|
 6a. 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
(1)
HELP IN SOLVING Ratios - 'a ghastly scribble!' (I'm
typing up some 'neater' examples at the moment!)
(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
AND see also 7.
Reacting masses using moles and
molar ratio problem solving
AND see also 14.5
How
much of a reactant is needed?, which is essentially a particular
application of a reacting mass calculation
CALCULATING REACTING MASSES IN
CHEMICAL
REACTIONS
How to use reacting mass ratios
from a balanced chemical equation
How do we calculate mass of
products formed? How do we calculate mass of reactants needed?
Atomic masses are obtained from a
suitable copy of the periodic table
-
Reacting
mass calculation Example 6a.1
-
Burning magnesium to
form magnesium oxide
-
What mass of
magnesium oxide is formed on burning 2.00g of magnesium?
-
(atomic masses Mg =24, O = 16)
-
converting the equation into
reacting masses
gives ...
-
and this gives a
basic reacting mass ratio of
...
-
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.
-
|
You can solve reacting mass
problems (i.e. solve the ratio problem) with a series of
logical steps set out in a table illustrated and
explained below ... |
|
Pick out the particular
ratio you need to solve the reacting mass problem
e.g. the particular reactant and product |
comments |
|
2Mg |
==> |
2MgO |
only these bits of the
equation are needed to solve the problem, you can ignore
the rest of the equation |
|
2 x 24 = 48g |
==> |
2 x 40 = 80g |
basic reacting mass ratio
from the balanced symbol equation and the relevant
atomic or formula masses |
|
1g |
==> |
80/48 = 1.667g |
divide by 48 to scale down
to 1g of Mg reactant |
|
2 x 1 = 2g |
==> |
2 x 1.667 = 3.334g |
then scale up to 2g of Mg
reactant, x 2 factor |
|
Therefore on burning 2g of
magnesium you make 3.33g of magnesium oxide |
-
This reaction can be carried out
in a school or college laboratory using a ceramic crucible and lid.
-
The crucible & lid are weighed
(m1), a piece of magnesium ribbon added and the crucible weighed
again (m2).
-
The difference in weights gives
the mass of magnesium (m2 - m1).
-
You can then do a theoretical
calculation of how much magnesium oxide should be formed.
-
The crucible is placed on a clay
pipe triangle resting on a tripod above a bunsen burner.
-
The crucible is heated so the
magnesium reacts with the oxygen in air, removing the lid to allow
air to come in.
-
The crucible is cooled and
reweighed with the lid on (m3).
-
Strictly speaking, the procedure
should be repeated until no further gain in weight is observed.
-
The gain in weight is due to the
solid magnesium combining with oxygen gas from the air to give the
solid magnesium oxide.
-
The mass of magnesium oxide = m3
- m1, to compare with the theoretical prediction.
-
Another approach to this
experiment is to pretend you don't know the formula and deduce it
from the results.
-
Reacting mass calculation Example 6a.2
-
The neutralisation of
sulfuric acid with sodium hydroxide.
-
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 5.00g 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
-
|
You can solve reacting mass problems
(i.e. solve the ratio problem) with a series of logical
steps set out in a table illustrated which I've
explained below ... |
|
Pick out the particular
ratio you need to solve the reacting mass problem
e.g. the particular reactant and product |
comments |
|
2NaOH |
==> |
Na2SO4 |
only these bits of the
equation are needed to solve the problem, you can ignore
the rest of the equation |
|
2 x 40 = 80 g |
==> |
142 g |
basic reacting mass ratio
from the balanced symbol equation and the relevant
atomic or formula masses |
|
80/142 = 0.563 g |
==> |
142/142 = 1.0 g |
divide by 142 to scale down
to 1g of Na2SO4 product |
|
5.0 x 0.563 = 2.82g |
==> |
5.0 x 1.0 = 5.00g |
then scale up to the 5g of
Na2SO4 product by multiplying x5 |
|
Therefore you need 2.82g
of sodium hydroxide to make 5.00g of sodium sulfate |
-
(b) calculate how much water is formed when 10g of sulphuric
acid reacts with sodium hydroxide.
