masses and ratios in chemical calculations
(not using moles)
You can use the ideas of relative atomic,
molecular or formula mass AND the law of conservation of mass
to do quantitative calculations in chemistry. Underneath an equation you can add the appropriate atomic or formula
masses. This enables you to see what mass of what,
reacts with what mass of other reactants. It also allows you to predict what mass of
products are formed (or to predict what is needed to make so much of a
particular product). You must take into account the balancing numbers in the
equation (e.g. 2Mg), as well of course, the numbers in the formula (e.g. O2).
HELP IN SOLVING RATIO'S and
see also section 7. using moles
(2) the symbol equation must be
correctly balanced to get the right answer!
(3) 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.
See discussion in section 14.2
(4) See 6b.
for solution concentration and titration
calculations based on reacting masses NOT involving moles
Example 6a.1: 2Mg + O2 ==> 2MgO
(atomic masses Mg =24, O = 16)
converting the equation into
gives ... (2 x 24) + (2 x 16) ==> 2 x (24 + 16)
and this gives a
basic reacting mass ratio of
48g Mg + 32g O2 ==> 80g MgO
The ratio can be used, no matter what the
units, to calculate and predict quite a lot! and you don't necessarily
have to work out and use all the numbers in the ratio.
What you must be able
to do is solve a ratio!
e.g. 24g Mg will make 40g MgO, why?, 24 is half of
48, so half of 80 is 40.
Example 6a.2: 2NaOH + H2SO4 ==> Na2SO4 + 2H2O
(atomic masses Na = 23, O = 16, H = 1, S = 32)
mass ratio is: (2 x 40) + (98)
==> (142) + (2 x 18) = (80) + (98)
==> (142) + (36),
(a) calculate how much sodium hydroxide is needed
to make 5g of sodium sulphate.
from the reacting
mass equation: 142g Na2SO4 is formed
from 80g of NaOH
5g Na2SO4 is formed from
5g x 80 / 142 = 2.82 g of NaOH by scaling down from 142 => 5
(b) calculate how much water is formed when 10g of sulphuric
from the reacting
mass equation: 98g of H2SO4 forms
36g of H2O
10g of H2SO4 forms
10g x 36 / 98 = 3.67g of H2O by scaling down from 98 =>
Example 6a.3: 2CuO(s) + C(s)
==> 2Cu(s) + CO2(g)
(atomic masses Cu=64, O=16,
Formula Mass ratio is 2 x (64+16) + (12)
==> 2 x (64) + (12 + 2x16)
= Reacting mass
ratio 160 + 12 ==> 128 + 44 (in the calculation, impurities are
(a) In a copper smelter, how many tonne of
carbon (charcoal, coke) is needed to make 16 tonne of copper?
from the reacting
mass equation: 12 of C makes 128 of Cu
scaling down numerically: mass of carbon needed
= 12 x 16 / 128 = 1.5 tonne of C
(b) How many tonne of copper can be made from
640 tonne of copper oxide ore?
from the reacting
mass equation: 160 of CuO makes 128 of Cu (or direct from
formula 80 CuO ==> 64 Cu)
scaling up numerically: mass copper formed =
128 x 640 / 160 = 512 tonne Cu
Example 6a.4: What
mass of carbon is required to reduce 20 tonne of iron(II)
oxide ore if carbon monoxide is formed in the process as
well as iron?
masses: Fe = 56, O = 16)
+ 3C ==> 2Fe + 3CO
formula mass Fe2O3
= (2x56) + (3x16) = 160
160 mass units of iron oxide
reacts with 3 x 12 = 36 mass units of carbon
So the reacting mass ratio is 160
So the ratio to solve is 20 :
scaling down, x = 36 x 20/160 = 4.5 tonne carbon needed.
+ 3CO ==> 2Fe + 3CO2 is the other most likely reaction that
reduces the iron ore to iron.
6a.5: (a) Theoretically how much copper can be obtained from
2000 tonne of pure chalcopyrite ore, formula CuFeS2 ?
is a copper-iron sulphide compound
and one of the most important and common ores containing copper.
Atomic masses: Cu
= 64, Fe = 56 and S = 32
For every one CuFeS2
==> one Cu can be extracted, f. mass of ore = 64 + 56 + (2x32) =
Therefore the reacting
mass ratio is: 184 ==> 64
so, solving the
ratio, 2000 CuFeS2 ==> 2000 x 64 / 184 Cu = 695.7 tonne copper
(max. can be
(b) If only
670.2 tonne of pure copper is finally obtained after further
purification by electrolysis, what is the % yield of
the overall process?
% yield = actual
yield x 100/theoretical yield
% yield = 670.2 x
100 / 695.7 = 96.3%
More on % yield in calculations section 14.1
6a.6: A sample of magnetite iron ore contains 76%
of the iron oxide compound Fe3O4 and 24% of
waste silicate minerals. (a) What is the maximum theoretical mass of
iron that can be extracted from each tonne (1000 kg) of magnetite
ore by carbon reduction? [ Atomic masses: Fe = 56, C = 12 and O
= 16 ]
equation is: Fe3O4 + 2C ==> 3Fe + 2CO2
Before doing the
reacting mass calculation, you need to do simple calculation to take
into account the lack of purity of the ore.
76% of 1 tonne
is 0.76 tonne (760 kg).
