7. Incidencia en las ODS de la OMS
7.2. Efecto del seguro propuesto en los distintos ODS
Revise section 7. moles and mass before proceeding in this section 11 and eventually you may need to be familiar with the use of the apparatus illustrated above, some of which gives great accuracy when dealing with solutions and some do not.
It is very useful to be know exactly how much of a dissolved substance is present in a solution of particular concentration or volume of a solution.
o So we need a standard way of comparing the concentrations of solutions in.
o The more you dissolve in a given volume of solvent, or the smaller the volume you dissolve a given amount of solute in, the more concentrated the solution.
The concentration of an aqueous solution is usually expressed in terms of moles of dissolved substance per cubic decimetre (reminder mole formula triangle on the right),
with units mol dm-3 (or mol/dm3)and this is called molarity, sometimes denoted in shorthand as M.
o Note: 1dm3 = 1 litre = 1000ml = 1000 cm3, so dividing cm3/1000 gives dm3, which
is handy to know since most volumetric laboratory apparatus is calibrated in cm3 (or ml), but solution concentrations are usually quoted in molarity, that is mol/dm3 (mol/litre).
o Concentration is also expressed in a 'non-molar' format of mass per volume e.g. g/dm3
Equal volumes of solution of the same molar concentration contain the same number of moles of solute i.e.
the same number of particles as given by the chemical formula.
You need to be able to calculate
o the number of moles or mass of substance in an aqueous solution of given volume and concentration o the concentration of an aqueous solution given the amount of substance and volume of water, for this you
use the equation .... (reminder molarity formula triangle on the right), o (1a) molarity (concentration) of Z = moles of Z / volume in dm3
you need to be able to rearrange this equation ... therefore ...
(1b) moles = molarity (concentration) x volume in dm3 and ...
(1c) volume in dm3 = moles / molarity (concentration) o You may also need to know that ...
(2) molarity x formula mass of solute = solute concentration in g/dm3
and dividing this by 1000 gives the concentration in g/cm3, and
(3) concentration in g/dm3 / formula mass = molarity in mol/dm3
both equations (2) and (3) result from equations (1) and (4), work it out for yourself.
o and don't forget by now you should know:
(4) moles Z = mass Z / formula mass of Z
(5) 1 mole = formula mass in grams
(reminder molarity formula triangle top right)
Molarity calculation Example 11.1
o What mass of sodium hydroxide (NaOH) is needed to make up 500 cm3 (0.500 dm3) of a 0.500 mol dm-3 (0.5M) solution? [Ar's: Na = 23, O = 16, H = 1]
o 1 mole of NaOH = 23 + 16 + 1 = 40g
o molarity = moles / volume, so mol needed = molarity x volume in dm3 o mol NaOH needed = 0.500 x 0.500 = 0.250 mol NaOH
o therefore mass = mol x formula mass o = 0.25 x 40 = 10g NaOH required
Molarity calculation Example 11.2
o (a) How many moles of H2SO4 are there in 250cm3 of a 0.800 mol dm-3 (0.8M) sulphuric acid solution?
o (b) What mass of acid is in this solution? [Ar's: H = 1, S = 32, O = 16]
(a) molarity = moles / volume in dm3, rearranging equation for the sulfuric acid
mol H2SO4 = molarity H2SO4 x volume of H2SO4 in dm3
mol H2SO4 = 0.800 x 250/1000 = 0.200 mol H2SO4
(b) mass = moles x formula mass
formula mass of H2SO4 = 2 + 32 + (4x16) = 98
0.2 mol H2SO4 x 98 = 19.6g of H2SO4
Molarity calculation Example 11.3
o 5.95g of potassium bromide was dissolved in 400cm3 of water. Calculate its molarity. [Ar's: K = 39, Br
= 80]
o moles = mass / formula mass, (KBr = 39 + 80 = 119) o mol KBr = 5.95/119 = 0.050 mol
o 400 cm3 = 400/1000 = 0.400 dm3
o molarity = moles of solute / volume of solution o molarity of KBr solution = 0.050/0.400 = 0.125M
Molarity calculation Example 11.4
o What is the concentration of sodium chloride (NaCl) in g/dm3 and g/cm3 in a 1.50 molar solution?
