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what mass of co is required to reduce the iron

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Check out what is available? Study the different examples then try the Quizzes! REACTING MASS CALCULATIONS

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

study examples carefully 6a. Reacting mass chemical adding methods using equations (but Not using moles)

Also some examples of % yield and cantlet economic system calculations

Quantitative Chemistry calculations online Calculations involving reacting masses of reactants and calculating masses of products formed based on counterbalanced 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 information, 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 computing the mass of reactants or the mass of products involved in a chemical reaction using the ratio information from the counterbalanced symbol equation. Online practice examination chemistry CALCULATIONS and solved issues for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level Equally/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 chemic calculations and worked examples should evidence useful for the new AQA, Edexcel and OCR GCSE (9�1) chemical science science courses.

Spotted whatever careless error? EMAIL query ? comment or asking a type of GCSE calculation not covered?

QUIZ on reacting mass ratio calculations of reactants and products

study examples carefully 6a. Reacting masses and ratios in chemical calculations (not using moles)study examples carefully

You can use the ideas of relative atomic, molecular or formula  mass AND the law of conservation of mass to do quantitative calculations in chemical science. Underneath an equation you can add the appropriate diminutive or formula masses. This enables you to see what mass of what, reacts with what mass of other reactants. It also allows y'all to predict what mass of products are formed (or to predict what is needed to brand so much of a particular production). You must take into account the balancing numbers in the equation (e.one thousand. 2Mg), as well of course, the numbers in the formula (east.one thousand. O2 ).

Annotation

(1) Aid IN SOLVING Ratios - 'a ghastly scribble !' (I'g typing up some 'neater' examples at the moment!)

(two) the symbol equation must exist correctly counterbalanced to get the correct answer!

(3) At that place are good reasons why, when doing a real chemical preparation-reaction to make a substance you will not become 100% of what you theoretically calculate. See word in section 14.2

(4) See 6b. for solution concentration and titration calculations based on reacting masses NOT involving moles

AND encounter also 7. Reacting masses using moles and tooth ratio problem solving

AND see besides 14.5 How much of a reactant is needed?, which is substantially a detail awarding of a reacting mass calculation

CALCULATING REACTING MASSES IN CHEMICAL REACTIONS

How to use reacting mass ratios from a balanced chemical equation

How practise we calculate mass of products formed? How do nosotros summate mass of reactants needed?

Atomic masses are obtained from a suitable re-create of the periodic tabular array

  • Reacting mass calculation Instance 6a.one

    • Burning magnesium to form magnesium oxide

    • What mass of magnesium oxide is formed on burning 2.00g of magnesium?

      • east.thou. by heating magnesium metallic ribbon in a crucible.

      • 2Mg + O2 ==> 2MgO

    • (diminutive masses Mg =24, O = 16)

    • converting the equation into reacting masses gives ...

      • (two x 24) + (ii ten sixteen) ==> 2 x (24 + xvi)

    • and this gives a basic reacting mass ratio of ...

      • 48g Mg + 32g O2 ==> 80g MgO

    • The ratio tin can be used, no affair what the units, to summate and predict quite a lot! and you don't necessarily have to piece of work out and use all the numbers in the ratio.

      • BUT all units must be the aforementioned 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 volition make 40g MgO, why?, 24 is half of 48, so half of 80 is twoscore.

    • Y'all tin can solve reacting mass problems (i.e. solve the ratio problem) with a series of logical steps set up out in a table illustrated and explained beneath ...
      Pick out the detail ratio you demand to solve the reacting mass problem e.g. the particular reactant and production comments
      2Mg ==> 2MgO only these $.25 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 = one.667g divide by 48 to scale downwardly to 1g of Mg reactant
      2 10 ane = 2g ==> 2 x 1.667 = 3.334g and so calibration upwardly to 2g of Mg reactant, ten two cistron
      Therefore on burning 2g of magnesium y'all make 3.33g of magnesium oxide

    • This reaction can exist carried out in a school or college laboratory using a ceramic crucible and hat.

    • The crucible & lid are weighed (m1), a slice of magnesium ribbon added and the crucible weighed again (m2).

    • The divergence in weights gives the mass of magnesium (m2 - m1).

    • You can then practise a theoretical adding of how much magnesium oxide should exist formed.

    • The crucible is placed on a clay pipe triangle resting on a tripod to a higher place a bunsen burner.

    • The crucible is heated so the magnesium reacts with the oxygen in air, removing the lid to permit air to come up in.

