LAW
of CONSERVATION of MASS
Doc
Brown's Chemistry  GCSE/IGCSE/GCE (basic A level)
O Level
Online Chemical Calculations
3.
Law of Conservation of Mass
Experiments and Reacting Mass Calculations
Quantitative Chemistry
calculations online A demonstration experiment
of the Law of Conservation of Mass is described and explained. Help for problem solving
in doing law of conservation of mass calculations, using experiment
data, making predictions. Practice revision
questions on the law of conservation of mass in chemical reactions using
the balanced equation. The Law of Conservation of
Mass is defined and explained using examples of reacting mass calculations using the law
are fully explained with worked out examples using the balanced symbol
equation. The method involves reacting masses deduced from the balanced
symbol equation. Online practice exam chemistry CALCULATIONS and solved
problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical
calculations for A level AS/A2/IB courses. These revision notes and practice
questions on the law
of conservation of mass chemical calculations and worked examples should prove useful for
the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.
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3.
Law of Conservation of mass
calculations
 What is the Law of Conservation of Mass?
 When elements and compounds react to form
new products, mass cannot be lost or gained.
 "The Law of Conservation of Mass"
definition states that "mass cannot be created
or destroyed, but changed into different forms".
 So, in a chemical change, the total mass of reactants must equal the total mass of products.
 The law of conservation of mass can also be stated "no
atoms can be lost or made in a chemical reaction", which is why the
total mass of products must equal the total mass of reactants you started
with.
 By using this law, together with atomic and formula masses, you can calculate the quantities of reactants and products involved in a
reaction and the simplest formula of a compound
 One consequence of the law of conservation of
mass is that In a balanced chemical symbol equation, the total of
relative formula masses of the reactants is equal to the total relative
formula masses of the products.
 You can see this in the examples worked out for
you ...
 ... so, this page just explains how to do simple
reacting mass calculations based on the reaction equation and applying the Law of
Conservation of Mass,
 but first, by at least one
clearly observed experiment, that the Law of Conservation of Mass holds
good, even in the humble school or college laboratory! see diagram below
and read on. ...
 See also Section 5.
which shows how to use this law to get to a compound's
formula too
... 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 and represents an
enclosed system, where nothing can escape !
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 cm^{3} 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
CuSO_{4} + 2NaOH ===> Cu(OH)_{2}
+ Na_{2}SO_{4}
Teacher note
50 cm^{3} of 1 molar copper sulfate = 1.0
x 50 / 1000 = 0.05 mol CuSO_{4}, M_{r}(NaOH) = 40, you need 2 x
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: If you carry out an experiment that produces
a gas in a nonenclosed system (i.e. the gas can escape), you will observe a
mass loss, BUT, if you could somehow weigh the gas, you would find that the
total mass of reactants and products had remained constant. So, even in
reactions producing a gas, the law of conservation still holds good.
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 reacting masses.
(2) There are good reasons why, when
doing a real chemical preparationreaction to make a substance you will not get
100% of what you theoretically calculate.
See discussion in section 14.2
 Law of conservation of mass
calculation Example 3.1
 Magnesium + Oxygen ==> Magnesium oxide
 2Mg + O_{2}
==> 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.
 CaCO_{3}
==>
CaO + CO_{2} (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 ...
Selfassessment Quizzes
[com] type in answer
for
F and H or
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F and H
OTHER CALCULATION PAGES

What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass

Calculating relative
formula/molecular mass of a compound or element molecule

Law of Conservation of Mass and simple reacting mass calculations
(this page)

Composition by percentage mass of elements
in a compound

Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)

Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination

Introducing moles: The connection between moles, mass and formula mass  the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)

Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)

Moles and the molar volume of a gas, Avogadro's Law

Reacting gas volume
ratios, Avogadro's Law
and GayLussac's Law (ratio of gaseous
reactantsproducts)

Molarity, volumes and solution
concentrations (and diagrams of apparatus)

How to do acidalkali
titration calculations, diagrams of apparatus, details of procedures

Electrolysis products calculations (negative cathode and positive anode products)

Other calculations
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy

Energy transfers in physical/chemical changes,
exothermic/endothermic reactions

Gas calculations involving PVT relationships,
Boyle's and Charles Laws

Radioactivity & halflife calculations including
dating materials
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