Which is the correct equation to use when performing calculations involving the dilution of a solution?

calculating dilutions - volumes involved etc.

• In conjunction with this page it be important to study

• It is important to know how to accurately dilute a more concentrated solution to a specified solution of lower concentration. It involves a bit of logic using ratios of volumes. The diagram above illustrates some of the apparatus that might be used when dealing with solutions.

• Dilution calculation Example 14.3.1

• A purchased standard solution of sodium hydroxide had a concentration of 1.0 mol/dm3.

  • (a) How would you prepare 100 cm3 of a 0.1 mol/dm3 solution to do a titration of an acid?

    • The required concentration is 1/10th of the original solution.

    • To make 1dm3 (1000 cm3) of the diluted solution you would take 100 cm3 of the original solution and mix with 900 cm3 of water.

    • The total volume is 1dm3 but only 1/10th as much sodium hydroxide in this diluted solution,

      • so the concentration is 1/10th, 0.1 mol/dm3.

    • To make only 100 cm3 of the diluted solution you would dilute 10cm3 by mixing it with 90 cm3 of water.

    • How to do this in practice is described at the end of Example 14.3.2 below and a variety of accurate/'less inaccurate' apparatus is illustrated above.

  • 
    (b) Given a standard concentrated solution of hydrochloric acid of concentration 2.0 mol/dm3, how much of this solution in cm3, is needed to make up 250 cm3 of a more dilute standard solution of concentration 0.20 mol/dm3? Also, briefly describe the procedure to make this solution.

    • molarity = moles / volume, mol = molarity x volume, volume = moles / molarity

    • 1st consider the solution to be made up:

      • 250 cm3 = 250/1000 = 0.25 dm3

      • mol HCl needed = 0.20 x 0.25 = 0.05 mol HCl

    • 2nd consider the original standard solution:

      • volume = moles / molarity = 0.05 / 2.0 = 0.025 dm3

      • Therefore, volume of original standard solution needed is 0.025 x 1000 = 25.0 cm3.

    • This would be measured out with a 50 ml burette or better, a calibrated 25.0 cm3 pipette, and carefully transferred into a 250 cm3 calibrated flask and topped up with deionised water to the calibration mark.

    • For more details see steps

      
      and
      
      in section 11(d) How to make up a standard solution

  • -

• Dilution calculation Example 14.3.2

  • Given a stock solution of sodium chloride of 2.0 mol/dm3, how would you prepare 250cm3 of a 0.5 mol/dm3 solution?

  • The required 0.5 mol/dm3 concentration is 1/4 of the original concentration of 2.0 mol/dm3.

  • To make 1dm3 (1000 cm3) of a 0.5 mol/dm3 solution you would take 250 cm3 of the stock solution and add 750 cm3 of water.

  • Therefore to make only 250 cm3 of solution you would mix 1/4 of the above quantities

    • i.e. mix 62.5 cm3 of the stock solution plus 187.5 cm3 of pure water.

  • This can be done, but rather inaccurately, using measuring cylinders and stirring to mix the two liquids in a beaker.

  • It can be done much more accurately by using a burette or a pipette to measure out the stock solution directly into a 250 cm3 graduated-volumetric flask.

  • Topping up the flask to the calibration mark (meniscus should rest on it). Then putting on the stopper and thoroughly mixing it by carefully shaking the flask holding the stopper on at the same time!

  • For picture details see  11(d) How to make up a standard solution

  • -

• Dilution calculation Example 14.3.3

• 
  In the analytical laboratory of a pharmaceutical company a laboratory assistant was asked to make 250 cm3 of a 2.0 x 10-2 mol dm-3 (0.02M) solution of paracetamol (C8H9NO2).

  • (a) How much paracetamol should the laboratory assistant weigh out to make up the solution?

