What is the Law of Conservation of Mass?
...
before tackling the first calculations based on the Law of Conservation of
mass, its worth describing a simple experiment to demonstrate the validity of
the law. The experiment is illustrated in the diagram above. You prepare
solutions of copper sulfate (blue) and sodium hydroxide (colourless, light grey
in diagram!). The most impressive way to demonstrate this is to use a sealed
system on an accurate electronic one pan balance. You can use 50 cm3
of 1 molar copper sulfate solution and pour into conical flask. The
concentrated sodium hydroxide solution is suspended by a string in a suitable
container - small test tube or weighing/sample bottle. The whole lot is weighed
(fictitiously 67.25g) with the rubber bung on sealing the 'system'. Then,
releasing the bung and string, the sodium hydroxide container is lowered into
the copper sulfate solution and shaken gently to thoroughly mix the reactants.
The reaction is immediate and a dark blue precipitate of copper hydroxide is
formed and the solution eventually turns colourless because only colourless
sodium sulfate is left in solution. The recorded mass will still be 67.25g
showing that no mass was created or destroyed in the chemical reaction, though
to observe the law in action, you must do the experiment in a sealed system
where nothing can get in or get out i.e. no atoms have been gained or lost.
The
equation for this reaction is ...
copper
sulfate + sodium hydroxide ==> copper hydroxide + sodium sulfate
CuSO4
+ 2NaOH ===> Cu(OH)2 + Na2SO4
Teacher
note
50 cm3
of 1 molar copper sulfate = 1.0 x 50 / 1000 = 0.05 mol CuSO4, Mr(NaOH)
= 40, you need 2x 0.05 = 0.10 mol NaOH,
which
equals 0.10 x 40 = 4.0g NaOH pellets dissolved in the minimum volume of water,
4.1g should complete the precipitation.
NOTE that
in calculations ...
(1) the
symbol equation must be correctly balanced to get the right answer!
(2) You
convert all the formula in the equations into their formula masses AND take
into account any balancing numbers to get the true theoretical reactiing
masses.
(2) There
are good reasons why, when doing a real chemical preparation-reaction to make a
substance you will not get 100% of what you theoretically calculate.
- Law of conservation of mass
calculation Example 3.1
- Magnesium + Oxygen ==> Magnesium oxide
- 2Mg + O2 ==> 2MgO (atomic
masses required: Mg=24 and O=16)
- think of the ==> as an =
sign, so the mass changes in the reaction are:
- (2 x 24) + (2 x 16) = 2 x
(24 + 16)
- 48 + 32 = 2 x 40 and so 80
mass units of reactants = produces 80 mass units of products.
- You can work with any mass
units such as g, kg or tonne (1 tonne = 1000 kg), as long as you use the
same units for all the masses involved.
- Law of conservation of mass
calculation Example 3.2
- iron + sulphur ==> iron sulphide
(see the diagram at the top of the page!)
- Fe + S ==> FeS (atomic masses: Fe = 56, S
= 32)
- If 59g of iron is heated
with 32g of sulphur to form iron sulphide, how much iron is left
unreacted? (assuming all the sulphur reacted)
- From the atomic masses, 56g
of Fe combines with 32g of S to give 88g FeS.
- This means 59 - 56 = 3g Fe
unreacted.
- Law of conservation of mass
calculation Example 3.3
- When limestone (calcium
carbonate) is strongly heated, it undergoes thermal decomposition
to form lime (calcium oxide) and carbon dioxide gas.
- CaCO3 ==> CaO + CO2
(relative atomic masses: Ca = 40, C = 12 and O = 16)
- Calculate the mass of
calcium oxide and the mass of carbon dioxide formed by decomposing 50
tonnes of calcium carbonate.
- (40 + 12 + 3x16) ==> (40
+ 16) + (12 + 2x16)
- 100 ==> 56 + 44
- scaling down by a factor of
two
- gives
- 50 ==> 28 + 22
- so decomposing 50 tonnes of
limestone produces 28 tonnes of lime and 22 tonnes of carbon dioxide
gas.
- Example 3.4:
- For more complicated examples and more practice of calculations based on reacting masses in accordance with the Law of Conservation of Mass ...
credit to : doc brown chemistry as level
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