3.
Law of Conservation of mass
calculations
Reminder! What is chemistry?
Chemistry is basically taking 'stuff' (the
reactants) and changing it and separating out 'different stuff' (the
products).
In a chemical
change, the atoms of the reactants are rearranged to give the products.
The atoms remain the same elements BUT are
arranged or bonded in a different way in the products.
In the 'picture equation' above, just look at how the
copper, carbon, oxygen, hydrogen and sulfur atoms are changed in their
arrangement from reactants on the left, to products to the right of the arrow 
indicating the direction of chemical change.
So, what about the relative mass of all the products
compared to the total mass of the original reactants?
Read on .... !!!
 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
whatever the physical state of the reactants and products.
 The law of conservation of mass can also be stated as "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 if all the precipitate
settles out, because only colourless
sodium sulfate is left in solution.
copper sulfate + sodium hydroxide ===> copper
hydroxide + sodium sulfate
CuSO_{4}(aq) + 2NaOH(aq) ===> Cu(OH)_{2}(s)
+ Na_{2}SO_{4}(aq)
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(aq)} + 2NaOH_{(aq)} ===>
Cu(OH)_{2(s)}
+ Na_{2}SO_{4}_{(aq)}
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.
You can do exactly the same sort of experiment using the
same apparatus and procedure as above using lead nitrate solution and potassium
iodide solution. It doesn't matter which solution is in the little test tube or
the conical flask.
Both are colourless solutions BUT on mixing you get a
bright yellow precipitate of lead iodide, an obvious chemical change has
taken place.
And, again, you find the total mass at the start (unmixed
solutions) equals the total mass at the end (including the residual solution and
precipitate).
The chemistry of this demonstration is ..
lead nitrate + potassium iodide ===>
lead iodide + potassium nitrate
Pb(NO_{3})_{2}(aq) + 2KI(aq)
===> PbI_{2}(s) + 2KNO_{3}(aq)
Note: (i) Here I've included the state symbols, (aq)
meaning an aqueous solution (solvent water), and (s) to denote the solid
precipitate formed.
(ii) The two 2s are needed to balance the equation, a
consequence of the law of conservation of mass and the rules on balancing
chemical equations.
TOP OF PAGE
Experiments in which a mass change is observed

where there doesn't seem to be conservation of mass, we need an explanation!
Reactions that do not go to completion might give false
results and if the reaction involves a reacting gas OR a gaseous product
it is difficult to make accurate measurements to confirm the validity of the law
of conservation of mass.
Three typical situations you will encounter:
(i) If a solid reacts with a gas to give a solid
product you appear to gain mass e.g. heating a metal in air to form an
oxide, there is no direct way to measure the mass of oxygen used from the
air outside of the crucible to show the law of conservation of mass is
obeyed. You can only measure directly the mass of one of the reactants, but
not the other.
(ii) If a solid decomposes on strong heating to give a
solid and a gas e.g. decomposition of a metal carbonate to a metal oxide and
carbon dioxide, there is no direct way to measure the mass of carbon dioxide
lost to the air outside of the crucible/test tube to show the law of
conservation of mass is obeyed. This means you can only measure directly the
mass of one of the products, but not the other.
(iii) If solid reacts with an acid to form a gas e.g.
a metal + acid ==> hydrogen or a carbonate + acid ==> carbon dioxide, there
is no direct way to measure the mass of the gaseous products lost to the air
outside of the conical flask to show the law of conservation of mass is
obeyed. All you can measure is the mass of the products that were not gases.
BUT, as you will see, theoretical calculations
based on the law of conservation of mass can get round the situation!
Some reactions may appear to involve a change in mass as
measured with limited school laboratory apparatus. However, these can usually be
explained because a reactant or product is a gas and its mass not been taken
into account.
For example: when a metal reacts with oxygen the mass of
the oxide produced is greater than the mass of the metal, so when you heat
magnesium ribbon in a crucible there is gain in mass because the 'weightless'
oxygen in air has combined with the magnesium to form magnesium oxide. It is
impossible to directly measure the mass of the oxygen used from the air.
2Mg(s) + O_{2}(g) ====>
2MgO(s) mass gained
BUT, using the law of conservation of mass, you can do
a reacting mass calculation from the mass of magnesium oxide formed to find
out the mass of oxygen used in making the magnesium oxide.