    • Atomic masses: C = 12, H = 1, N = 14, O = 16

    • method (i): Mr(paracetamol) = (8 x 12) + (9 x 1) + (1 x 14) + (2 x 16) = 151

      • 1000 cm3 of 1.0 molar solution needs 151g

      • 1000 cm3 of 2.0 x 10-2 molar solution needs 151 x 2.0 x 10-2/1 = 3.02g

      • (this is just scaling down the ratio from 151g : 1.0 molar)

      • Therefore to make 250 cm3 of the solution you need 3.02 x 250/1000 = 0.755 g

      • -

    • method (ii): Mr(paracetamol) = 151

      • moles = molarity x volume in dm3

      • mol paracetamol required = 2.0 x 10-2 x 250/1000 = 5.0 x 10-3 (0.005)

      • mass = mol x Mr = 5.0 x 10-3 x 151 = 0.755 g

      • -

  • (b) Using the 2.0 x 10-2 molar stock solution, what volume of it should be added to a 100cm3 volumetric flask to make 100 cm3 of a 5.0 x 10-3 mol dm-3 (0.005M) solution?

    • The ratio of the two molarities is stock/diluted = 2.0 x 10-2/5.0 x 10-3 = 4.0 or a dilution factor of 1/4 (0.02/0.005).

    • Therefore 25 cm3 (1/4 of 100) of the 2.0 x 10-2 molar solution is added to the 100 cm3 volumetric flask prior to making it up to 100 cm3 with pure water to give the 5.0 x 10-3 mol dm-3 (0.005M) solution.

    • There are more questions involving molarity in

      •  

        
        section 7. introducing molarity and
        
        section 12. on dilution

      • -

• Dilution calculation Example 14.3.4

  • You are given a stock solution of concentrated ammonia with a concentration of 17.9 mol dm-3 (conc. ammonia! ~18M!)

  • (a) What volume of the conc. ammonia is needed to make up 1dm3 of 1.0 molar ammonia solution?

    • Method (i) using simple ratio argument.

      • The conc. ammonia must be diluted by a factor of 1.0/17.9 to give a 1.0 molar solution.

      • Therefore you need (1.0/17.9) x 1000 cm3 = 55.9 cm3 of the conc. ammonia.

      • If the 55.9 cm3 of conc. ammonia is diluted to 1000 cm3 (1 dm3) you will have a 1.0 mol dm-3 (1M) solution.

      • -

    • Method (ii) using molar concentration equation - a much better method that suits any kind of dilution calculation involving molarity.

    • molarity = mol / volume (dm3), therefore mol = molarity x volume in dm3

    • Therefore you need 1.0 x 1.0 = 1.0 moles of ammonia to make 1 dm3 of 1.0M dilute ammonia.

    • Volume = mol / molarity

    • Volume of conc. ammonia needed = 1.0 / 17.9 = 0.0559 dm3 (55.9 cm3) of the conc. ammonia is required,

    • and, if this is diluted to 1 dm3, it will give you a 1.0 mol dm-3 dilute ammonia solution.

    • -

  • (b) What volume of conc. ammonia is needed to make 5 dm3 of a 1.5 molar solution?

    • molarity = mol / volume (dm3), therefore mol = molarity x volume in dm3

    • Therefore you need 1.5 x 5 = 7.5 moles of ammonia to make 5 dm3 of 1.5M dilute ammonia.

    • Volume (of conc. ammonia needed) = mol / molarity

    • Volume of conc. ammonia needed = 7.5 / 17.9 = 0.419 dm3 (419 cm3) of the conc. ammonia is required,

    • and, if this is diluted to 5 dm3, it will give you a 1.5 mol dm-3 dilute ammonia solution.

    • -

• Dilution calculation 14.3.5

  • Oleic has the structural formula CH3(CH2)7CH=CH(CH2)7COOH, molecular formula C18H34O2

  • It is not very soluble in water, and solutions of it are often prepared using an organic solvent like petroleum ether.

  • (a) Calculate the molecular mass of oleic acid (Relative atomic masses C = 12, H = 1, O =16)

    • Mr = (18 x 12) + (34 x 1) + (2 x 16) = 282

  • (b) A stock solution of oleic acid in a solvent has a molarity of 4.0 x 10-4 mol dm-3.

    • 2 cm3 of this solution is added to 8 cm3 of the solvent to make a more dilute solution.

    • 1 cm3 of this diluted solution is further diluted and mixed with 9.0 cm3 of the same solvent.

    • What is the final molarity of the doubly diluted solution?