    • The crucible is cooled and reweighed with the lid on (m3).

    • Strictly speaking, the procedure should be repeated until no farther 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.

      • For this come across empirical formula and formula mass of a compound from reacting masses

  • Reacting mass calculation Example 6a.2

    • The neutralisation of sulfuric acid with sodium hydroxide.

    • 2NaOH + HtwoSOfour ==> Na2So4 + 2HiiO

    • (atomic masses Na = 23, O = 16, H = i, South = 32)

    • mass ratio is: (2 x 40) + (98) ==> (142) + (2 x xviii) = (80) + (98) ==> (142) + (36),

      • simply pick the ratio needed to solve the item trouble

        • e.m. reacting mass ratio of 2NaOH : Na2SO4 is fourscore (ii 10 40) : 142

    • (a) calculate how much sodium hydroxide is needed to make five.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 tin solve reacting mass problems (i.e. solve the ratio problem) with a serial of logical steps fix out in a tabular array illustrated which I've explained beneath ...
      Pick out the particular ratio you need to solve the reacting mass problem e.m. the particular reactant and product comments
      2NaOH ==> Na2And sofour only these bits of the equation are needed to solve the trouble, y'all tin can ignore the balance of the equation
      2 x twoscore = 80 grand ==> 142 thousand basic reacting mass ratio from the counterbalanced symbol equation and the relevant atomic or formula masses
      lxxx/142 = 0.563 g ==> 142/142 = 1.0 1000 split up by 142 to scale down to 1g of Na2SO4 product
      5.0 x 0.563 = 2.82g ==> v.0 x 1.0 = v.00g then calibration upwards to the 5g of NaiiSO4 product past multiplying x5
      Therefore you lot need 2.82g of sodium hydroxide to brand 5.00g of sodium sulfate

    • (b) summate how much h2o is formed when 10g of sulphuric acid reacts with sodium hydroxide.

    • from the reacting mass equation: 98g of  HtwoSO4 forms 36g of H2O

    • 10g of  H2SO4 forms 10g ten 36 / 98 = three.67g of H2O past scaling down from 98 => 10

      You tin can solve reacting mass problems (i.east. solve the ratio problem) with a series of logical steps set out in a table illustrated and explained beneath ...
      Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product comments
      H2And so4 ==> 2HiiO just these bits of the equation are needed to solve the problem, you can ignore the rest of the equation
      98g ==> two x xviii = 36g basic reacting mass ratio from the counterbalanced symbol equation and the relevant diminutive or formula masses
      1g ==> 36/98 = 0.367g divide past 98 to scale downwardly to 1g of HiiSO4 reactant
      10 x 1 = teng ==> 10 ten 0.367 = three.67g and then scale upward to the 10g of HtwoSOiv reactant, factor x10
      Therefore 3.67g of water is formed when 10g of sulfuric acid reacts with sulfuric acid

    • -

  • Reacting mass adding Example 6a.3

    • The reduction of copper(2) oxide by heating with carbon

    • 2CuO(s) + C(s) ==> 2Cu(due south) + CO2(g)

    • (diminutive masses Cu=64, O=16, C=12)

    • Formula Mass ratio is 2 x (64+16) + (12) ==> two 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 brand sixteen.00 tonne of copper?

    • from the reacting mass equation: 12 of C makes 128 of Cu

    • scaling downwards numerically: mass of carbon needed = 12 ten 16 / 128 = 1.5 tonne of C

      You can solve reacting mass problems (i.due east. solve the ratio problem) with a series of logical steps fix out in a table illustrated and explained beneath ...
      Pick out the particular ratio you lot need to solve the reacting mass problem e.g. the detail reactant and product comments
      C ==> 2Cu just these bits of the equation are needed to solve the problem, you tin ignore the residuum of the equation
      12 tonne ==> two x 64 = 128 tonne basic reacting mass ratio from the balanced symbol equation and the relevant diminutive or formula masses
      12/128 = 0.09375 tonne ==> 128/128 = one.0 tonne split by 128 to calibration downwards to i tonne of copper production
      sixteen x 0.09375 = i.l tonne ==> xvi ten 1.0 = xvi tonne then scale upwards to the 16 tonne of Cu production, the factor is x 16
      Therefore 1.50 tonne of carbon is needed to reduce xvi tonne of copper oxide to copper

    • (b) How many tonne of copper tin exist made from 800 tonne of copper oxide ore?