o At. masses: Na = 23, Cl = 35.5, formula mass NaCl = 23 + 35.5 = 58.5 o since mass = mol x formula mass, for 1 dm3
o concentration = 1.5 x 58.5 = 87.8 g/dm3, and o concentration = 87.75 / 1000 = 0.0878 g/cm3
Molarity calculation Example 11.5
o A solution of calcium sulphate (CaSO4) contained 0.500g dissolved in 2.00 dm3 of water.
o Calculate the concentration in (a) g/dm3, (b) g/cm3 and (c) mol/dm3.
(a) concentration = 0.500/2.00 = 0.250 g/dm3, then since 1dm3 = 1000 cm3
(b) concentration = 0.250/1000 = 0.00025 g/cm3
(c) At. masses: Ca = 40, S = 32, O = 64, f. mass CaSO4 = 40 + 32 + (4 x 16) = 136
moles CaSO4 = 0.5 / 136 = 0.00368 mol
concentration CaSO4 = 0.00368 / 2 = 0.00184 mol/dm3
SECTION 14. % ACTUAL AND THEORETICAL YIELD, DILUTION CALCULATIONS, ATOM ECONOMY, VOLUMETRIC TITRATION APPARATUS, WATER OF CRYSTALLISATION, HOW MUCH REACTANT IS
NEEDED?
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.
% purity is the percentage of the material which is the actually desired chemical in a sample of it.
MASS of USEFUL PRODUCT
PERCENT PURITY = 100 x
in TOTAL MASS of SAMPLE
Example 14.1.1
o A 12.00g sample of a crystallised pharmaceutical product was found to contain 11.57g of the active drug.
o Calculate the % purity of the sample of the drug.
o % purity = actual amount of desired material x 100 / total amount of material o % purity = 11.57 x 100 / 12 = 96.4% (to 1dp)
o
- Example 14.1.2
o Sodium chloride was prepared by neutralising sodium hydroxide solution with dilute hydrochloric acid. The solution was evaporated to crystallise the salt.
o The salt is required to be completely anhydrous, that is, not containing any water.
o The prepared salt was analysed for water by heating a sample in an oven at 110oC to measure the evaporation of any residual water.
o The following results were obtained and from them calculate the % purity of the salt.
o Mass of evaporating dish empty = 51.32g.
o Mass of impure salt + dish = 56.47g o Mass of dish + salt after heating = 56.15g
o Therefore the mass of original salt = 56.47 - 51.32 = 5.15g o and the mass of pure salt remaining = 56.15 - 51.32 = 4.83g o % salt purity = 4.83 x 100 / 5.15 = 93.8% (to 1dp)
14.2a Percentage yield of the product of a reaction
The % yield of a reaction is the percentage of the product obtained compared to the theoretical maximum (predicted) yield calculated from the balanced equation.
o You get the predicted maximum yield from a reacting mass calculation (see examples further down).
ACTUAL YIELD (e.g. in grams)
PERCENTAGE YIELD = 100 x PREDICTED YIELD (g)
In carrying out a chemical preparation, the aim is to work carefully and recover as much of the desired reaction product as you can, and as pure as is possible and practicable.
o Despite the law of conservation of mass, i.e. no atoms lost or gained, in real chemical preparations things cannot work out completely according to chemical theory, often for quite simple, physical or sometimes chemical reasons,
o and it doesn't matter with its a small scale school laboratory preparation of a large scale industrial manufacturing process, the percent yield is never 100%.
So, in any chemical process, it is almost impossible to get 100% of the product because of several reasons:
1.
The reaction might not be 100% completed because it is reversible reaction and an equilibrium is established (note the sign in the equation below. Both reactants and products co-exist in the same reaction mixtures (solutions or gases) i.e. the reaction can never go to completion. A good example of this is the preparation of an ester
ethanoic acid + ethanol ethyl ethanoate + water
+ + H2O
2.