For the reacting
mass ratio: 1 Fe3O4 ==> 3 Fe (you can
ignore rest of equation)
reacting mass units: (3 x 56) + (4 x 16) ==> 3 x 56
so, from the
reacting mass equation: 232 Fe3O4
==> 168 Fe
tonne ==> x tonne Fe
ratio, x = 0.76 x 168/232 = 0.55
= 0.55 tonne
Fe (550 kg)/tonne (1000 kg) of magnetite ore
(b) What is
the atom economy of the carbon reduction reaction?
You can use some
of the data from part (a).
% atom economy =
total mass of useful product x 100 / total mass of reactants
= 168 x 100 /
(232 + 2x12) = 168 x 100 / 256 = 65.6%
(c) Will the
atom economy be smaller, the same, or greater, if the reduction
involves carbon monoxide (CO) rather than carbon (C)? explain?
The atom economy
will be smaller because CO is a bigger molecular/reactant
mass than C and 4 molecules would be needed per 'molecule' of Fe3O4,
so the mass of reactants is greater for the same product mass of
iron (i.e. bottom line numerically bigger, so % smaller). This is
bound to be so because the carbon in CO is already chemically bound
to some oxygen and can't remove as much oxygen as carbon itself.
+ 4CO ==> 3Fe + 4CO2
so the atom
economy = 168 x 100 / (232 + 4x28) = 48.8 %
Atom economy is fully explained
in calculations section 14.2b
On analysis, a sample of hard water was found to contain 0.056 mg of
calcium hydrogen carbonate per cm3 (0.056 mg/ml). If the
water is boiled, calcium hydrogencarbonate Ca(HCO3)2,
decomposes to give a precipitate of calcium carbonate CaCO3,
water and carbon dioxide.
(a) Give the
symbol equation of the decomposition complete with state symbols.
the mass of calcium carbonate in grammes deposited if 2 litres (2 dm3,
2000 cm3 or ml) is boiled in a kettle. [ atomic
masses: Ca = 40, H = 1, C = 12, O = 16 ]
ratio is based on: Ca(HCO3)2 ==> CaCO3
masses are 162 (40x1 + 1x2 + 12x2 + 16x6) and 100 (40
+ 12 + 16x3) respectively
reacting mass ratio is 162 units of Ca(HCO3)2
==> 100 units of CaCO3
the mass of
Ca(HCO3)2 in 2000 cm3 (ml) = 2000 x
0.056 = 112 mg
solving the ratio 162 : 100 and 112 : z mg CaCO3
where z =
unknown mass of calcium carbonate
z = 112 x
100/162 = 69.1 mg CaCO3
since 1g = 1000
mg, z = 69.1/1000 = 0.0691 g CaCO3, calcium carbonate
on the result, its consequences and why is it often referred to as
of calcium carbonate will cause a white/grey deposit to be formed on
the side of the kettle, especially on the heating element. Although
0.0691 g doesn't seem much, it will build up appreciably after many
cups of tea! The precipitate is calcium carbonate, which occurs
naturally as the rock limestone, which dissolved in rain
water containing carbon dioxide, to give the calcium hydrogen
carbonate in the first place. Since the deposit of 'limestone'
builds up in layers it is called 'limescale'.
This is a much more elaborate reacting mass calculation involving
solution concentrations and extended ideas from the results. In
this exemplar Q I've used the formulae a lot for short-hand.
A solution of
hydrochloric contained 7.3 g HCl/dm3. A solution of a
metal hydroxide of formula MOH was prepared by dissolving 4.0g of
MOH in 250 cm3 of water. M is an unknown metal but
it is known that the ionic formula of the hydroxide is M+OH-.
25cm3 samples of the MOH solution were pipetted into a
conical flask and titrated with the hydrochloric solution using a
burette and a few drops of phenolphthalein indicator. All the MOH is
neutralised as soon as the pink indicator colour disappears (i.e.
the indicator becomes colourless). On average 19.7 cm3 of
the HCl acid solution was required to completely neutralise 25.0 cm3
of the MOH solution. [Atomic masses: H = 1, Cl = 35.5, O = 16, M =
the equation for the reaction between the metal hydroxide and the
+ HCl(aq) ==> MCl(aq) + H2O(l)
You may or may
not be required to give the state symbols in (), or you may be just
asked to complete the equation given part of it.
Calculate the mass of HCl used in each titration.
Calculate the mass of MOH that reacts with the mass of HCl
calculated in (b).
Calculate the formula mass of HCl.
Calculate the mass in g of MOH that reacts with 36.5g of HCl and
hence the formula mass of MOH.
0.1438 g HCl
reacts with 0.40 g MOH
36.5g HCl reacts with z g of MOH
ratio for z, z = 36.5 x 0.40 / 0.1438 = 101.5 g MOH
formula mass of HCl is 36.5 and from the equation, 1 MOH reacts with
1 HCl the experimental formula mass of MOH is found to be 101.5
is the atomic mass of the metal?
the formula information on the metal hydroxide deduce the following
Appendix 1 Solving Ratios
type in answer
F and H or
F and H
OTHER CALCULATION PAGES
What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass
formula/molecular mass of a compound or element molecule
Law of Conservation of Mass and simple reacting mass calculations
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
moles), mention of actual percent % yield and theoretical yield,
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
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 Gay-Lussac's Law (ratio of gaseous
Molarity, volumes and solution
concentrations (and diagrams of apparatus)
do volumetric titration calculations e.g. acid-alkali titrations
(and diagrams of apparatus)
Electrolysis products calculations (negative cathode and positive anode products)
e.g. % purity, % percentage & theoretical yield, volumetric titration
apparatus, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
Energy transfers in physical/chemical changes,
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
Radioactivity & half-life calculations including
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