      • The first dilution is 2 ==> 10 (or 1 ==> 5 by ratio)

      • so concentration = 4.0 x 10-4 / 5 = 0.8 x 10-4 = 8.0 x 10-5 mol dm-3

      • The 2nd dilution is 1 ==> 10 by ratio

      • so concentration = 8.0 x 10-5 / 10  = 8.0 x 10-6 mol dm-3 (final molarity)

      • -

  • (c) What mass of oleic acid would be in 5.0 cm3 of this final diluted solution?

    • Molarity is mol per dm3 (litre) or mol per 1000 cm3 (ml)

    • Therefore in each cm3 of the final dilution there are 8.0 x 10-6 / 1000 = 8.0 x 10-9 mol

    • Therefore in 5.0 cm3 there are 5 x 8.0 x 10-9 =  40.0 x 10-9 = 4.0 x 10-8 mol

    • mol = mass (g) / Mr,   so mass = mol x Mr,

    • Therefore mass of oleic acid in the 5.0 cm3 of solution

      • = 4.0 x 10-8 x 282 = = 1128 x 10-8 = 1.128 x 10-5 = 1.1 x 10-5 g (= 11 µg, 2 s.f.)

Above is typical periodic table used in GCSE science-chemistry specifications in doing chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

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

2. Calculating 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 compound

5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

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 calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

9. Moles and the molar volume of a gas, Avogadro's Law

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

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

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

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

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

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

    [SEARCH BOX]

Page 2

See also

14.1 % purity of a product and assay calculations

14.2a % reaction yield and theoretical yield calculations and why you can't actually get 100% yield in practice

14.3 dilution of solutions calculations  *  14.4 water of crystallisation calculations

14.5 how much of a reactant is needed? calculation of quantities required, limiting reactant quantities

Chemical & Pharmaceutical Industry Economics & Sustainability, Life Cycle Assessment, Recycling

14.2b The Atom economy of a chemical reaction

The atom economy (a measure of atom utilisation or efficiency) is a measure of the amount of starting materials that end up as useful products. It is important for sustainable development and for good economic reasons to use reactions with high atom economy.

A chemical reaction may give, and often does, more than one product, but of the mixture of products, perhaps only one of them is the desired useful product.

The percentage atom economy of a reaction is readily calculated using the balanced equation for the reaction expressed in reacting masses.

You need to be able to calculate the atom economy of a reaction to form a desired product from the balanced equation and ...

... perhaps explain why a particular reaction pathway is chosen to produce a specified product given appropriate information such as atom economy, percent yield, rate of reaction, equilibrium position and usefulness of by-products.

The atom economy of a reaction is a theoretical percentage measure of the amount of starting materials that ends up as the 'desired' useful reaction products. Its sometimes referred to as atom utilisation.

	MASS of desired USEFUL PRODUCT
ATOM ECONOMY   =  100  x	-------------------------------------------------------------------------------
	TOTAL MASS of all REACTANTS or PRODUCTS

In atom economy calculations you can say REACTANTS or PRODUCTS because of the law of conservation of mass.

The greater the % atom economy of a reaction, the more 'efficient' or 'economic' it is likely to be, though this is a gross simplification when complex and costly chemical synthesis are looked at.

Many reactions give more than one product, and not all of them are useful, so it is useful to calculate what % of the products is theoretically useful, and we call this the atom economy of the reaction.

The reactions that only give one product, have the maximum atom economy of 100% and these are the most economic reactions e.g. synthesis of ammonia and reacting ethene with water to make ethanol.

N2 + 3H2 ==> 2NH3    and    CH2=CH2  +  H2O  ===>  CH3CH2OH

These are simple addition reactions where two reactants give one product e.g. synthesis of ammonia from nitrogen and hydrogen and the synthesis of ethanol from ethene and water.

100 minus the atom economy gives you the % waste, but reactions with only one product will always give the highest atom economy

If a reaction gives more than one product it may be possible to use and sell these other products, thereby making the process more economic overall.

Whatever, waste products have to be dealt with and disposed of in some safe way, and this costs money!

Reactions with a low atom economy are very wasteful and use up resources at faster rate than high atom economy reactions.

They are usually less sustainable in the long run and the cost of raw materials will increase over time.

e.g. the process may involve non-renewable raw materials which will increase in costs as reserves become depleted.

BUT take care on this 'low economy' point, e.g. many products can be recycled and, as already mentioned, the waste products may have some economic value after further separation and processing, known as useful by-product.