You always get losses of the desired product as it is separated from the reaction mixture by filtration, distillation, crystallisation or whatever method is required. e.g. bits of solid or droplets are left behind on the sides of the apparatus or reactor vessel.
Small amounts of liquid will be left in distillation units or solid particles on the surface of filtration units.
You cannot avoid losing traces of product in all stages of the manufacturing process.
3.
Some of the reactants may react in another way to give a different product to the one you want (so-called by products). By products are very common in organic chemistry reactions
(i) A + B ==> C + D
(i) The main reaction to give the main desired products A + B
(ii) A + B ==> E + F
(ii) A con-current reaction, maybe just involving a few % of the reactants to give the minor, and often undesirable, by products of E + F.
Sometimes the by products can be separated as a useful product and sold to help the economics of the chemical process.
4.
The aim is to work carefully and recover as much of the desired reaction product, and as pure as is possible and practicable % yield = actual amount of desired chemical obtained x 100 / maximum theoretical amount formed o If the reaction doesn't work the yield is zero or 0%.
o If the reaction works perfectly and you obtain all the product, the yield is 100%, BUT this never happens in reality (as already discussed above).
Example 14.2a.1
o Magnesium metal dissolves in hydrochloric acid to form magnesium chloride.
o Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g)
o Atomic masses : Mg = 24 and Cl = 35.5, and formula mass MgCl2 = 24 + (2 x 35.5) = 95
o (a) What is the maximum theoretical mass of anhydrous magnesium chloride which can be made from 12g of magnesium?
Reacting mass ratio calculation from the balanced equation:
1 Mg ==> 1 MgCl2, so 24g ==> 95g or 12g ==> 47.5g MgCl2
o (b) If only 47.0g of purified magnesium chloride was obtained after crystallising the salt from the solution and heating it to drive off the water of crystallisation, what is the % yield from the salt preparation?
% yield = actual amount obtained x 100 / maximum theoretical amount possible
% yield = 47.0 x 100 / 47.5 = 98.9% (to 1dp)
o More examples of % yield and atom economy calculations in section 6.
Example 14.2a.2o
-o 2.8g -of ir-on was heated with excess sulphur t-o f-orm ir-on sulphide.
o Fe + S ==> FeS
o The excess sulphur was dissolved in a solvent and the iron sulphide filtered off, washed with clean solvent and dried.
o If 4.1g of purified iron sulphide was finally obtained, what was the % yield of the reaction?
o 1st a reacting mass calculation of the maximum amount of FeS that can be formed:
Relative atomic/formula masses: Fe =56, FeS = 56 + 32 = 88
This means 56g Fe ==> 88g FeS, or by ratio, 2.8g Fe ==> 4.4g FeS
because 2.8 is 1/20th of 56, so theoretically you can get 1/20th of 88g of FeS.
o 2nd the % yield calculation itself.
% yield = actual amount obtained x 100 / maximum theoretical amount possible
% yield = 4.1 x 100 / 4.4 = 93.2% (to 1dp)
o More examples of % yield and atom economy calculations in section 6.
o
- Example 14.2a.3
o (a) Theoretically how much iron can be obtained from 1000 tonne of pure haematite ore, formula Fe2O3 in a blast furnace?
o If we assume the iron(III) oxide ore (haematite) is reduced by carbon monoxide, the equation is:
o Fe2O3(s) + 3CO(g) ==> 2Fe(l) + 3CO2(g)
o (atomic masses: Fe = 56, O = 16)
o For every Fe2O3 ==> 2Fe can be extracted, formula mass of ore = (2 x 56) + (3 x 56) = 160 o Therefore reacting mass ratio is: 160 ==> 112 (from 2 x 56)
o so, solving the ratio, 1000 ==> 112 / 160 = 700 tonne copper = max. can be extracted
o (b) If in reality, only 670 tonne of iron is produced what is the % yield of the overall blast furnace process?
