extranet.barneyschool.org.uk · web viewthe mass number of an isotope is equal to the number of...

THE MASSES OF ATOMS AND MOLECULESRELATIVE ATOMIC MASSBecause of the incredibly small size of atoms, it is not possible to weigh them directly. Instead the masses of atoms are related to an arbitrary standard; for this purpose the mass of one atom of 12C is taken to be 12.0000. The mass of any other atom compared to that of the carbon atom is called it’s RELATIVE ATOMIC MASS and is given the symbol Ar.

N.B. Relative atomic mass is a ratio, so it is simply a number and has no units.

The 12C atom contains 6 protons, 6 neutrons and 6 electrons. The hydrogen atom contains 1 proton and 1 electron. Since the mass of an electron is negligible compared to that of a proton or a neutron, the hydrogen atom has only 1/12 the mass of a carbon atom; therefore the relative atomic mass of hydrogen is 1.

Similarly, the relative mass of an oxygen atom, which contains 8 protons, 8 neutrons and 8 electrons, is 16.

MASS NUMBER AND RELATIVE ATOMIC MASSThe mass number of an isotope is equal to the number of protons plus neutrons in one atom and is, therefore, a whole number.

In round numbers, mass number and relative atomic mass are equal, but the relative atomic mass is not an exact whole number, partly because protons and neutrons do not have exactly the same mass and partly because of the existence of isotopes.

THE COMPLICATION OF ISOTOPESIn their natural form many elements are mixtures of isotopes. The relative atomic mass of such an element is an average mass for the atom and will depend on the mass of each isotope and the relative proportion of each. In very many cases, the relative atomic mass approximates closely to a whole number, but in a few cases the deviation is considerable.

MODA./Mole 12.1.2

TOPIC 12.2: AMOUNT OF SUBSTANCE 1

The relative atomic mass of an element is defined as the average mass of oneatom of the element compared with one twelfth the mass of an atom of 12C.

Example: Natural chlorine consists of 75% 35Cl and 25% 37Cl. Calculate the relative atomic mass of chlorine.

In every 100 atoms of chlorine, there are: 75 of mass 35, total mass 2625 25 of mass 37, total mass 925

The mass of 100 atoms = 2625 + 925 = 3550The average mass of one chlorine atom = 3550 = 35.5

100The relative atomic mass of chlorine = 35.5

RELATIVE MOLECULAR MASS & RELATIVE FORMULA MASS

It is calculated by adding together the relative atomic masses of all the atoms present in one molecule.

For ionic or macromolecular compounds, which have infinite three-dimensional structures, the term relative molecular mass is replaced by RELATIVE FORMULA MASS. It is also given the symbol Mr. It is obtained by adding together the relative atomic masses of all the atoms present in one formula unit.

Like relative atomic mass, relative molecular/formula mass is simply a number and has no units.

The setting out of these calculations is important and should follow the method given in the example.


The average mass of one molecule of an element (diatomic gases) or a simple covalent compound compared with one twelfth mass of an atom of 12C is called

its RELATIVE MOLECULAR MASS and is given the symbol Mr.

Example: Calculate the relative molecular mass of phosphoric acid, H3PO4.

Mr (H3PO4) = (3x1) + 31 + (4x16) = 98

WORKSHEET 12.1.1

RELATIVE ATOMIC MASS AND RELATIVE MOLECULAR MASS

1. Natural copper comprises 69% 63Cu and 31% 65Cu.29 29

Calculate the relative atomic mass of copper.

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2. Natural potassium comprises 93% 39K and 7% 41K 19 19

Calculate the relative atomic mass of potassium.

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3. Natural gallium comprises 60.4% 69Ga and 39.6% 71Ga. 31 31

Calculate the relative atomic mass of gallium.

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4. Calculate the relative molecular/formula masses of the following compounds, using the relative atomic masses given on your periodic table. Make sure you set out the calculation correctly.

a) H2SO4

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b) AlCl3

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c) HNO3

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d) K2CO3

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e) Cu(NO3)2

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COUNTING ATOMS AND MOLECULESTHE MOLE CONCEPT

Atoms and molecules are such minute particles that they are difficult to count. Nevertheless, one of the most important things chemists need to know is the NUMBER of particles they are dealing with.

If 1g of hydrogen is weighed out, it will contain a certain number of atoms (let this no. be x). Therefore 1g of any other element will contain fewer atoms, because each atom is heavier.