    • from the reacting mass equation: 160 of CuO makes 128 of Cu (or direct from formula lxxx CuO ==> 64 Cu)

    • scaling up numerically: mass copper formed = 800 10 128 / 160 = 640 tonne Cu

      You can solve reacting mass issues (i.due east. solve the ratio problem) with a series of logical steps set out in a tabular array illustrated and explained below ...
      Pick out the particular ratio you demand to solve the reacting mass problem east.thousand. the particular reactant and product comments
      2CuO ==> 2Cu only these bits of the equation are needed to solve the problem, you can ignore the residual of the equation
      2 10 80 = 160 tonne ==> 2 10 64 = 128 tonne basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses
      i tonne ==> 128/160  = 0.80 tonne carve up past 160 to calibration down to 1 tonne of CuO reactant
      800 x 1 = 800 tonne ==> 800 ten 0.80 = 640 tonne and so scale by a gene 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 past 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 fe?

      • (diminutive masses: Fe = 56, O = xvi)

      • reaction equation: FetwoO3 + 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 downward, x = 36 ten 20/160 = 4.5 tonne carbon needed.

      • Note: Fe2O3 + 3CO ==> 2Fe + 3CO2 is the other most probable reaction that reduces the atomic number 26 ore to fe.

      You can solve reacting mass problems (i.due east. solve the ratio trouble) with a series of logical steps prepare out in a table illustrated and explained below ...
      Option out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product comments
      Iron2O3 + 3C just these bits of the equation are needed to solve the problem, yous can ignore the remainder of the equation
      160 tonne + 3 10 12 = 36 tonne basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses
      160/160 = one.0 tonne + 36/160  = 0.225 tonne divide past 160 to scale down to one tonne of fe oxide reactant
      20 x 1.0 = twenty.0 tonne + 20 x 0.225 = iv.5 tonne then calibration up past factor of 20
      Therefore 4.5 tonne of carbon is needed to reduce 20 tonne of the iron oxide

    • -

  • Reacting mass adding 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 CuFeStwo ?

      • Chalcopyrite is a copper-fe sulphide compound and one of the most important and common ores containing copper.

      • Atomic masses: Cu = 64, Iron = 56 and S = 32

      • For every 1 CuFeS2 ==> one Cu can be extracted, formula mass of ore = 64 + 56 + (2x32) = 184

      • Therefore the reacting mass ratio is: 184 ==> 64

      • then, solving the ratio ...

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

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

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

      You can solve reacting mass bug (i.due east. solve the ratio problem) with a serial 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 eastward.g. the particular reactant and product comments
      CuFeStwo ==> Cu only these bits of the equation are needed to solve the problem, you tin ignore the residual of the equation
      184 tonne ==> 64 tonne basic reacting mass ratio from the counterbalanced symbol equation and the relevant diminutive or formula masses
      184/184 = i.0 tonne ==> 64/184  = 0.3478 tonne carve up past 184 to scale down to 1 tonne of copper ore reactant
      2000 x one.0 = 2000 tonne ==> 2000 x 0.0.3478 = 695.six tonne then scale upward by factor of 2000 for the initial 2000 tonne of ore
      Therefore 695.half dozen tonne of copper can be extracted from 2000 tonne of chalcopyrite copper ore

    • (b) If only 670.two tonne of pure copper is finally obtained later on further purification of the extracted copper by electrolysis, what is the % yield of the overall procedure?

      • % yield = actual yield x 100/theoretical yield

      • % yield = 100 ten 670.2 / 695.7 = 96.3%

      • More on % yield and atom economy in calculations section 14.

      • -

  • 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 Fe3Ofour and 24% of waste material 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? [ Diminutive masses: Atomic number 26 = 56, C = 12 and O = 16 ]

      • The reduction equation is: IronthreeO4 + 2C ==> 3Fe + 2CO2

      • Before doing the reacting mass calculation, you lot demand to do simple adding to take into account the lack of purity of the ore.

      • 76% of i tonne is 0.76 tonne (760 kg).

      • For the reacting mass ratio:  one Fe3Ofour ==> 3 Iron (you can ignore rest of equation)

      • Therefore in reacting mass units: (three ten 56) + (4 x 16) ==> iii x 56

      • so, from the reacting mass equation:  232 Fe3O4 ==> 168 Fe

      • 0.76 Fe3O4 tonne  ==> 10 tonne Fe

      • solving the ratio, x = 0.76 x 168/232 = 0.55

      • = 0.55 tonne Iron (550 kg)/tonne (m kg) of magnetite ore

      • -

    • (b) What is the cantlet economy of the carbon reduction reaction?