o % yield = actual yield x 100 / theoretical yield o % yield = 670 x 100 / 700 = 95.7%
o In other words, 4.3% of the iron is lost in waste e.g. in the slag.
o More examples of % yield and atom economy calculations in section 6.
o
- Example 14.2a.4
o Given the atomic masses: Mg = 24 and O = 16,
o and the reaction between magnesium to form magnesium oxide is given by the symbol equation o 2Mg(s) + O2(g) ==> 2MgO(s)
o (a) What mass of magnesium oxide can be made from 1g of magnesium?
2Mg ==> 2MgO
in terms of reacting masses (2 x 24) ==> {2 x (24 +16)}
so 48g Mg ==> 80g MgO (or 24g ==> 40g, its all the same)
therefore solving the ratio
1g Mg ==> w g MgO, using the ratio 48 : 80
w = 1 x 80 / 48 = 1.67g MgO
o (b) Suppose the % yield in the reaction is 80%. That is only 80% of the magnesium oxide formed is actually recovered as useful product. How much magnesium needs to be burned to make 30g of magnesium oxide?
This is a bit tricky and needs to done in two stages and can be set out in several ways.
Now 48g Mg ==> 80g MgO (or any ratio mentioned above)
so y g Mg ==> 30g MgO
y = 30 x 48 / 80 = 18g Mg
BUT you only get back 80% of the MgO formed,
so therefore you need to take more of the magnesium than theoretically calculated above.
Therefore for practical purposes you need to take, NOT 18g Mg, BUT ...
... since you only get 80/100 ths of the product ...
... you need to use 100/80 ths of the reactants in the first place
therefore Mg needed = 18g x 100 / 80 = 22.5g Mg
CHECK: reacting mass calculation + % yield calculation CHECK:
22.5 Mg ==> z MgO, z = 22.5 x 80 / 48 = 37.5g MgO,
but you only get 80% of this,
so you actually get 37.5 x 80 / 100 = 30g
This means in principle that if you only get x% yield ...
... you need to take 100/x quantities of reactants to compensate for the losses.
14.2b The Atom economy of a chemical reaction
The atom economy of a reaction is a theoretical measure of the amount of starting materials that ends up as 'desired' reaction product.
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.
Quite simply, the larger the atom economy of a reaction, the less waste products are produced.
It can be defined numerically in words in several ways, all of which amount to the same theoretical % number!
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.
MASS of desired USEFUL PRODUCT ATOM ECONOMY = 100 x
TOTAL MASS of REACTANTS
MASS of desired USEFUL PRODUCT ATOM ECONOMY = 100 x
TOTAL MASS of PRODUCTS
TOTAL FORMULA MASSES of USEFUL PRODUCT ATOM ECONOMY = 100 x
TOTAL FORMULA MASS of REACTANTS
TOTAL FORMULA MASSES of USEFUL PRODUCT ATOM ECONOMY = 100 x
TOTAL FORMULA MASS of PRODUCTS THEY ALL GIVE THE SAME ANSWER!
Example 14.2b.1
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 from the law of conservation of mass
Example 14.2b.2
The fermentation of sugar to make ethanol ('alcohol')
e.g. 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 = 46 (2x12 + 5x1 + 1x16 + 1x1)
relative mass of desired useful product = 2 x 46 = 92 Atom economy = 100 x 92/180 = 51.1%
14.3 Dilution of solutions calculations calculating dilutions - volumes involved etc.
In conjunction with this page it be important to study
o Calculations Part 11 Introducing Molarity, volumes and the concentration of solutions o before tackling this section and note the triangle of relationships you need to know!
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.
Example 14.3.1
A purchased standard solution of sodium hydroxide had a concentration of 1.0 mol/dm3. How would you prepare 100 cm3 of a 0.1 mol/dm3 solution to do a titration of an acid?
o The required concentration is 1/10th of the original solution.
o 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.
o 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.
o To make only 100 cm3 of the diluted solution you would dilute 10cm3 by mixing it with 90 cm3 of water.
o 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.