If 1g hydrogen contains x atomsthen 1g carbon contains x/12 atoms

1g oxygen contains x/16 atoms etc.

Since chemical reactions take place between whole numbers of particles (atoms, ions or molecules), it is more useful to know the number of particles present than the mass. Thus, continuing the above example,

If 1g hydrogen contains x atomsthen 12g carbon contains x atoms

16g oxygen contains x atoms etc.

These numbers correspond to the relative atomic masses of these elements expressed in grammes. Therefore, if elements are measured out in the same proportions as their relative atomic masses, we are dealing with the same number of atoms, i.e. numbers of atoms can be controlled by weighing.

The number of atoms contained in 1g hydrogen, 12g carbon, 16g oxygen etc. has been determined very accurately and has a value of 6.02x1023

This is called the AVOGADRO NUMBER or AVOGADRO CONSTANT and is given the symbol L. This number is of considerable importance in chemistry and has been given a special unit called the MOLE.

A mole of a substance is its Ar (or Mr) in grams; this is also known as the MOLAR MASS.

Pay attention to the type of particle when calculating molar masses. For example, O=16, so 1mol. of oxygen atoms(O) has a mass of 16 g, but 1mol. of oxygen molecules (O2) has a mass of 32g. 1mol. of water molecules (H2O) has a mass of 1+1+16 = 18g


A MOLE OF A SUBSTANCE IS AN AMOUNT CONTAINING THE SAME NUMBER OF PARTICLES

AS THERE ARE ATOMS IN 12g OF 12C

(This number of particles is 6.02x1023)

For most elements:

For diatomic elements (e.g. O2) and compounds:


no. of moles = mass in g Ar

no. of moles = mass in g Mr

MOLES

MASS

Ar

MOLES

MASS

Mr

THE SPECIAL SITUATION OF GASESIt is usually more important to know the volumes of gases involved in a reaction, rather than their masses. The volume occupied by a gas depends on:

It’s massIts temperatureIts pressure

In 1811, Avogadro, an Italian chemist, put forward the following law:

Thus it follows that one mole of any gas under a given set of temperature and pressure conditions occupies the same volume. This is called the MOLAR VOLUME. At 293K and 1 atmosphere pressure (referred to as r.t.p. or room temperature and pressure), the molar volume is 24 litres (24,000 cm3). This value can be used in calculations.

In general, the relationship between the pressure, volume, temperature and no. of moles for an ideal gas is given by the ideal gas equation:

Which can usefully be re-written as:

Where p is the pressure of the gas in N.m-2 (or Pa)V is the volume occupied by the gas in m3

n is the no. of moles of the gasT is the temperature of the gas in KR is the gas constant ( =8.314 J.K-1.mol-1 )m is the mass of the gas in gMr is the relative molecular mass of the gas

Remember always to check that the units you are using are correct before substituting values into these equations.

Using the above equation to calculate the volume of 1mole of a gas at 293K and atmospheric pressure ( 101325 N.m-2 )

V = nRT = 1x 8.314 x 293 = 0.024m3 = 24000cm3

P 101325


EQUAL VOLUMES OF ALL GASES UNDER THE SAME CONDITIONS OF TEMPERATURE AND PRESSURE

CONTAIN EQUAL NUMBERS OF MOLECULES.

pV = nRT

pV = mRT Mr

WORKSHEET 12.1.2

CALCULATIONS INVOLVING MASSES, VOLUMES & MOLES

A) MASSES & MOLES

Calculations of this type are based on these simple relationships.

For most elements:

no. of moles = mass in g Ar

For diatomic elements (e.g. O2) and compounds:

no. of moles = mass in g Mr

Note that the abbreviation for moles is mol.

1. Calculate the mass of :a) 3.1 moles of sodium chlorideb) 2.5 moles of MgSO4.7H2Oc) 1.5mol. of chlorine gas (remember that chlorine is diatomic)d) 0.25mol. of sodium carbonate

2. Calculate the number of moles contained in:a) 4.98g of potassium iodideb) 2.37g of potassium manganate (VII)c) 2.5g of sodium hydroxided) 1.00g of magnesium


B) VOLUMES AND MOLES

These calculations are based on the relationships:

pV = nRT pV = mRT Mr

1. Calculate the volume occupied by 2.1g of chlorine at 150,000Pa and 25oC.

2. What pressure is exerted by 0.25 mol. of oxygen at 100oC in a container with a capacity of 1dm3 ?

3. Calculate the relative molecular weight of a gas, 2.2g of which occupy a volume of 1050cm3 at a temperature of 298K and a pressure of 118,000Pa.

