Reacting 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).
NOTE
the symbol equation must be correctly balanced to get the right answer!
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.
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Example 6a.1: 2Mg + O2 ==> 2MgO
o
(atomic masses Mg =24, O = 16)
o
converting the equation into reacting masses gives ... (2 x 24) + (2 x 16) ==> 2 x (24 +
16)
o
and this gives a basic reacting mass ratio of 48g Mg + 32g O2 ==> 80g MgO
o
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.
o
What you must be able to do is solve a ratio!
o
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
o
(atomic masses Na = 23, O = 16, H = 1, S = 32)
o
mass ratio is: (2 x 40) + (98) ==> (142) + (2 x 18) = (80) + (98) ==> (142) + (36),
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but pick the ratio needed to solve the particular problem
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o
e.g. reacting mass ratio of 2NaOH : Na2SO4 is 80 (2 x 40) : 142
(a) calculate how much sodium hydroxide is needed to make 5g of sodium
sulphate.
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
o
from the reacting mass equation: 142g Na2SO4 is formed from 80g of NaOH
o
5g Na2SO4 is formed from 5g x 80 / 142 = 2.82 g of NaOH by scaling down from 142
=> 5
o
(b) calculate how much water is formed when 10g of sulphuric acid reacts.
o
from the reacting mass equation: 98g of H2SO4 forms 36g of H2O
o
10g of H2SO4 forms 10g x 36 / 98 = 3.67g of H2O by scaling down from 98 => 10
Example 6a.3: 2CuO(s) + C(s) ==> 2Cu(s) + CO2(g)
o
(atomic masses Cu=64, O=16, C=12)
o
Formula Mass ratio is 2 x (64+16) + (12) ==> 2 x (64) + (12 + 2x16)
o
= Reacting mass ratio 160 + 12 ==> 128 + 44 (in the calculation, impurities are
ignored)
o
(a) In a copper smelter, how many tonne of carbon (charcoal, coke) is needed to
make 16 tonne of copper?
o
from the reacting mass equation: 12 of C makes 128 of Cu
o
scaling down numerically: mass of carbon needed = 12 x 16 / 128 = 1.5 tonne of C
o
(b) How many tonne of copper can be made from 640 tonne of copper oxide ore?
o
from the reacting mass equation: 160 of CuO makes 128 of Cu (or direct from
formula 80 CuO ==> 64 Cu)
o
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?
o
(atomic masses: Fe = 56, O = 16)
o
reaction equation: Fe2O3 + 3C ==> 2Fe + 3CO
o
formula mass Fe2O3 = (2x56) + (3x16) = 160
o
160 mass units of iron oxide reacts with 3 x 12 = 36 mass units of carbon
o
So the reacting mass ratio is 160 : 36
o
So the ratio to solve is 20 : x, scaling down, x = 36 x 20/160 = 4.5 tonne
carbon needed.
o
Note: Fe2O3 + 3CO ==> 2Fe + 3CO2 is the other most likely reaction that reduces the
iron ore to iron.
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Example 6a.5: (a) Theoretically how much copper can be obtained from 2000 tonne of
pure chalcopyrite ore, formula CuFeS2 ?
o
Chalcopyrite is a copper-iron sulphide compound and one of the most important
and common ores containing copper.
o
Atomic masses: Cu = 64, Fe = 56 and S = 32
o
For every one CuFeS2 ==> one Cu can be extracted, f. mass of ore = 64 + 56 + (2x32)
= 184
o
Therefore the reacting mass ratio is: 184 ==> 64
o
so, solving the ratio, 2000 CuFeS2 ==> 2000 x 64 / 184 Cu = 695.7 tonne
copper (max. can be extracted)
o
(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?
o
% yield = actual yield x 100/theoretical yield
o
% yield = 670.2 x 100 / 695.7 = 96.3%
Example 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 ]
o
The reduction equation is: Fe3O4 + 2C ==> 3Fe + 2CO2
o
Before doing the reacting mass calculation, you need to do simple calculation to
take into account the lack of purity of the ore.
o
76% of 1 tonne is 0.76 tonne (760 kg).
o
For the reacting mass ratio: 1 Fe3O4 ==> 3 Fe (you can ignore rest of equation)
o
Therefore in reacting mass units: (3 x 56) + (4 x 16) ==> 3 x 56
o
so, from the reacting mass equation: 232 Fe3O4 ==> 168 Fe
o
0.76 Fe3O4 tonne ==> x tonne Fe
o
solving the ratio, x = 0.76 x 168/232 = 0.55
o
= 0.55 tonne Fe (550 kg)/tonne (1000 kg) of magnetite ore
o
(b) What is the atom economy of the carbon reduction reaction?
o
You can use some of the data from part (a).
o
% atom economy = total mass of useful product x 100 / total mass of reactants
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o
= 168 x 100 / (232 + 2x12) = 168 x 100 / 256 = 65.6%
o
(c) Will the atom economy be smaller, the same, or greater, if the reduction
involves carbon monoxide (CO) rather than carbon (C)? explain?
o
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.
o
Fe3O4 + 4CO ==> 3Fe + 4CO2
o
so the atom economy = 168 x 100 / (232 + 4x28) = 48.8 %
Example 6a.7: 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.
o
(a) Give the symbol equation of the decomposition complete with state symbols.
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o
o
Ca(HCO3)2(aq) ==> CaCO3(s) + H2O(l) + CO2(g)
(b) Calculate 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 ]
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the relevant ratio is based on: Ca(HCO3)2 ==> CaCO3
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The formula masses are 162 (40x1 + 1x2 + 12x2 + 16x6) and 100 (40 + 12 +
16x3) respectively
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there the reacting mass ratio is 162 units of Ca(HCO3)2 ==> 100 units of
CaCO3
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the mass of Ca(HCO3)2 in 2000 cm3 (ml) = 2000 x 0.056 = 112 mg
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therefore solving the ratio 162 : 100 and 112 : z mg CaCO3