Using the atomic mass: Mg = 24 and O = 16, you can
write the equation in terms of mass ...
(2 x 24) + (2 x 16) ====> 2 x (24 + 16),
both sides of the equation arrow should add up to 80, check it out!
In thermal decompositions of metal carbonates, the carbon
dioxide is produced and escapes into the atmosphere leaving the metal oxide as
the only solid product. Again, it is impossible to directly measure the carbon
dioxide evolved from the decomposing carbonate. However, you can clearly observe
the chemical change because of the colour changes.
green copper carbonate ====> black copper
oxide + colourless carbon dioxide.
CuCO_{3}(s) ====> CuO(s) +
CO_{2}(g) lost mass
BUT, again, using the law of conservation of mass, you
can do a reacting mass calculation from the residual mass of the oxide to
find out the mass of carbon dioxide produced by the thermal decomposition of
the carbonate e.g. copper carbonate decomposing heating to give solid copper
oxide and gaseous carbon dioxide.
Using atomic masses: Cu = 63.5, C = 12 and
O = 16 to show mass is conserved writing the equation in masses
{63.5 + 12 + (3 x 16)} ====> (63.5 + 16) +
{12 + (2 x 16)}, both sides of the equation add up to 123.5
An
experiment in observing mass loss  quantitative measurements
When an acid reacts with a metal, hydrogen gas is
produced, or reacting an acid with a carbonate produces carbon dioxide gas. The
gas escapes into the air. You can follow the mass loss by carrying out the
reaction in a flask placed on an electronic balance. You can even use the mass
loss to see how fast the reaction is going e.g.
hydrogen from a metalacid reaction
carbon dioxide from a carbonateacid reaction
illustration of a nonenclosed system
In the case above, carbon dioxide is produced when
limestone reacts with acid forming carbon dioxide gas which escapes through
the cotton wool plug.
The other products of the reaction remain in the
flask.
You can monitor the mass of the whole 'system' and
plot a graph of the decrease of mass with time (graph above).
You can then work out the actual mass loss by
subtracting the changing and decreasing total mass from the initial mass
of the 'system' at the start.
This gives you the lost mass of carbon dioxide gas
with time e.g. the graph below.
If you could weigh the CO_{2} formed, all would add up to
comply with the law of conservation of mass!
If a gas escapes from a nonenclosed reaction
vessel you cannot observe the law of conservation of mass, BUT, since we
have a verified scientific law (of mass conservation)
we can calculate mass
changes that we cannot observe directly.
No matter what happens, the law of conservation of
mass must be obeyed!
You need to be able to explain any observed changes in
mass in nonenclosed systems during a chemical reaction given the balanced
symbol equation for the reaction and explain these changes in terms of the
particle model.
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. The same
arguments applies to when a gas is a reactant producing a solid product.
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 i.e. as ratios to enable calculations to be
done with any given 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
of % yield
TOP OF PAGE
Examples of 'atom counting' to illustrate the law of
conservation of mass
(these are from my page on
how to balance chemical equations)
Any correctly balanced equation, especially in
diagrammatic form, illustrates the law of conservation of mass, ie the number of
atoms for each element MUST be the same on both sides of the equation. No atoms
lost or gained and neither do atoms change their atomic mass in a reaction.
ATOMS at the START = ATOMS at the END (atoms
conserved, just arranged differently!) and
TOTAL MASS REACTANTS = TOTAL MASS of PRODUCTS
(the law of conservation of mass)
You also need to be able to read chemical formula and
balance chemical equations, at least appreciate why an equation is balanced 
which after all is a symbolic or diagrammatic representation of the Law of
Conservation of Mass.
The 'Law of Conservation of Mass' means you can do
theoretical calculations on the relative amounts of reactants and products
involved in a chemical reaction, as long as you have the correct formulae and a
correctly balanced chemical equation.
Atomic masses are quoted and listed on a periodic table at
the bottom of the page. Incidentally, the examples are worked out in terms of
atomic masses, but in real calculations you may use g, kg or tonnes, as long as
you use the same mass units for each mass value involved.
AND you must be able to
working out formula masses
You will follow the arguments for the more complex
examples if you can already balance awkward equations!