4. A sample of gallium chloride weighing 0.1g was vaporised, and it was found to occupy 16cm3 at a temperature of 415oC and a pressure of 101325Pa.

a) Calculate a value for the relative formula mass of gallium chloride under the conditions of the experiment.

b) Suggest a molecular formula for gallium chloride in the vapour state, indicating how you arrive at your answer.(Ga = 69.7, Cl = 35.5)


CALCULATIONS OF CONCENTRATION

The concentration of a standard solution is usually expressed as the number of moles of the solute contained in 1dm3 of the solution. The unit is mol.dm-3, which can be abbreviated to M. Strictly speaking the concentration given in mol dm-3 should be called the molarity but at A-level the terms concentration and molarity are used interchangeably.

Thus, 0.100M NaOH (said as “point one molar NaOH”) is a solution containing 0.100 mole of NaOH dissolved in every dm3 (litre) of the solution.

If no. of moles = mass and concentration = no. of moles Mr volume in dm3

These two expressions can be combined to give:

This expression can be used to determine the value of any one term given enough information to calculate the other three terms.

Examples:

1. What mass of solute would be needed to prepare the following standard solutions?a) 250cm3 of 0.1M sodium chloride, NaClb) 200cm3 of 0.017M potassium chromate(VI), K2Cr2O7

c) 2dm3 of 0.15M ammonium iron(II) sulphate, (NH4)2SO4.FeSO4.6H2Od) 750cm3 of 0.100M ethanedioic acid, H2C2O4.2H2O

2. Calculate the concentration of the following standard solutions.a) 4.0g of sodium hydroxide, NaOH, is dissolved to make 1000cm3 of solution.b) 4.25g of silver nitrate, AgNO3 is dissolved to make 250cm3 of solution.c) 20.75g of potassium iodide, KI, is dissolved to make 100cm3 of solution.d) 0.79g of potassium manganate(VII), KMnO4 is dissolved to make 500cm3 of

solution.

3. From the data, calculate the relative molecular mass, Mr, of the following substances.a) 1.325g of substance A is used to make up 250cm3 of a 0.05M solution.b) 53.5g of substance B is used to make up 500cm3 of a 2M solution.c) 124.75g of substance C is used to make up 1000cm3 of a 0.5M solution.d) 3.15g of substance D is used to make up 500cm3 of a 0.05M solution.

4. To what volume of solution must each of the following masses of solutes be made up to give a solution of the specified concentration? Give your answer in cm3.

a) 6.9g of potassium carbonate to give a 0.100M solutionb) 0.117g of sodium chloride to give a 0.02M solutionc) 20.75g of potassium iodide to give a 0.5M solutiond) 9.35g of sodium tetraborate, Na2B4O7.10H2O to give a 0.100M solution


concentration (moldm-3) = mass of solute (g) volume of solution in dm3 x Mr of

solute

VOLUMETRIC ANALYSIS

STANDARD SOLUTIONS AND PRIMARY STANDARDSA standard solution is a solution of known concentration; i.e. it contains a known mass of solid in a known volume of solution at a particular temperature. The concentration of a standard solution is usually expressed in moles per dm3 (abbreviation mol.dm-3 or M). A concentration can be determined either by direct weighing, subject to the constraints described below, or by titration.

A primary standard is a substance which can be used to make up a standard solution by direct weighing; i.e. a solid is weighed out and made up to a known volume of solution. However to be a primary standard, a substance must be

a) pureb) stablec) non-volatiled) of fixed compositione) not hygroscopic, deliquescent or efflorescent

The concentration of an unknown solution can be found (this is called standardising the solution) by titration. This involves finding what volume of it is equivalent to (i.e. reacts completely with) a known volume of standard solution; an indicator is used to show when the reaction is complete. The number of moles of each substance reacting can be deduced from the balanced equation for the reaction, and the unknown concentration can then be calculated. Alternatively, the appropriate numbers can be substituted into the formula:

SOURCES OF ERRORS IN TITRATIONS1. If the standard solution is not shaken thoroughly to ensure that it is homogeneous, very

large errors can arise.

2. There is an error in the titre, because the minimum volume which can be added at one time from a burette (1 drop) is about 0.05cm3. This error is reduced by averaging 3 titres. N.B. burette readings are usually taken to 2 d.p.