Example 14.3.2
o Given a stock solution of sodium chloride of 2.0 mol/dm3, how would you prepare 250cm3 of a 0.5 mol/dm3 solution?
o The required 0.5 mol/dm3 concentration is 1/4 of the original concentration of 2.0 mol/dm3.
o 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.
o 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.
o This can be done, but rather inaccurately, using measuring cylinders and stirring to mix the two liquids in a beaker.
o 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.
o 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!
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).
o (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
There are more questions involving molarity in section 7. introducing molarity and section 12. on dilution
Example 14.3.4
o You are given a stock solution of concentrated ammonia with a concentration of 17.9 mol dm-3 (conc.
ammonia! ~18M!)
o (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.
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.
14.4 Water of crystallisation in a crystallised salt
Example 14.4.1: Calculate the % of water in hydrated magnesium sulphate MgSO4.7H2O salt crystals o Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1
o relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246 o 7 x 18 = 126 is the mass of water
o so % water = 126 x 100 / 246 = 51.2%
Example 14.4.2 The % water of crystallisation and the formula of the salt are calculated as follows:
o Suppose 6.25g of blue hydrated copper(II) sulphate, CuSO4.xH2O, (x unknown) was gently heated in a crucible until the mass remaining was 4.00g. This is the white anhydrous copper(II) sulphate.
o The mass of anhydrous salt = 4.00g, mass of water (of crystallisation) driven off = 6.25-4.00 = 2.25g o The % water of crystallisation in the crystals is 2.25 x 100 / 6.25 = 36%
o [ Ar's Cu=64, S=32, O=16, H=1 ]
o The mass ratio of CuSO4 : H2O is 4.00 : 2.25
o To convert from mass ratio to mole ratio, you divide by the molecular mass of each 'species' o CuSO4 = 64 + 32 + (4x18) = 160 and H2O = 1+1+16 = 18
o The mole ratio of CuSO4 : H2O is 4.00/160 : 2.25/18
o which is 0.025 : 0.125 or 1 : 5, so the formula of the hydrated salt is CuSO4.5H2O
Example 14.4.3 How to calculate the theoretical % of water in a hydrated salt o eg magnesium sulphate MgSO4.7H2O salt crystals
o Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1 o relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246 o 7 x 18 = 126 is the mass of water
o so % water = 126 x 100 / 246 = 51.2%
Example 14.4.4 Determination and calculation of salt formula containing 'water of crystallisation'.
o Some salts, when crystallised from aqueous solution, incorporate water molecules into the structure. This is known as 'water of crystallisation', and the 'hydrated' form of the compound.
o e.g. magnesium sulphate MgSO4.7H2O. The formula can be determined by a simple experiment (see the copper sulphate example below).
o A known mass of the hydrated salt is gently heated in a crucible until no further water is driven off and the weight remains constant despite further heating.
The mass of the anhydrous salt left is measured.
The original mass of hydrated salt and the mass of the anhydrous salt residue can be worked out from the various weighings.
o The % water of crystallisation and the formula of the salt are calculated as follows:
Suppose 6.25g of blue hydrated copper(II) sulphate, CuSO4.xH2O, (x unknown) was gently heated in a crucible until the mass remaining was a constant4.00g.
When the mass on subsequent weighings stays constant, you know all the water of crystallisation has driven off by the heat.
This is the white anhydrous copper(II) sulphate.
The mass of anhydrous salt = 4.00g, mass of water (of crystallisation) driven off = 6.25-4.00 = 2.25g
The % water of crystallisation in the crystals is 2.25 x 100 / 6.25 = 36%
[ Ar values: Cu=64, S=32, O=16, H=1 ]
The mass ratio of CuSO4 : H2O is 4.00 : 2.25
To convert from mass ratio to mole ratio, you divide by the molecular mass of each 'species'
To convert from mass ratio to mole ratio, you divide by the molecular mass of each 'species'