The less waste there is, the higher the atom economy, the less materials are wasted, less energy used, so making the process more economic, 'greener' and more sustainable.

Quite simply, the larger the atom economy of a reaction, the less waste products are produced and hopeful the waste is easy and cheap to deal with, or even better, some use can be found for these waste 'by-products'.

It can be defined numerically in words in several ways, all of which amount to the same theoretical % number!

The chemical industry is continuously working on chemical synthetic routes that have the highest atom economy and give the highest % yield (see section 14.2a % reaction yield)

If a chemical reaction that has a low atom economy or gives a low yield of useful product research would be undertaken to find ways the reaction can been improved to increase the yield of useful product.

Note that there are other factors when considering industrial production.

% reaction yield and theoretical yield calculations and why you can't actually get 100% yield in practice

The factors affecting the rate of a reaction - is the chemical process fast enough to be economic?

Is the reaction reversible?  and 

Does it involve a chemical equilibrium? - can you alter reaction conditions to increase the yield of the desired product?

CALCULATING the ATOM ECONOMY of a chemical reaction

You can do the calculation in any mass units you want, or non at all by simply using the atomic/formula masses of the reactants and products as appropriate , and I suggest you just think like that.

The formula to calculate atom economy can be written in several different ways and they are ALL equivalent to each other because of the law of conservation of mass e.g.

THEY ALL GIVE THE SAME ANSWER!

As long as you take into account ANY balancing numbers in the equation, which itself should be balanced!

but they are theoretical values, you don't get these results in real life chemistry - see % purity of a product  & % reaction yield

Examples of how to work out atom economy calculations

Atom economy calculation Example 14.2b (1) See Extraction of Iron and Steel Making for detailed chemistry

This is illustrated by using the blast furnace reaction from example 14.2a.3 above.

Fe2O3(s) + 3CO(g) ===> 2Fe(l) + 3CO2(g)

Using the atomic masses of Fe = 56, C = 12, O = 16, we can calculate the atom economy for extracting iron.

the reaction equation can be expressed in terms of theoretical reacting mass units

[(2 x 56) + (3 x 16)] + [3 x (12 + 16)] ===> [2 x 56] + [3 x (12 + 16 + 16)]

[160 of Fe2O3] + [84 of CO] ===> [112 of Fe] + [132 of CO2]

so there are a total of 112 mass units of the useful/desired product iron, Fe,

out of a total mass of reactants or products of 160 + 84 = 112 + 132 = 244.

Therefore the atom economy = 100 x 112 / 244 = 45.9%

Note: It doesn't matter whether you use the total mass of reactants or the total mass products in the calculations, they are the same due to the law of conservation of mass

Atom economy calculation Example 14.2b (2) See Ethanol Chemistry

The fermentation of sugar to make ethanol ('alcohol') and (b) converting ethanol to ethene

(a) glucose (sugar) == enzyme ==> ethanol + carbon dioxide

C6H12O6(aq) ==> 2C2H5OH(aq) + 2CO2(g) 

atomic masses: C = 12, H = 1 and O = 16

formula mass of glucose reactant = 180 (6x12 + 12x1 + 6x16)

formula mass of ethanol product = 46 (2x12 + 5x1 + 1x16 + 1x1)

relative mass of desired useful product in the equation = 2 x 46 = 92

Atom economy = 100 x 92/180 = 51.1%

(b) ethanol === heat/catalyst ===> ethene + water

It possible to use ethanol from fermentation to produce ethene for plastics (polymers) manufacture instead of relying on the cracking of crude oil fractions (e.g. a country like Brazil with a huge agriculture system but no oil deposits.

CH3CH2OH  ===>  CH2=CH2  +  H2O

formula masses: ethanol = 46, ethene = 28, water = 18

% atom economy = 100 x 28/46 = 60.9%

(c) ethene + water ===> ethanol

 CH2=CH2  +  H2O ====>  CH3CH2OH

BUT, it is also possible to manufacture ethanol by catalytically reacting steam with ethene from cracking oil fractions.

Since this is a 'simple' addition reaction, the atom economy is 100%.

This is a much cleaner and efficient process than fermentation which only has a 51% atom economy.