-
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 =>
10
|
You can solve reacting mass
problems (i.e. solve the ratio problem) with a series of
logical steps set out in a table illustrated and
explained below ... |
|
Pick out the particular
ratio you need to solve the reacting mass problem
e.g. the particular reactant and product |
comments |
|
H2SO4 |
==> |
2H2O |
only these bits of the
equation are needed to solve the problem, you can ignore
the rest of the equation |
|
98g |
==> |
2 x 18 = 36g |
basic reacting mass ratio
from the balanced symbol equation and the relevant
atomic or formula masses |
|
1g |
==> |
36/98 = 0.367g |
divide by 98 to scale down
to 1g of H2SO4 reactant |
|
10 x 1 = 10g |
==> |
10 x 0.367 = 3.67g |
then scale up to the 10g of
H2SO4 reactant, factor x10 |
|
Therefore 3.67g of water
is formed when 10g of sulfuric acid reacts with sulfuric
acid |
-
-
-
Reacting
mass calculation Example 6a.3
-
The reduction of
copper(II) oxide by heating with carbon
-
2CuO(s) + C(s)
==> 2Cu(s) + CO2(g)
-
(atomic masses Cu=64, O=16,
C=12)
-
Formula Mass ratio is 2 x (64+16) + (12)
==> 2 x (64) + (12 + 2x16)
-
= Reacting mass
ratio 160 + 12 ==> 128 + 44
-
(a) In a copper smelter, how many tonne of
carbon (charcoal, coke) is needed to make 16.00 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
|
You can solve reacting mass
problems (i.e. solve the ratio problem) with a series of
logical steps set out in a table illustrated and
explained below ... |
|
Pick out the particular
ratio you need to solve the reacting mass problem
e.g. the particular reactant and product |
comments |
|
C |
==> |
2Cu |
only these bits of the
equation are needed to solve the problem, you can ignore
the rest of the equation |
|
12 tonne |
==> |
2 x 64 = 128 tonne |
basic reacting mass ratio
from the balanced symbol equation and the relevant
atomic or formula masses |
|
12/128 = 0.09375 tonne |
==> |
128/128 = 1.0 tonne |
divide by 128 to scale down
to 1 tonne of copper product |
|
16 x 0.09375 = 1.50
tonne |
==> |
16 x 1.0 = 16 tonne |
then scale up to the 16
tonne of Cu product, the factor is x 16 |
|
Therefore 1.50 tonne of
carbon is needed to reduce 16 tonne of copper oxide
to copper |
-
(b) How many tonne of copper can be made from
800 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 = 800 x 128 / 160 = 640 tonne Cu
|
You can solve reacting mass
problems (i.e. solve the ratio problem) with a series of
logical steps set out in a table illustrated and
explained below ... |
|
Pick out the particular
ratio you need to solve the reacting mass problem
e.g. the particular reactant and product |
comments |
|
2CuO |
==> |
2Cu |
only these bits of the
equation are needed to solve the problem, you can ignore
the rest of the equation |
|
2 x 80 = 160 tonne |
==> |
2 x 64 = 128 tonne |
basic reacting mass ratio
from the balanced symbol equation and the relevant
atomic or formula masses |
|
1 tonne |
==> |
128/160 = 0.80 tonne |
divide by 160 to scale down
to 1 tonne of CuO reactant |
|
800 x 1 = 800 tonne |
==> |
800 x 0.80 = 640
tonne |
then scale by a factor of
800 for the copper oxide reactant |
|
Therefore 640 tonne of
copper can be extracted from 800 tonne of copper(II)
oxide |
-
-
-
Reacting mass calculation Example 6a.4
-
The reduction of iron
oxide ore in a furnace by heating with carbon
-
What
mass of carbon is required to reduce 20.0 tonne of iron(III)
oxide ore if carbon monoxide is formed in the process as
well as iron?
-
(atomic
masses: Fe = 56, O = 16)
-
reaction equation:
Fe2O3
+ 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
: 36
-
So the ratio to solve is
20 :
x,
scaling down, x = 36 x 20/160 = 4.5 tonne carbon needed.
