Sulphate
Then re-weigh when the salt had been dehydrated to form Anhydrous Copper Sulphate
Yet more maths!!!!
What's weighed? | Mass(g) | |
Empty Crucible | 35.45 | A |
Crucible and Hydrated salt | 37.95 | B |
Crucible and Anhydrous salt | 37.05 | C |
Mass of hydrated salt = B - A= 37.95 - 35.45 = 2.5 g
Mass of water lost = B - C = 37.95 - 37.05 = 0.9g
Moles
Moles = Mass ÷ Molar Mass
Molar Mass of Copper Sulphate (CuSO4)= 64 + 32 + (4 x 16) = 160
Molar Mass of Water (H2O) = (2 x 1) + 16 = 18
Moles of Water = 0.9 ÷ 18 = 0.05
Moles of CuSO4 = 1.6 ÷ 16 = 0.01
Ratio CuSO4 : H2O = 0.05:0.01 which is the same as 5:1
Formula of the Hydrated salt is therefore, CuSO4•5H2O