Atom economy calculation Example 14.2b (3) All about making hydrogen See Ammonia Synthesis

In these three examples, Reactions 1 to 3: Relative atomic masses used: C = 12.0,  H = 1.0,  O = 16.0

Hydrogen is used in synthesising ammonia and making margarine, and is made on a large scale from reacting methane with water

methane + water  ==>  hydrogen + carbon monoxide

Reaction equation 1.: CH4(g) + H2O(g)  ==>  3H2(g)  +  CO(g)

using formula masses gives the ratios gives

16 + 18 ==> (3 x 2) + 28

34 mass units of reactants ==> 6 mass units of useful product

Atom economy = 100 x 6 / 34 = 17.6%

BUT, the 82.4% of waste toxic carbon monoxide must be dealt with in some way!

It seems a very inefficient process BUT ...

(i) hydrogen has very small molecular mass and 75% of the molecules are the desired product

(ii) you can actually burn the carbon monoxide as a fuel to provide energy for power generation.

(iii) methane is readily available from the petrochemical industry and is much cleaner to work with than impure coal (see reactions 2. and 3.), which must be converted to coke first - an extra process.

So, a low atom economy doesn't always mean the process is not economically viable or necessarily produces undue waste.

Hydrogen can be made from reacting water with coke (a form of carbon made by roasting coal)

A second reaction used to manufacture hydrogen

carbon  +  water  ===> hydrogen + carbon monoxide

Reaction equation 2.: C(s)  + H2O(g)  ===>  H2(g) +  CO(g)

formula masses:  12 + 18 (= 30) ==> 28 + 2 (= 30)

% atom economy = 100 x 2 / 30 = 6.67 %

This is a much lower atom economy than reaction, and only 50% of the molecules are useful product.

However, the carbon monoxide can be made to react further with water to form more hydrogen

carbon monoxide  +  water  ===>  hydrogen  +  carbon dioxide

Reaction equation 3.: CO(g)  +  H2O(g)  ===>  H2(g)  +  CO2(g)

formula masses:  28 + 18 (= 46)  ===> 2 + 44 (= 46)

% atom economy = 100 x 2 / 46 = 4.35 %

This is an even lower atom economy than reaction, and again, only 50% of the molecules are useful product.

However: (a) it does take care of poisonous carbon monoxide, but there is also something else you can do ...

(b) You can combine these process, so combining equations 2. and 3. we get ...

C(s)  + H2O(g)  ===>  H2(g) +  CO(g)   plus

CO(g)  +  H2O(g)  ===>  H2(g)  +  CO2(g)   gives

------------------------------------------------------------------

C(s)  + 2H2O(l)  ====>  2H2(g)  +  CO2(g)

------------------------------------------------------------------

formula masses:  12 + 2x18 (= 48)  ===> 4 + 44 (= 48)

% atom economy = 100 x 4 / 48 = 8.33 %

This is a higher atom economy than reactions 2 or 3, and 67% of the molecules are useful product.

Only 50% of the molecular products are the desired product in reactions 2. and 3.

Therefore employing a secondary process improves the efficiency of this particular process for making hydrogen.

It should be noted that carbon monoxide is a water insoluble toxic gas and not easy to deal with.

As already mentioned, you can burn it as a fuel for power generation or heating a reactor vessel.

BUT, carbon dioxide is not toxic (as long as plenty of air is around!) and importantly, it readily dissolves in sodium hydroxide solution in gas flow 'scrubbers' to leave hydrogen as the only remaining gas.

This greatly reduces the cost of purifying the hydrogen to use e.g. in the Haber Synthesis of ammonia and the hydrogenation of unsaturated plant fats to make spreadable margarine.

Also, the reaction between steam and carbon monoxide is exothermic, this reduces the energy needs of the overall process.

Making hydrogen using electrolysis

(I'm only interested in atom economy here - see other notes on electrolysis for details)

Relative atomic masses: Na = 23.0,  Cl = 35.5,  H = 1.0,  O = 16.0

Process A. Electrolysis of aqueous sodium chloride solution (brine)

Overall change: 2NaCl(aq)  +  2H2O(l)  ====>  2NaOH(aq)  +  Cl2(g)  +  H2(g)

total masses for reactants or products = (2 x 58.5) + (2 x 18) =

mass of desired product 2

% atom economy = 100 x 2 / = 153 = 1.31%

Very low atom economy and much waste to deal with?, BUT, sodium hydroxide and chlorine are useful saleable chemicals.