-
Note: Fe2O3
+ 3CO ==> 2Fe + 3CO2 is the other most likely reaction that
reduces the iron ore to iron.
|
You can solve reacting mass
problems (i.e. solve the ratio problem) with a series of
logical steps set out in a table illustrated and
explained below ... |
|
Pick out the particular
ratio you need to solve the reacting mass problem
e.g. the particular reactant and product |
comments |
|
Fe2O3 |
+ |
3C |
only these bits of the
equation are needed to solve the problem, you can ignore
the rest of the equation |
|
160 tonne |
+ |
3 x 12 = 36 tonne |
basic reacting mass ratio
from the balanced symbol equation and the relevant
atomic or formula masses |
|
160/160 = 1.0 tonne |
+ |
36/160 = 0.225 tonne |
divide by 160 to scale down
to 1 tonne of iron oxide reactant |
|
20 x 1.0 = 20.0 tonne |
+ |
20 x 0.225 = 4.5 tonne |
then scale up by factor of
20 |
|
Therefore 4.5 tonne of
carbon is needed to reduce 20 tonne of the iron
oxide |
-
-
-
Reacting
mass calculation Example
6a.5
-
The production of
copper from a copper ore
-
(a) Theoretically how much copper can be obtained from
2000 tonne of pure chalcopyrite ore, formula CuFeS2 ?
-
Chalcopyrite
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, formula mass of ore = 64 + 56 + (2x32) =
184
-
Therefore the reacting
mass ratio is: 184 ==> 64
-
so, solving the
ratio ...
|
You can solve reacting mass
problems (i.e. solve the ratio problem) with a series of
logical steps set out in a table illustrated and
explained below ... |
|
Pick out the particular
ratio you need to solve the reacting mass problem
e.g. the particular reactant and product |
comments |
|
CuFeS2 |
==> |
Cu |
only these bits of the
equation are needed to solve the problem, you can ignore
the rest of the equation |
|
184 tonne |
==> |
64 tonne |
basic reacting mass ratio
from the balanced symbol equation and the relevant
atomic or formula masses |
|
184/184 = 1.0 tonne |
==> |
64/184 = 0.3478 tonne |
divide by 184 to scale down
to 1 tonne of copper ore reactant |
|
2000 x 1.0 = 2000 tonne |
==> |
2000 x 0.0.3478 = 695.6
tonne |
then scale up by factor of
2000 for the initial 2000 tonne of ore |
|
Therefore 695.6 tonne of
copper can be extracted from 2000 tonne of
chalcopyrite copper ore |
-
(b) If only
670.2 tonne of pure copper is finally obtained after further
purification of the extracted copper by electrolysis, what is the % yield of
the overall process?
-
Reacting
mass calculation Example
6a.6
-
Calculating the
theoretical yield of iron from an impure iron oxide ore.
-
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 ]
-
The reduction
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)
-
Therefore in
reacting mass units: (3 x 56) + (4 x 16) ==> 3 x 56
-
so, from the
reacting mass equation: 232 Fe3O4
==> 168 Fe
-
0.76 Fe3O4
tonne ==> x tonne Fe
-
solving the
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 =
100 x total mass of useful product / total mass of products
-
Because of the law on
conservation of mass, total mass reactants = total mass of reactants
-
% atom economy = 100 x
total mass of useful product / total mass of products
-
It doesn't matter which
version you use for the atom economy calculation, you should get the
same answer!
-
= 100 x 168 /
(232 + 2x12) = 100 x 168/256 = 65.6%
-
More on % yield and atom economy in calculations section 14.
-
-
-
(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.
-
Fe3O4
+ 4CO ==> 3Fe + 4CO2
-
so the atom
economy = 100 x 168 / (232 + 4x28) = 48.8 %
-
Note reactants mass (232
+ 4x28) = (3x56 + 4x44) products mass
-
More on % yield and atom economy in calculations section 14.
-
-
-
Reacting mass
calculation 6a.7, including calculating atom economy and percentage
yield
-
The displacement
reaction between iron and copper(II) sulfate solution
-
(a) Write out the
equation for this reaction
-
You can write the
equation out in several different ways, and any can be used to do a
reacting mass calculation
-
Fe + CuSO4
==> Cu + FeSO4 (simple 'molecular'
equation)
-
Fe(s) + CuSO4(aq)
==> Cu(s) + FeSO4(aq) (with state symbols)
-
Fe + Cu2+
==> Cu + Fe2+ (ionic equation)
-
Fe(s) + Cu2+(aq)
==> Cu(s) + Fe2+(aq) (ionic equation with state
symbols)
-
-
-
(b) Calculate the
maximum amount of copper that can be displaced by 2.8g of iron?