3. A graduated flask is calibrated at a certain temperature (usually 20oC). It must not be used for hot solutions otherwise large errors arise.

4. Failure to ensure that the volume below the burette tap is full before taking the first burette reading will lead to an overestimate in the volume of solution added.

5. Failure to swirl the flask during the titration can lead to premature indication of the end-point.

6. Failure to remove the funnel from the top of the burette before the first burette reading is taken can permit solution to drain down into the burette during the titration.


PRESENTATION OF RESULTSa. WeighingsAll weighings taken must be recorded (to 3 d.p.) and should be written down while still at the balance.

e.g. Mass of weighing bottle + (named solid) = g Mass of weighing bottle = g

Mass of (named solid) taken = g

b. Titration figuresThe burette is read to the bottom of the meniscus to two decimal places; it is necessary to estimate the second place. The total volume of solution added from the burette is called the titre. It is usual to obtain three concordant titres (i.e. titres which agree within a range of 0.2cm3). The results are recorded in a table.

An average titre is calculated, ignoring results which are obviously spurious, and is quoted to two decimal places.

Titration no.cm3 1 2 3 4

Final burette reading

Initial burette reading

Volume of (named solution) added

Average of titres (specified numbers) = …………………cm3

Before beginning the calculation, the balanced equation for the reaction taking place in the titration should be written.


WORKSHEET 12.1.9

VOLUMETRIC ANALYSIS CALCULATIONS

1. What mass of sodium carbonate dihydrate (Na2CO3.2H2O) is required to react completely with 50.0cm3 of 0.100M sulphuric acid?

2. What volume of 0.200M potassium hydroxide will react with 50.0cm3 of 0.100M sulphuric acid?

3. What volume of 0.125M sodium hydroxide solution is needed to titrate 25.0cm3 of 0.085M sulphuric acid?

4. Soda lime is a mixture containing 85.0% NaOH and 15.0% CaO. What volume of 0.500M nitric acid is needed to neutralise 2.50g of soda lime?

5. A 1.00g sample of limestone is allowed to react with 100cm3 of 0.200M hydrochloric acid. The excess acid required 24.8cm3 of 0.100M sodium hydroxide for titration. Calculate the percentage of calcium carbonate in the limestone.

6. 0.500g of impure ammonium chloride is warmed with an excess of sodium hydroxide solution. The ammonia liberated is absorbed in 25.0cm3 of 0.200M sulphuric acid. The excess sulphuric acid requires 5.64cm3 of 0.200M sodium hydroxide solution for titration. Calculate the percentage ammonium chloride in the original sample.

7. A solution of a metal carbonate, M2CO3, was prepared by dissolving 7.46g of the anhydrous solid in water to give one litre of solution. 25cm3 of this solution reacted with 27.0cm3 of 0.100M hydrochloric acid. Calculate the relative formula mass of M2CO3 and hence the relative atomic mass of the unknown metal M.

8. 10cm3 of concentrated hydrochloric acid were diluted to 100cm3 with water. 10cm3 of the diluted acid was found to react exactly with 20cm3 of 1.0M sodium hydroxide solution.

a) Calculate the concentration of the diluted acid solution.b) Calculate the concentration of the undiluted acid.

9. 1.500g of a sample of limestone were dissolved in 50cm3 of 1M hydrochloric acid. The resulting solution was made up to 250cm3 with distilled water. 25cm3 of this solution required 21.05cm3 of 0.100M sodium hydroxide for neutralisation. Assuming all the basic material in the rock to be calcium carbonate, calculate the percentage of calcium carbonate in the limestone.

10.100cm3 of concentrated hydrochloric acid were diluted to 1dm3 with distilled water. 26.8cm3 of the diluted acid were needed to neutralise 25.0cm3 of 0.500M sodium carbonate solution. Calculate the molarity of the concentrated hydrochloric acid and hence its concentration in g.dm-3.

11.8.58g of washing soda were made up to 250cm3 of aqueous solution. 25cm3 of this solution required 30.0cm3 of 0.2M HCl for neutralisation with methyl orange as indicator. Calculate x in the formula for washing soda, Na2CO3.xH2O.

THE PERCENTAGE COMPOSITION OF COMPOUNDSTOPIC 12.2: AMOUNT OF SUBSTANCE 13

A pure chemical compound has a fixed chemical composition and is represented by a fixed formula. This formula conveys information about:

a) the elements present in the substanceb) the mass of 1 mole of the substance (Mr expressed in g)c) the number of moles of each element present in 1 mole of the substance

The percentage composition of a compound is the % contribution by mass which each component element makes to the substance.