Atom counting and balancing the
reactant mass units and product mass units from the equation ratios
I am assuming you can work out the
relative
formula mass of an element or compound.
(1) iron + sulfur ==>
iron sulfide
Fe + S ====> FeS
Atomic masses: Fe = 56, S = 32
One atom of each element on each side of the equation
Law of conservation of mass balance: 56 +
32 ====> 88, (calculation check 56 + 32 = 88)
Important note on units:
I haven't
specified the units of mass in these 7 examples BUT ...
... it
doesn't matter whether you are working in mass units of g, kg or tonnes!
as
long as you 'work' the final ratio numbers from the equation in the SAME
mass units.
(2)
sodium hydroxide + hydrochloric
acid ===> sodium chloride + water
NaOH + HCl ====> NaCl +
H_{2}O
Atomic masses: Na = 23, O = 16, H = 1, Cl = 35.5
one atom of Na, one atom of oxygen, two atoms of
hydrogen and one atom of chlorine on both sides of the equation
Law of conservation of mass balance: 40 +
36.5 ====> 58.5 + 18, (both sides equal 76.5 mass
units)
Note the subscript 2 after the H in water
means two atoms of that element.
(3) magnesium + hydrochloric acid
====> magnesium chloride + hydrogen
Mg +
2HCl ====> MgCl_{2} + H_{2}
Atomic masses: Mg =
24, H = 1, Cl = 35.5
one atom of Mg, 2 atoms of H and 2 atoms
of Cl on both sides of the equation
Law of conservation of mass balance: {24 + (2 x (1 +
35.5)} ===> {24 + (2 x 35.5)} + (2 x 1), (both sides equal 97)
Note the subscript 2 after the Cl in
magnesium chloride or the 2 after the H in the hydrogen molecule, means two
atoms of that element.
The 2 before the HCl doubles the number
of hydrochloric acid molecules.
(4) methane + oxygen ====>
carbon dioxide + water
CH_{4} + 2O_{2} ====> CO_{2} + 2H_{2}O
Atomic masses: C = 12, H = 1, O = 16
One atom of carbon, 4 atoms of hydrogen and four atoms
of oxygen on both sides of the equation
Law of conservation of mass balance: 16 + (2 x 32)
===> 44 + (2 x 18), both sides of equation equal 80 mass units
Note the 2 before the O_{2} and H_{2}O
doubles the number of these molecules to balance the equation.
The subscript 4 in methane means 4 atoms
of hydrogen in the methane molecule.
(5) copper carbonate + sulfuric acid
===> copper sulfate + water + carbon dioxide
CuCO_{3} +
H_{2}SO_{4} ====> CuSO_{4} + H_{2}O
+ CO_{2}
Atomic masses: Cu = 63.5, C = 12, O = 16,
H = 1, S = 32
Law of conservation of mass balance: one atom of
copper, one atom of carbon, one atom of sulfur and seven atoms of oxygen on
both sides of the equation, no atoms lost or gained, no mass lost or gained!
Law of conservation of mass balance: I'm assuming you
can work out formula masses by now.
(123.5) + (98) ====> (159.5) + (18) + (44),
both sides of equation equal 221.5 mass units in total
Reminders on reading a formula:
The formula H_{2}SO_{4} means 2 atoms
of H, 1 atom of S and four atoms of O
The formula CuCO_{3} means 1 atom of Cu, 1
atom of C and 3 atoms of O.
(6) magnesium hydroxide + nitric acid ====>
magnesium nitrate + water
Mg(OH)_{2}
+ 2HNO_{3} ====> Mg(NO_{3})_{2} + 2H_{2}O
Atomic masses: Mg = 24, O = 16, H = 1. N = 14
Law of conservation of mass balance: 1 atom of Mg, 8
atoms of O, 4 atoms of H and 2 atoms of N on both sides of the equation, no
atoms lost or gained, no mass lost or gained!
Work this one out for yourself using the formula
masses for practice.
Note that here the subscript 2 after the (NO_{3})
in the magnesium nitrate means everything in the brackets is doubled.