-
Atomic masses: Fe = 56,
Cu = 64, S = 32, O =16
-
From the equation 1 atom
of iron displaces 1 atom of copper,
-
therefore 56g of iron
will displace 64g of copper
-
therefore 2.8g of iron
will displace x g of copper
-
by scaling down
-
x = 2.8 x 64/56 = 3.2
g Cu can be theoretically displaced
-
-
-
(c) What is the atom
economy of the reaction based on the ionic equation?
-
(d) If only 2.7 g of
pure copper was recovered from the experiment, what was the % yield?
-
Reacting mass
calculation 6a.8, including calculating atom economy and percentage
yield
-
The displacement
reaction between copper and silver nitrate solution.
-
The equation for this
reaction is
-
Cu + 2AgNO3
==> Cu(NO3)2 + 2Ag
-
atomic masses: Cu = 64,
Ag = 108, N = 14, O = 16
-
-
-
(a) What is the atom
economy of the reaction?
-
You can add up the total
mass of reactants or total mass of products, its all the same
(because of the law of conservation of mass!)
-
Based on atomic mass
units
-
mass of reactants = 64 +
2 x [108 + 14 + (3 x 16)] = 64 + (2 x 170) = 404
-
mass of useful product =
2 x 108 = 216
-
atom economy =
100 x 216/ = 100 x 216/404 = 53.5 %
-
More on % yield and atom economy in calculations section 14.
-
-
-
(b) 500g of copper was
used to displace silver from a silver nitrate solution.
-
(i) What is the
maximum
amount of silver that could be obtained from the process?
-
From the equation 1 atom
Cu ==> 2 atoms silver
-
therefore 64g Cu ==> 2 x
108 g Ag
-
64g Cu ==> 216g Ag
-
500 g Cu ==> x g Ag
-
x = 500 x 216/64 =
1687.5 g Ag
-
-
-
(ii) If 1.5 kg g of pure
silver was extracted, what was the percentage yield of silver?
-
Reacting mass
calculation 6a.9 The thermal decomposition of a carbonate.
-
You can do simple experiments
with copper carbonate, magnesium carbonate or zinc carbonate, both
of which readily decompose on strong heating in a crucible to leave
an oxide residue ('MO') and give off carbon dioxide gas (CO2).
-
You can make predictions as to
how much mass loss will occur as the carbonate compounds
decompose and the carbon dioxide driven off.
-
The detailed chemistry is covered
on
thermal
decomposition of compounds
-
The three similar equations are,
with the corresponding relative reacting masses (atomic masses from
periodic table) ...
-
CuCO3 ===>
CuO + CO2
-
MgCO3 ===>
MgO + CO2
-
ZnCO3 ===>
ZnO + CO2
-
You can then weigh a crucible
(m1).
-
Add a few grams of the carbonate
and reweigh the crucible (m2)
-
Mass of carbonate = m2 - m1.
-
The crucible is placed on a clay
pipe triangle resting on a tripod above a bunsen burner.
-
Heat the crucible strongly with a
bunsen burner, cool and reweigh (m3)
-
Mass of oxide left = m2 - m3
-
Your predictions are not likely
to be that accurate, particularly with copper carbonate, because its
quite difficult to get these three compounds in a purely carbonate
form.
TOP OF PAGE
Appendix 1 Solving Ratios
(a)
spotting the multiplying factor - another version of the method above
For ... Ex 6a1. 2Mg + O2 ==> MgO
and Ex 6a2a./2b. 2NaOH + H2SO4 ==> Na2SO4
+ 2H2O
(b)
cross-multiplying - apparently, according to maths departments, the naughty way'
to solve ratios!
As pupil in the late 1950s and (very) early1960s I was taught to solve ratios by cross-multiplying, 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 cross-multiplying you
get 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!
TOP OF PAGE
Self-assessment Quizzes on reacting mass calculations
(i)
QUIZ on reacting mass ratio calculations of
reactants and products
Other related calculation
pages
AND see also 7.
Reacting masses using moles and
molar ratio problem solving
AND see also 14.5
How
much of a reactant is needed?, which is essentially a particular
application of a reacting mass calculation
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
-
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), mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination
(this page, see also section 14)
-
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 Gay-Lussac's Law (ratio of gaseous
reactants-products)
-
Molarity, volumes and solution
concentrations (and diagrams of apparatus)
-
How to do acid-alkali
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 & half-life calculations including
dating materials
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