CALCULATION OF % COMPOSITION

WORKSHEET 12.1.3


Example: Calculate the % composition by mass of ammonium nitrate.

1. Work out the formula of the compound. The formula of ammonium nitrate is NH4NO3.

2. Calculate the relative molecular/formula mass of the compound.

Mr (NH4NO3) = 14 + (4x1) + 14 + (3x16) = 80

80g NH4NO3 contains 28g nitrogen 4g hydrogen 48g oxygen

3. Express the mass of each element as a percentage of the molar mass.

%N = 28 x 100 = 35.0%80

%H = 4 x 100 = 5.0% 80

%O = 48 x 100 = 60.0%80

Answers are usually given to 1 d.p.

4. Check that the percentages add up to 100%.

N.B. Errors from rounding up or down may give answers between 99.8 and 100.2%, which is satisfactory.

THE PERCENTAGE COMPOSITION OF COMPOUNDSCalculate the percentage composition by mass of the following compounds, giving your answers to 1 d.p. Use the relative atomic masses given on your periodic table.

a) Na2SO4

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b) Ca(NO3)2

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c) K2CO3

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d) (NH4)2SO4

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CALCULATION OF EMPIRICAL FORMULAE

The two are related by the expression:

Molecular formula = Empirical formula x n

where n is a whole number.

For example:The molecular formulae of methane and ethane are CH4 and C2H6 respectively.The empirical formula of methane is CH4; in the above expression n = 1.The empirical formula of ethane is CH3, which is the simplest whole number ratio of carbon to hydrogen atoms. In the above expression n = 2.

CALCULATION OF AN EMPIRICAL FORMULA


The EMPIRICAL FORMULA is the simplest WHOLE number ratio of the atoms of each element present in a compound.

The MOLECULAR FORMULA is the actual number of atoms of each element present in one molecule of the compound.

Example: Calculate the empirical formula of a compound having the composition: lead 8.32g; sulphur 1.28g; oxygen 2.56g

If its molecular mass is 303, what is its molecular formula?

PROCEDURE:

1) Convert the mass of each element into moles: this is done by dividing the mass by the relative atomic mass.

Moles Pb 8.32 = 0.040 moles 207

Moles S = 1.28 = 0.040 moles 32

Moles O = 2.56 = 0.160 moles 16

This transformation counts the combining atoms; therefore, in the compound,0.040 moles (i.e. 0.040 x 6 x 1023 atoms ) of Pb0.040 moles (i.e. 0.040 x 6 x 1023 atoms) of S0.160 moles (i.e. 0.160 x 6 x 1023 atoms) of O

have combined together.

WORKSHEET 12.1.5TOPIC 12.2: AMOUNT OF SUBSTANCE 17

2) This ratio of combining atoms is converted to a whole number by dividing through by the lowest number:

0.040 moles Pb 0.040

0.040 moles S combine together 0.040

0.160 moles O 0.040

1 mole Pb, 1 mole S and 4 moles O combine together.

The empirical formula is PbSO4

3) To find the molecular formula:

Molecular formula = Empirical formula x n

Molecular mass = Empirical mass x n

The molecular mass (Mr) = 303The empirical mass (of PbSO4) = 303

n = 1

Molecular formula is PbSO4

The calculation of an empirical formula can be set out more briefly and more conveniently in table form.

Element Mass Ar Moles Simplest ratio Empirical formula

Pb 8.32 207 8.32 = 0.040207

0.040 = 10.040 1

S 1.28 32 1.28 = 0.04032

0.040 = 10.040 1

O 2.56 16 2.56 = 0.1616

0.16 = 40.040

4

The empirical formula is PbSO4

PERCENTAGE COMPOSITION AND EMPIRICAL FORMULAE

1. Calculate the percentage composition by mass of the following compounds, giving your answers to 1 d.p. Use the relative atomic masses given on your periodic table.

a) aluminium chloride

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b) sodium sulphite

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c) lead (II) nitrate

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d) iron (III) oxide

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2. Calculate the empirical formulae of the substances having the following percentage compositions.

a) magnesium 28.57% b) potassium 26.53% c) sodium 36.51% carbon 14.29% chromium 35.37% sulphur 25.40% oxygen 57.14% oxygen 38.10% oxygen 38.10%

a) …………………………………………………………………………………………………………

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b) …………………………………………………………………………………………………………

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c) …………………………………………………………………………………………………………

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3. A hydrocarbon has the composition carbon 85.71%; hydrogen 14.29%

a) Calculate its empirical formula

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b) If its molecular mass is 56, calculate its molecular formula.

