REACTING MASS CALCULATIONS

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

6a. Reacting mass chemical calculation methods and equations (NOT using moles)

Also some examples of % yield and atom economy calculations

Quantitative Chemistry calculations online Calculations involving reacting masses of reactants and calculating masses of products formed based on balanced equations for chemical reactions. How to solve problems using reacting mass and product mass ratio. Here revision help for problem solving in doing reacting mass calculations, using experiment data, making predictions. Exemplar practice revision questions on reacting masses using balanced chemical equations are fully described. This page describes and explains, with fully worked out examples, the methods of calculating the mass of reactants or the mass of products involved in a chemical reaction using the ratio information 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 reacting mass calculations in chemistry and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses. These revision notes and practice questions on how to do reacting mass ratio chemical calculations and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.

Spotted any careless error? EMAIL query ? comment or request a type of GCSE calculation not covered?

Self-assessment Quizzes on reacting mass calculations: type in answer QUIZ  or  multiple choice QUIZ

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?

• e.g. by heating magnesium metal ribbon in a crucible.

• 2Mg + O2 ==> 2MgO

• (atomic masses Mg =24, O = 16)

• converting the equation into reacting masses 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.

• BUT all units must be the same e.g. all masses in grams, all masses in kg or all masses in tonnes.

• 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),

• but pick the ratio needed to solve the particular problem

• e.g. reacting mass ratio of 2NaOH : Na2SO4 is 80 (2 x 40) : 142

• (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 and 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 by 5 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 x 10 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

• (in the calculation, impurities are ignored)

• (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 ...

• 2000 CuFeS2 ==> 2000 x 64 / 184 Cu = 695.7 tonne copper

• This is the maximum amount of copper that can be theoretically extracted from the 'pure' ore.

• In reality there are impurities in the ore (e.g. other minerals) and in the extracted molten copper.

 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 = 00 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?

• % yield = 100 x actual yield/maximum theoretical yield

• % yield = 100 x 2.7/3.2 = 84.4%

• -

• 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

• The three similar equations are, with the corresponding relative reacting masses (atomic masses from periodic table) ...

• CuCO3  ===>  CuO  +  CO2

• In terms of reacting formula masses

• (63.5 + 12 + 3 x 16)  ===> (63.5 + 16)  +  (12 + 2 x 16)

•  123.5  ===>  79.5  +  44

• MgCO3  ===>  MgO  +  CO2

• In terms of reacting formula masses

• (24 + 12 + 3 x 16)  ===> (24 + 16)  +  (12 + 2 x 16)

•  84  ===>  40  +  44

• ZnCO3  ===>  ZnO  +  CO2

• In terms of reacting formula masses

• (65 + 12 + 3 x 16)  ===> (65 + 16)  +  (12 + 2 x 16)

•  125  ===>  81  +  44

• 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.

• Quite often, these carbonates, are a mixture of the carbonate and hydroxide.

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 early (very) 1960s 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

 A C – = – B D

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!

Self-assessment Quizzes on reacting mass calculations

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

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