Process B. Electrolysis of acidified water

Overall change: 2H2O(l)  ===> 2H2(g)  +  O2(g)

total masses for reactants or products = (2 x 18) = (4 + 32) = 36

mass of desired product 2

% atom economy = 100 x 4 / = 36 = 11.1%

This has a much greater atom economy than process B and no waste products, no pollution.

This is an ideal process, especially if you can use green energy e.g. from wind turbine, solar cell or hydroelectric sources of electricity.

BUT, both processes have their place in the chemical industry - from Process A we do need sodium hydroxide and chlorine to manufacture other products e.g. soap from NaOH + plant oil and PVC from ethene and chlorine.

Note 1: These reactions contrast with the 100% atom economy of ammonia production (for which much of the hydrogen is made),

because its an addition reaction with no extra waste products

N2 + 3H2 ==> 2NH3

mass of reactants = mass of useful products = 100% atom economy

Note 2: In organic chemistry, these three types of reaction can never have a 100% atom economy because there are always at least two products - desired useful product and often a waste product e.g.

(i) condensation reaction - a small molecule formed in joining two molecules to make a larger molecule

e.g. n HOOC-[][][][]-COOH + n H2N-[][][][]-NH2 ==> -(-OC[][][][]-CONH-[][][][]-NH-)n-+  2n H2O

a schematic example of condensation polymerisation

(ii) elimination reaction - a group of atoms eliminated from a molecule

e.g. CH3CH2OH  ===>  CH2=CH2  +  H2O

water eliminated from ethanol to form ethene

(iii) substitution reaction - one or more atoms substituted by another atom or group of atoms

e.g. CH3CH2Br  +  NaOH  ===>  CH3CH2OH  +  NaBr

-Br group substituted by an -OH group

Atom economy calculation Example 14.2b (4) See Ammonia Synthesis

You can use either (a) hydrogen or (b) a hydrocarbon gas like methane to reduce the oxides of metals of low reactivity to obtain the metal itself.

e.g. the reduction of copper(II) oxide.

Using the atomic masses: Cu = 63.5,  H = 1,  O =16,  C = 12

(a) CuO  + H2  ==>  Cu  +  H2O

Calculate the atom economy of the reaction.

Formula masses: CuO = 79.5,  H2 = 2,  Cu = 63.5,  H2O = 18

Total mass of reactants = total mass of products 79.5 + 2 = 63.5 + 18 = 81.5

atom economy of desired product Cu = 100 x 63.5/81.5 = 77.9%

(b) 4CuO  +  CH4  ==>  4Cu  +  2H2O  +  CO2

Calculate the atom economy of the reaction.

Formula masses: CuO = 79.5,  CH4 = 16,  Cu = 63.5,  H2O = 18,  CO2 = 44

Total mass of reactants = total mass of products

(4 x 79.5) + 16 = (4 x 63.5) + (2 x 18) + 44 = 334

atom economy of desired product Cu = 100 x (4 x 63.5)/334 = 76.0%

(c) Which reaction has the higher atom economy? BUT, would this be the preferential method used?

Reaction (a) has the higher atom economy, BUT, hydrogen is probably more costly to produce than cheap methane gas from crude oil. Therefore method (b) is probably more economic.

Rather than here, I've added more atom economy calculations to the Reacting mass ratio calculations of reactants and products from equations page.

See also

14.1 % purity of a product and assay calculations

14.2a % reaction yield and theoretical yield calculations and why you can't actually get 100% yield in practice

14.3 dilution of solutions calculations  *  14.4 water of crystallisation calculations

14.5 how much of a reactant is needed? calculation of quantities required, limiting reactant quantities

Chemical & Pharmaceutical Industry Economics & Sustainability, Life Cycle Assessment, Recycling

TOP OF PAGE

Above is typical periodic table used in GCSE science-chemistry specifications in doing chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

OTHER CALCULATION PAGES

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

2. Calculating 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 compound

5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

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 calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

9. Moles and the molar volume of a gas, Avogadro's Law

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

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

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

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

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

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

    [SEARCH BOX]

Website content © Dr Phil Brown 2000+. All copyrights reserved on revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries & references to science course specifications are unofficial.