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WORKSHEET 12.1.6

EMPIRICAL & MOLECULAR FORMULAE

1. Calculate the empirical formulae of the substances having the following percentage compositions.

a) potassium 28.2% b) sodium 33.3% chlorine 25.6% nitrogen 20.3% oxygen 46.2% oxygen 46.4%

a) …………………………………………………………………………………………………………

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b) …………………………………………………………………………………………………………

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2. What is the empirical formula of a compound containing:4.04% hydrogen, 24.24% carbon and 71.72% chlorine

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The compound has a relative molecular mass of 99; calculate its molecular formula

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3. a) 5g of an oxide (A) of chromium were found on reduction to give 2.6g of chromium. Calculate the empirical formula of this oxide.

Ar (Cr) = 52 Ar (O) = 16

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b) Oxide (A) on strong heating forms oxygen and a second oxide of chromium (B), which is found to contain 24g of oxygen combined with the molar mass of chromium. Calculate the empirical formula of oxide (B).

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c) Using the empirical formula for the two oxides, deduce an equation for the formation of oxide (B) from oxide (A).

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WORKSHEET 12.1.7

CALCULATION OF EMPIRICAL FORMULAE

1. Find the empirical formulae of the compounds formed in the reactions described below.a) 10.800g of magnesium form 18.000g of an oxideb) 3.400g of calcium from 9.435g of a chloridec) 3.528g of iron from 10.237g of a chloride

2. A piece of tin of mass 1.527g was placed in a solution of iodine in trichloromethane and left for several days in a stoppered flask. At the end of that time, the tin remaining in excess was removed, washed with tridhloromethane and dried; its mass was 1.170g. The trichloromethane washings were added to the original solution, then the trichloromethane was evaporated to leave an orange solid, a compound of tin and iodine, of mass 1.881g. Calculate the empirical formula of the compound of tin and iodine.

3. 0.5g of X on complete combustion gave 0.6875g of CO2 and 0.5625g of H2O as the only products. Calculate the masses of C and H in the sample of X and hence find the empirical formula of X.

4. An organic acid contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen; calculate its empirical formula. If the Mr = 60, deduce its molecular formula.

5. An organic compound has a molar mass of 150g.mol-1 and contains 72.0% carbon, 6.67% hydrogen and 21.33% oxygen. What is its molecular formula?


CALCULATION OF REACTING MASSES AND VOLUMESSince chemical equations are quantitative as well as qualitative statements, it is possible to calculate the quantities of reactants necessary to produce a given amount of product, or to calculate the quantity of product(s) given by a specified amount of reactants.

PROCEDURE:1) Write the balanced equation for the reaction concerned.2) Write the ratio of reacting moles for the chemicals in question.3) Convert the masses into moles using moles = mass/(Ar or Mr)4) Use the ratio of moles from step 2 to find number of moles of product formed.5) Calculate mass of product from the number of moles formed.


Example 1:What mass of carbon dioxide will be given off when 9g of calcium carbonate is decomposed by heating?

1) CaCO3(s) CaO(s) + CO2(g) 2) 1 mole 1 mole3)

9g = 9/100 Mr (CO2) = 44 =0.09mol Mr (CaCO3) = 100 4) If 1 mole CaCO3 1 mole CO2

then 0.09 mol 0.09mol

5) mass = moles x Mr

Mass = 0.09 x 44 =3.96g

9g calcium carbonate will produce 3.96g of carbon dioxide.

Example 2:If 4g phosphorus is burned in oxygen, what mass of phosphorus(V) oxide will be formed?

1) 4P(s) + 5O2(g) 2P2O5(s)2) 4 mole 2 moles3)

4g = 4/31 A r (P) = 31 =0.129mol Mr (P2O5) = 142 4) If 4 mole of P 2 mole P2O5

then 0.129 mol 0.0625mol

5) mass = moles x Mr

Mass = 0.09 x 142 =9.16g

4g phosphorus will burn to give 9.16g phosphorus(V) oxide.

THE SPECIAL SITUATION OF GASES:

It is usually more important to know the volumes of gases involved in a reaction rather than their masses. Therefore, molar volume is often used in the calculation in place of molar mass.