(7) aluminium oxide + sulfuric acid
====> aluminium sulfate + water
Al_{2}O_{3} + 3H_{2}SO_{4}
====> Al_{2}(SO_{4})_{3} + 3H_{2}O
Atomic masses: Al = 27, O = 16, H = 1, S = 32
Law of conservation of mass balance: 2 atoms of Al, 15
atoms of O, 6 atoms of H and 3 atoms of S on both sides of the equation, no
atoms lost or gained, no mass lost or gained!
Work this one out for yourself using the formula
masses for practice.
TOP OF PAGE
Examples of using the 'Law of Conservation of Mass' in
reacting mass calculations.
By converting the atoms or formulae into atomic masses or
formula/molecular masses you can then use the law of conservation of mass to
reacting mass calculations. (More reacting mass
calculations in section 6.)
 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 = 80, so 80 mass units of
reactants = producing 80 mass units of products.
 You can then state things like ..
 48 g of Mg gives in excess oxygen, 80 g of MgO
 or 24 tonnes of Mg produces 40 tonnes of MgO
 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.
 You are quite simply 'working' the
equation in terms of mass units, but taking into account the
balancing numbers in the symbol equation.
 
 Law of conservation of mass
calculation Example 3.2
 sodium + chlorine ===> sodium
chloride
 2Na + Cl_{2}
===> 2NaCl
(atomic masses required: Na=23 and
Cl=35.5)
 If 23 g of sodium is burned in excess
chlorine gas, theoretically, what mass of sodium chloride is
produced?
 The equation in terms of mass units is ...
 (2 x 23) + (2 x 35.5) ==> 2(23 +35.5) =
(2 x 58.5)
 46 + 71 = 117
 therefore, dividing by two: 23 + 35.5
==> 58.5
 So 58.5 g of sodium chloride is
formed.
 When you get more experienced, you can just
pick out the formula mass ratio required
 e.g 1 Na ==> 1 NaCl giving a
working ratio of 23 : 58.5
 
 Law of conservation of mass
calculation Example 3.3
 Law of conservation of mass
calculation Example 3.4
 iron +
sulphur ==> iron sulphide (see the diagram at the top of
the page!)
 Fe + S ==> FeS (atomic
masses: Fe = 56, S = 32)
 If 59 g of iron is heated with
32 g of sulphur to form iron sulphide, how much iron is left unreacted?
(assuming all the sulphur reacted)
 From the atomic masses, 56 g of
Fe combines with 32 g of S to give 88g FeS.
 This means 59  56 = 3 g Fe
unreacted.
 This is an example of a 'limiting reactant'
calculation  where the reactant NOT in excess limits the maximum
amount of product.
 More
on
how
much of a reactant is needed? limiting reactant
 
 Law of conservation of mass
calculation Example 3.5
 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 + (3 x 16)}
===> (40 + 16) + {12 + (2 x 16)}
 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.
 
 For more complicated examples and
more practice of calculations based on reacting masses in accordance
with the Law of Conservation of
Mass ...
 See
Reacting mass ratio calculations of reactants
& products from
equations (NOT using moles)
Selfassessment Quiz
on the 'Law of Conservation of Mass' and simple reacting mass calculations
QUIZ on Law of Conservation of Mass and
simple reacting mass calculations
Crossmultiplying  apparently, according to math departments, the naughty way'
to solve ratios!
As pupil in the late 1950s and (very) early1960s I was taught to solve ratios by crossmultiplying, wrote
learning!
Suppose you have the ratio situation of A : B and
C : D as in the
reacting mass ratio questions on this page.
You can also express these ratios as
Therefore, logically, by crossmultiplying you can
say that arithmetically: A x D = B x C
and rearranging, as you do in simple algebra you
get the following relationship by dividing through by A, B, C or D appropriately
A = B x C / D, B = A
x D / C, C = A x D / B and D = B x C / A
and if you don't believe me, just put some
numbers in e.g 2 : 5 for A : B and 6 : 15 for C : D
2/5 = 6/15 and 2 x 15 = 5 x 6
I find its by far the quickest general route to
solving two ratios that match, but its frowned on!
It does actually amount to the same as the methods described
above, personally, I just find it quicker!
Above is typical periodic table used in GCSE sciencechemistry specifications in
doing reacting mass and conservation of mass chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
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
QUIZ on Law of Conservation of Mass and
simple reacting mass calculations
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