Page 3

Page 4

See also

14.2a % reaction yield and theoretical yield calculations and why you can't actually get 100% yield in practice

14.2b atom economy calculations  *  14.3 dilution of solutions calculations  *  14.4 water of crystallisation calculations

14.5 how much of a reactant is needed? calculation of quantities required, limiting reactant quantities

Chemical & Pharmaceutical Industry Economics & Sustainability, Life Cycle Assessment, Recycling

14.1 Percentage purity of a chemical reaction product

• % Purity is very important

  • e.g. for analytical standards in laboratories or pharmaceutical products where impurities could have dangerous side effects in a drug or medicine.

  • However in any chemical process it is almost impossible to get 100.00% purity and so samples are always analysed in industry to monitor the quality of the product.

  • The more a product is processed e.g. by distillation or crystallisation, the more costly the process, but the purer the product gets.

  • Somewhere there has to be a compromise, so it is important that before sale, the product is assayed or analysed as to its percentage purity.

  • It would not be acceptable e.g. in the pharmaceutical industry to manufacture a drug for treating us, with impurities in it, that may have harmful effects.

  • Similarly in fuels for road vehicles, which themselves have additives in to enhance engine performance, you wouldn't want other impurities that may cause engine damage.

  • You can apply the same sort of argument to thousands of domestic and industrial products from the chemical and pharmaceutical industries.

  • An assay is any procedure used to analyse and test for its purity of the % content of a specified component in a mixture of a % of an element or ion etc.

• % purity is the percentage of the material which is the actually desired chemical in a sample of it.

% PURITY CALCULATIONS and ASSAY CALCULATIONS Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Percent purity assay Chemical Calculations 14. Other GCSE chemical calculations % PURITY OF A PRODUCT from a chemical preparation reaction Index of all my online chemical calculation quizzes All my GCSE/IGCSE/US grade 8-10 Chemistry Revision notes Advanced UK A/AS level, IB and US grade 11-12 pre-university chemistry email doc brown - comment - query? Use your mobile phone or ipad etc. in 'landscape' orientation This is a BIG website, you need to take time to explore it [ SEARCH BOX] Keywords: Quantitative chemistry calculations Help for problem solving in doing % percentage purity calculations. To 'assay' means to analyse a sample for its purity. This page includes of fully worked examples of percent purity calculations. How do you do assay calculations? 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 how to calculate percent purity and assay chemical calculations with worked examples are fully worked out below and should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses as well as Advanced A Level Chemistry courses. Spotted any careless error? EMAIL query ? comment or request a type of GCSE calculation not covered?

• Example 14.1 (4) Purity calculation - an assay calculation is sketched out below for A Level students + link to others.

  • Even at pre-A level you can do a simple titration and analyse an aspirin sample without using the mole concept in the calculation e.g. the above assay calculation could be presented via a reacting mass calculation as follows ...

  • 0.300g of aspirin was titrated with sodium hydroxide solution of concentration 4.00g/dm3.

  • If the aspirin required 16.45 cm3 of the NaOH(aq) to neutralise it, calculate the percent purity of the aspirin.

    • The simplified equation for the reaction is ...

    • C6H4(OCOCH3)COOH + NaOH  ==> C6H4(OCOCH3)COONa + H2O

    • Mr(aspirin) = 180,  Mr(NaOH) = 40   (atomic masses: C = 12, H = 1, O = 16, Na = 23)

    • Therefore the reacting mass ratio is 180g aspirin reacts with 40g of sodium hydroxide.

    • The titration was 16.45 cm3, so, converting the cm3 to dm3,

    • the mass of NaOH used in the titration = 4.00 x 16.45/1000 = 0.0658g,

    • so we can scale this up to get the mass of aspirin titrated,

    • therefore the mass of aspirin titrated = 0.0658 x 180 / 40 = 0.296g

    • therefore the % purity = 100 x 0.296 / 0.300 = 98.7% (3 sf)

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See also

14.2a % reaction yield and theoretical yield calculations and why you can't actually get 100% yield in practice

14.2b atom economy calculations  *  14.3 dilution of solutions calculations  *  14.4 water of crystallisation calculations

14.5 how much of a reactant is needed? calculation of quantities required, limiting reactant quantities

Chemical & Pharmaceutical Industry Economics & Sustainability, Life Cycle Assessment, Recycling