      • You tin utilize some of the data from function (a).

      • % atom economy = 100 x full mass of useful product / total mass of products

      • Because of the law on conservation of mass, total mass reactants = full mass of reactants

      • % cantlet economic system = 100 ten full mass of useful production / total mass of products

      • It doesn't matter which version you utilise for the atom economy adding, yous should get the same reply!

      • = 100 x 168 / (232 + 2x12) = 100 x 168/256 = 65.half dozen%

      • More on % yield and atom economy in calculations section 14.

      • -

    • (c) Will the cantlet economy be smaller, the same, or greater, if the reduction involves carbon monoxide (CO) rather than carbon (C)? explain?

      • The atom economy volition exist smaller considering CO is a bigger molecular/reactant mass than C and four molecules would be needed per 'molecule' of Iron3Ofour, so the mass of reactants is greater for the same product mass of atomic number 26 (i.e. lesser line numerically bigger, and 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 equally carbon itself.

      • FeiiiOiv + 4CO ==> 3Fe + 4COii

      • and then the cantlet economy = 100 x 168 / (232 + 4x28) = 48.8 %

      • Note reactants mass (232 + 4x28) = (3x56 + 4x44) products mass

      • More on % yield and cantlet economy in calculations section fourteen.

      • -

  • Reacting mass calculation 6a.7, including calculating atom economic system and per centum yield

    • The displacement reaction betwixt iron and copper(Two) sulfate solution

    • (a) Write out the equation for this reaction

      • You can write the equation out in several different means, and whatever can be used to do a reacting mass calculation

      • Iron + CuSO4 ==> Cu + FeSO4    (unproblematic 'molecular' equation)

      • Fe(southward) + CuSO4(aq) ==> Cu(s) + FeSOfour(aq)  (with country symbols)

      • Fe + Cuii+ ==> Cu + Fetwo+   (ionic equation)

      • Fe(southward) + Cuii+(aq) ==> Cu(s) + Fe2+(aq)  (ionic equation with state symbols)

      • -

    • (b) Calculate the maximum corporeality of copper that can be displaced by 2.8g of fe?

      • Atomic masses: Iron = 56, Cu = 64, South = 32, O =16

      • From the equation 1 atom of fe displaces 1 cantlet of copper,

      • therefore 56g of iron volition displace 64g of copper

      • therefore 2.8g of iron volition displace x m of copper

      • by scaling down

      • x = 2.8 x 64/56 = 3.2 g Cu can be theoretically displaced

      • -

    • (c) What is the cantlet economy of the reaction based on the ionic equation?

      • Based on atomic mass units

      • total mass of products = 64 + 56 = 120

      • mass of useful product = 64

      • cantlet economy = 100 x 64/120 = 53.three %

      • More than on % yield and atom economy in calculations section 14.

      • -

    • (d) If only 2.7 1000 of pure copper was recovered from the experiment, what was the % yield?

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

      • % yield = 100 x 2.7/3.2 = 84.4%

      • -

  • Reacting mass calculation 6a.8, including computing atom economy and percentage yield

    • The displacement reaction between copper and silver nitrate solution.

    • The equation for this reaction is

      • Cu + 2AgNO3 ==> Cu(NO3)ii + 2Ag

      • atomic masses: Cu = 64, Ag = 108, N = xiv, O = 16

      • -

    • (a) What is the atom economy of the reaction?

      • You can add upward the total mass of reactants or total mass of products, its nonetheless (considering of the law of conservation of mass!)

      • Based on diminutive mass units

      • mass of reactants = 64 + two x [108 + 14 + (3 x 16)] = 64 + (2 x 170) = 404

      • mass of useful product = ii x 108 = 216

      • atom economic system = 100 x 216/  =  100 ten 216/404 = 53.5 %

      • More than on % yield and atom economy in calculations section 14.

      • -

    • (b) 500g of copper was used to displace silverish from a argent 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 chiliad Ag

        • 64g Cu ==> 216g Ag

        • 500 grand Cu ==> x g Ag

        • x = 500 x 216/64 = 1687.5 g Ag

        • -

      • (ii) If 1.5 kg g of pure argent was extracted, what was the percentage yield of silver?

        • one.5 kg = 1500g

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

        • % yield = 100 x 1500/1687.5 = 88.9 % yield of silverish

        • More than on % yield and cantlet economy in calculations section xiv.

