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)

This page describes and explains, with fully worked out examples, methods of calculating the mass of reactants or the mass of products involved in a chemical reaction using the information from the balanced symbol equation.

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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 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 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.

  • 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),

    • 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 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 acid reacts.

  • 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

• Example 6a.3: 2CuO_(s) + C_(s) ==> 2Cu_(s) + CO_2(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 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?

  • (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.

• Example 6a.5: (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, f. 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 (max. can be extracted)

  • (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

• 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 ]

  • 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 = 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.

  • Fe₃O₄ + 4CO ==> 3Fe + 4CO₂

  • so the atom economy = 168 x 100 / (232 + 4x28) = 48.8 %

  • Atom economy is fully explained in calculations section 14.2b

• 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.

  • (a) Give the symbol equation of the decomposition complete with state symbols.

    • Ca(HCO₃)_2(aq) ==> CaCO_3(s) + H₂O_(l) + CO_2(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 ]

    • the relevant ratio is based on: Ca(HCO₃)₂ ==> CaCO₃

    • The formula masses are 162 (40x1 + 1x2 + 12x2 + 16x6) and 100 (40 + 12 + 16x3) respectively

    • there the 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

    • therefore 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

  • (c) Comment on the result, its consequences and why is it often referred to as 'limescale'?

    • 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'.

• 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.

  • 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 = ?]

    • (a) Give the equation for the reaction between the metal hydroxide and the hydrochloric acid.

      • MOH_(aq) + 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.

    • (b) Calculate the mass of HCl used in each titration.

      • 1 dm³ = 1000 cm³, so in 19.7 cm³ of the HCl solution there will be

      • 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).

      • 25cm³ of the 250 cm³ MOH solution was used, so the mass of MOH titrated is

      • 25 x 4 / 250 = 0.40 g MOH

    • (d) Calculate the formula mass of HCl.

      • 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.

      • 0.1438 g HCl reacts with 0.40 g MOH

      • 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

      • 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?

      • 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:

      • What group of the periodic table is M likely to be from?

        • Group 1 Alkali Metals because they form singly charged positive ions (they readily lose their outer electron to do so).

      • Which metal is M likely to be?

        • 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 !

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