where z = unknown mass of calcium carbonate
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z = 112 x 100/162 = 69.1 mg CaCO3
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since 1g = 1000 mg, z = 69.1/1000 = 0.0691 g CaCO3, calcium carbonate
(c) Comment on the result, its consequences and why is it often referred to as
'limescale'?
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This precipitate 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'.
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Example 6a.8: 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.
o
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 = ?]
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(a) Give the equation for the reaction between the metal hydroxide and the
hydrochloric acid.
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MOH(aq) + HCl(aq) ==> MCl(aq) + H2O(l)
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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.
(b) Calculate the mass of HCl used in each titration.
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1 dm3 = 1000 cm3, so in 19.7 cm3 of the HCl solution there will be
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19.7 x 7.3 / 1000 = 0.1438 g HCl
(c) Calculate the mass of MOH that reacts with the mass of HCl calculated in
(b).
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25cm3 of the 250 cm3 MOH solution was used, so the mass of MOH
titrated is
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25 x 4 / 250 = 0.40 g MOH
(d) Calculate the formula mass of HCl.
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formula mass = 1 + 35.5 = 36.5
(e) Calculate the mass in g of MOH that reacts with 36.5g of HCl and hence
the formula mass of MOH.
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0.1438 g HCl reacts with 0.40 g MOH

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therefore 36.5g HCl reacts with z g of MOH

solving the ratio for z, z = 36.5 x 0.40 / 0.1438 = 101.5 g MOH
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Since the 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
(f) What is the atomic mass of the metal?
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
Since O = 16 and H = 1, total of 17, atomic mass of M = 101.5 - 17
= 84.5
(g) From the formula information on the metal hydroxide deduce the
following giving reasons:
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What group of the periodic table is M likely to be from?
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Which metal is M likely to be?
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Appendix 1 Solving Ratios
Group 1 Alkali Metals because they form singly charged
positive ions (they readily lose their outer electron to do so).
Rubidium. The atomic mass for rubidium is quoted as 85.5
and the experimental value of 84.5 is pretty close - don't
forget the experimental procedures will not be perfect,
especially at GCSE level !