        • -

  • Reacting mass adding 6a.9 The thermal decomposition of a carbonate.

    • You can exercise simple experiments with copper carbonate, magnesium carbonate or zinc carbonate, both of which readily decompose on strong heating in a crucible to exit an oxide rest ('MO') and requite off carbon dioxide gas (COtwo).

      • You can brand predictions as to how much mass loss volition occur every bit the carbonate compounds decompose and the carbon dioxide driven off.

      • The detailed chemistry is covered on thermal decomposition of compounds

    • The 3 similar equations are, with the corresponding relative reacting masses (atomic masses from periodic table) ...

      • CuCO3  ===>  CuO  +  CO2

        • In terms of reacting formula masses

        • (63.five + 12 + 3 x sixteen)  ===> (63.5 + 16)  +  (12 + ii x 16)

        •  123.five  ===>  79.5  +  44

      • MgCO3  ===>  MgO  +  CO2

        • In terms of reacting formula masses

        • (24 + 12 + three x sixteen)  ===> (24 + 16)  +  (12 + 2 x 16)

        •  84  ===>  40  +  44

      • ZnCO3  ===>  ZnO  +  CO2

        • In terms of reacting formula masses

        • (65 + 12 + 3 10 16)  ===> (65 + 16)  +  (12 + 2 10 16)

        •  125  ===>  81  +  44

    • You lot can and so weigh a crucible (m1).

    • Add together a few grams of the carbonate and reweigh the crucible (m2)

    • Mass of carbonate = m2 - m1.

    • The crucible is placed on a clay piping triangle resting on a tripod above a bunsen burner.

    • Heat the crucible strongly with a bunsen burner, absurd and reweigh (m3)

    • Mass of oxide left = m2 - m3

    • Your predictions are not probable 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.

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Elevation OF Folio

Appendix i Solving Ratios

(a) spotting the multiplying factor - another version of the method above

For ... Ex 6a1. 2Mg + Otwo ==> MgO

and Ex 6a2a./2b. 2NaOH + H2SOfour ==> NaiiSOiv + 2H2O

(b) cross-multiplying - evidently, according to maths departments, the naughty mode' to solve ratios!

As student in the late 1950s and (very) early1960s I was taught to solve ratios by cross-multiplying, wrote learning!

Suppose y'all take the ratio situation of A : B and C : D every bit in the reacting mass ratio questions on this page.

Y'all can likewise express these ratios as

Therefore, logically, by cross-multiplying y'all get A ten D = B x C

and rearranging, every bit you practise in simple algebra you get the following relationship by dividing through by A, B, C or D appropriately

A = B 10 C / D, B = A x D / C, C = A ten D / B  and D = B x C / A

and if you don't believe me, just put some numbers in east.yard two : v for A : B and half dozen : 15  for C : D

2/5 = half-dozen/15 and 2 x 15 = 5 x 6

I discover its by far the quickest general route to solving 2 ratios that match, only its frowned on!

Information technology does actually amount to the same as the methods described above, personally, I just find it quicker!

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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 trouble solving

AND run across also xiv.five How much of a reactant is needed?, which is essentially a detail awarding of a reacting mass calculation

OTHER CALCULATION PAGES

  1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

  2. Computing relative formula/molecular mass of a compound or element molecule

  3. Law of Conservation of Mass and simple reacting mass calculations

  4. Composition by percentage mass of elements in a chemical compound

  5. Empirical formula and formula mass of a compound from reacting masses (easy outset, non using moles)

  6. Reacting mass ratio calculations of reactants and products from equations (Non using moles), mention of actual percentage % yield and theoretical yield, atom economy and formula mass determination (this page, see also section fourteen)

    • Reacting masses, concentration of solution and volumetric titration calculations (Non using moles)

  7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

  8. Using moles to summate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % limerick)

  9. Moles and the tooth volume of a gas, Avogadro's Constabulary

  10. Reacting gas volume ratios, Avogadro'due south Law and Gay-Lussac'due south Constabulary (ratio of gaseous reactants-products)

  11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

  12. How to practice acid-alkali titration calculations, diagrams of apparatus, details of procedures

  13. Electrolysis products calculations (negative cathode and positive anode products)

  14. Other calculations e.1000. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

    • fourteen.one % purity of a product xiv.2a % reaction yield 14.2b atom economy xiv.3 dilution of solutions

    • 14.iv h2o of crystallisation adding  fourteen.five how much of a reactant is needed?

  15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

  16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

  17. Radioactivity & half-life calculations including dating materials

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