Example 3:What volume of hydrogen at r.t.p. will be produced when 3.25g zinc is dissolved in dilute sulphuric acid?

1) Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)2) 1 mole 1 mole3) Ar (Zn) = 65 Molar volume (H2) = 24000cm3

65g 24000cm3

4) 1g 24000 cm3

65

3.25g 24000 x 3.25 cm3

65

3.25g 1200cm3

3.25g zinc reacts to produce 1200cm 3 hydrogen.

WORKSHEET 12.1.8

CALCULATION OF REACTING MASSES AND VOLUMES

Use the relative atomic masses given on your periodic table. The molar volume of a gas at r.t.p. is 24000cm3.

1. If 4.5g of calcium is burned in oxygen according to the equation:2Ca + O2 2CaO

a) What mass of oxygen is needed?b) What mass of calcium oxide is produced?

2. Iron(III) oxide reacts with carbon monoxide to give iron and carbon dioxide according to the equation: Fe2O3 + 3CO 2Fe + 3CO2

If 160g of iron(III) oxide reacts, what volume of carbon dioxide (at r.t.p.) will be given off?

3. Limestone decomposes on heating according to the following equation:CaCO3 CaO + CO2

What mass of 94% pure limestone is needed to make 100 tonnes of quicklime (calcium oxide)?

4. Write an equation for the fermentation of glucose to give ethanol and carbon dioxide. What mass of ethanol could be produced by the fermentation of 1kg of glucose? What volume of carbon dioxide, measured at r.t.p., would be produced in this fermentation?

5. Iron reacts with sulphur to give iron(II)sulphide according to the equation:Fe + S FeS

What mass of sulphur must react to give 6.2g iron(II)sulphide?

6. Potassium reacts with water according to the equation:2K + 2H2O 2KOH + H2

What mass of potassium will give 150cm3 of hydrogen at r.t.p.?

7. Zinc dust burns brightly in air to form zinc oxide, according to the following equation:2Zn + O2 2ZnO

What mass of zinc oxide is formed when 1.626g of zinc reacts?

8. Methane burns in air to form water and carbon dioxide, according to the following equation:

CH4(g) + O2(g) CO2(g) + 2H2O(l)What mass of water and what volume of carbon dioxide (measured at r.t.p.) will be given off when 15000cm3 of methane burns?


9. Nickel carbonate decomposes when heated according to the following equation:NiCO3 NiO + CO2

Calculate the mass of nickel oxide produced when 2.5g of nickel carbonate decomposes.

10. Magnesium reacts with dilute sulphuric acid to form magnesium sulphate and hydrogen, according to the equation:

Mg + H2SO4 MgSO4 + H2

a) Calculate the mass of magnesium needed to produce 22.3g of magnesium sulphateb) Calculate the volume of hydrogen (measured at r.t.p.) given off when 22.3g of

magnesium sulphate are formed.

11. Write a balanced chemical equation for the reaction described in the following word equation:

copper(II) sulphate + sodium hydroxide sodium sulphate + copper(II) hydroxide

Calculate the masses of copper(II) sulphate and sodium hydroxide needed to produce 12.4g of copper(II) hydroxide.

12. Iron reacts when heated with sulphur to form iron(II) sulphide according to the following equation: Fe + S FeSIn one experiment, 5.2g of iron was heated with 4.5g of sulphur; one of the reactants was present in excess.

Calculate a) which reactant is present in excess b) how much of this reactant remains unreacted at the end of the experiment c) the mass of iron(II) sulphide which is formed

13. A mixture of calcium and magnesium carbonates weighing 10.0000g was heated until it reached a constant mass of 5.0960g. Calculate the percentage composition of the mixture by mass.

14. In the thermit reaction: 2Al(s) + Cr2O3(s) 2Cr(s) + Al2O3(s)

Calculate the percentage yield when 180g of chromium are obtained from a reaction between 100g of aluminium and 400g of chromium(III) oxide.


How efficient are reactions?We can gain an understanding of the efficiency of a reaction by considering the percentage yield or by looking at the atom economy of a reaction.

Percentage yieldThe amount of product obtained, usually measured in grams or kilograms, is known as the yield. Often, when we carry out a reaction, the starting materials do not fully react to form products. We say the reaction does not go to completion. This results in many reactions producing a lower yield than expected.

It is useful, therefore, to know the percentage yield of a reaction. This compares the amount of product that the reaction really produces with the maximum amount it could possibly produce if it went to completion.

Percentage yield = amount of product produced by the reaction x 100 maximum amount of product possible

Examples:

1. It was calculated that a reaction could produce 24g of iron. When it was carried out, the mass of iron produced was 15.6g. Calculate a percentage yield for this reaction.

2. 12g of magnesium were burned in oxygen. 15.8g of magnesium oxide were produced.

a) write a word equation and then a balanced symbol equation for the reaction

b) calculate the number of moles of magnesium that were used in the reaction

c) using the equation, state the number of moles of magnesium oxide that would be formed

d) calculate the mass of magnesium oxide that would be formed if the reaction went to completion

e) thus calculate the percentage yield of magnesium oxide for this reaction.


Atom EconomySome reactions could have very high percentage yields of product – the reaction could go to completion – but if a lot of unwanted materials are produced the process may still not be considered to be efficient.

If a reaction produces 3 different products but only one is useful, even a reaction with a 100% yield could produce a large amount of waste. A way to measure the amount of starting materials that end up as useful products is to calculate the atom economy (or atom utilisation) of the reaction. It is important for sustainable development and for economical reasons to use large scale reactions with a high atom economy.

% atom economy = (Mr of useful product) x its balancing number in the equation x 100(Mr of ALL reactants) x their balancing numbers

e.g. Iron oxide is converted into iron by reduction with carbon monoxide:

Fe2O3 + 3CO 2Fe + 3CO2

In this reaction, the carbon dioxide is a waste product. The iron is the useful product.

Percentage atom economy is calculated by:

M r of Fe x balancing number x 100 [(Mr of Fe2O3 x balancing number) + (Mr of CO x balancing number)]

So: (56 x 2) x 100 (160) + (28 x 3)

= 45.9%

Example 1:

Calcium carbonate decomposes on heating to form quicklime (calcium oxide), a useful product, according to the following equation:

CaCO3 CaO + CO2

Calculate the atom economy for this reaction.........................................................................................................................................................

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Example 2:

Ethanol can be manufactured by fermentation according to the following equation:

C6H12O6 2C2H5OH + 2CO2 Equation 1

It can also be manufactured by direct hydration of ethene:

C2H4 + H2O C2H5OH Equation 2

(a) Calculate the atom economy for the production of ethanol by fermentation.........................................................................................................................................................

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(b) What is the atom economy for the direct hydration of ethene? ........................................................................................................................................................

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(c) Suggest one advantage and one disadvantage of each method of manufacture.

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TOPIC 12.1: AMOUNT OF SUBSTANCE

LOWER SIXTH TEST: FORMULAE & MOLE CALCULTIONS

1. Write down the formulae of:

a) calcium chloride ………………………….

b) potassium sulphide ………………………….

c) iron (II) hydroxide ………………………….

d) magnesium nitrate ………………………….

e) aluminium oxide …………………………. (5)

2. Calculate the relative molecular masses of :

a) Na2CO3 .……………………………………………………………………………….……

……………………………………………………………………………………………………

b) Cu(NO3)2…………………………………………………………………………………….

……………………………………………………………………………………………… (4)

3. Natural magnesium comprises 78.6% 24Mg, 10.1% 25Mg and 11.3% 26Mg. Calculate the relative atomic mass of magnesium.

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………… (3)

4. Calculate the percentage composition ( to 1 d.p.) of MgCO3.

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

………………………………………………………………………………………………………. (4)


5. A solid is found to contain 31.8% K, 29.0% Cl, 39.2% O. Calculate its empirical formula.

(4)

6. Calculate the number of moles contained in:

a) 2.8g of sodium hydroxide

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

b) 22.5cm3 of 0.055M sodium carbonate

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

………………………………………………………………………………………………………. (4)

7. Calculate the relative molecular mass of a gas, 4.4g of which occupy a volume of 2100cm3 at a temperature of 25oC and a pressure of 118,000Pa.

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………… (3)

8. Potassium chlorate(V) decomposes on heating according to the following equation:2KClO3(s) 2KCl(s) + 3O2(g)

What mass of potassium chlorate(V) must be heated to give 1dm3 of oxygen at r.t.p.?(molar volume at r.t.p. = 24000cm3)

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

……………………………………………………………………………………………………………

………………………………………………………………………………………………………. (3)TOPIC 12.2: AMOUNT OF SUBSTANCE 31

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