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AS Foundation Chemistry - Atoms & Stoichiometry


Atoms & Stoichiometry - Definitions

(1) Isotope :

Isotopes of an element are atoms that have the same number of protons but a different number of neutrons.

N.B.: Some elements only have one isotope, for example fluorine, phosphorus and iodine.

(2) Relative Isotopic Mass :

This is the mass of an atom of an isotope of an element relative to 1/12 the mass of a 12C atom.

(3) Relative Atomic Mass :

This is the average mass of isotopes of an element relative to 1/12 the mass of a 12C atom.

(4) Relative Formula Mass :

This is the sum of the relative atomic masses of all the atoms present in a compound.

N.B.: This is sometimes called the relative molecular mass, though that term would properly only apply to a (covalent) molecule.

(5) The Mole :

A mole is defined as a particular number of particles of an element or compound. This number is called the Avogadro constant and has the value of 6.02x1023, a very large number.

"Particles of a compound" could mean atoms, diatomic molecules, covalent molecules to ionic compounds.

However, they must be particles of a particular compound or element and not a mixture of compounds or elements.

(6) Molar Mass :

This is defined as the mass of one mole of a substance, whether it be atoms or molecules of a substance.

(7) Empirical Formula :

The empirical formula for a compound (covalent or ionic) is the lowest whole number ratio of atoms (or ions) to each other.

(8) Molecular Formula :

The molecular formula of a compound is the actual number of each atom present.

Compound Empirical formula Molecular formula
ethane CH3 CH3CH3
sodium oxalate NaCO2 Na2C2O4
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Atoms & Stoichiometry - Mass Spectroscopy

Mass spectroscopy if the analytical technique where elements and compounds are turned into ions and the masses of the ions produced are detected, producing a graph of mass vs. intensity, for the element (or compound).

(1) Equipment :

 

(2) Process :

There are four different stages in mass spectroscopy,

(i) Ionisation -

In this stage the sample is injected, either as a gas or as a solution in a volatile solvent. If the sample isn't already a gas then it is heated until it vaporises.

The gaseous particles are then bombarded by high energy electrons, which push at least one electron out of the particle, creating positive ions.

M(g) → M+(g) + e-

N.B.: More than one electron could be removed, though at AS level it is limited to just one.

(ii) Acceleration -

The ions then pass through electrically charged plates. The first plate is negatively charged, which pulls the ions into the next plate which is positively charged, repelling the ions. This speeds up the positive ions as they pass into the deflection stage.

(iii) Deflection -

The high speed ions then pass through a variable electro-magnetic field. The mass of the ions determines whether they will pass through this stage to the detector.

For a particular magnetic strength, ions that are too heavy have too much momentum and do not bend. Ions that are too light are bent too much (see the diagram below).

(iv) Detection -

Once ions hit the detector, electrons are generated, which are multiplied and the potential produced recorded to produce the output.

The quantity measured is the mass of the ion / charge of the ion, and is given the symbol m/z.

N.B.: In AS only a charge of +1 is considered, so m/z values are just the mass of the ion.

Below is an example of a mass spectrum produced by chlorine gas, showing the details of the isotopes present in atoms and molecules,


(3) Calculations :

A mass spectrum can be used to determine the relative atomic mass for an element, or just to determine the elements present in a mixture of elements or compounds.

(i) Determining relative atomic mass -

The mass spectrum of an element has peaks at m/z for each individual atom of the element, as well as for different isotopic versions of the element's molecule (see the mass spectrum of chlorine above).

The relative atomic mass can be calculated by multiplying the intensity of each isotope by its mass, summing the numbers and dividing by the total intensity. For example, for chlorine -

(ii) Detecting elemental composition -

The appearance of certain lines in a mass spectrum enables the detection of particular elements. Some common atoms are given in the table below,

Atom Isotopic lines Relative abundances
C 12, 13 99 : 1
Cl 35, 37 25 : 75
Br 79, 81 50 : 50
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Atoms & Stoichiometry - Chemical Equations and Amount Calculations

Experimental sheet for balancing a chemical equation.

The importance of balancing the numbers of atoms present in a chemical equation was first mentioned in the GCSE introductory unit, and you should go back and look at this if you are unsure at all about balancing equations.

These balancing numbers have an additional meaning where these amount calculations are concerned. The balancing numbers give the numbers of moles present for each chemical involved in the reaction.

For example, sodium chloride may be prepared by the reaction of sodium hydroxide(aq) and hydrochloric acid, according to the following equation -

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Last year, all this equation showed was that sodium hydroxide reacts with hydrochloric acid to give a salt (sodium chloride) and water.

However, the equation says more than just this. The equation specifies what amounts of sodium hydroxide and hydrochloric acid will react together to give the products -

1NaOH(aq) + 1HCl(aq)1NaCl(aq) + 1H2O(l)

This equation states that 1 mol of sodium hydroxide and 1 mol of hydrochloric acid will react together to give of 1 mol sodium chloride and of 1 mol water.

The ratio of NaOH : HCl : NaCl : H2O is 1 : 1 : 1 : 1.

What this means is that if the amount of any one of the components in the above reaction is known then the others may be worked out from the ratio above.

Or, if the masses of chemicals reacting together are known then amounts can be calculated and the balancing numbers deduced from the amounts reacting together.

Exemplar calculations -

(i) What amount of sodium chloride will be produced from 0.25 mol of sodium hydroxide?

Ratio of NaOH : NaCl in the chemical equation is 1 : 1

amount of NaCl produced = 0.25 mol

(ii) What amount of hydrochloric acid is needed to make 0.5 mol if sodium chloride?

Ratio of NaCl : HCl in the chemical equation is 1 : 1

∴ amount of HCl needed = 0.5 mol

This is all well and good; however, measuring the amount of a compound present isn't possible directly. The mass of a compound is easy to measure though. Then the mass can be converted into an amount of a compound and the mathematical process as described above can be followed.

e.g. What amount of sodium chloride will be produced from 20 g of sodium hydroxide?

molar mass of sodium hydroxide = (23+16+1) g mol-1
= 40 g mol-1
∴ amount of sodium hydroxide = 20 g / 40 g mol-1
= 0.5 mol

Ratio of NaOH : NaCl in the chemical equation is 1 : 1

∴ amount of NaCl produced = 0.5 mol

It is then a simple matter to extend the question and ask for the mass of a compound formed/used and not its mass.

e.g. What mass of hydrochloric acid is need to give 19.5 g of sodium chloride?

molar mass of sodium chloride = (23+35.5) g mol-1
= 58.5 g mol-1
∴ amount of sodium chloride = 19.5 g / 58.5 g mol-1
= 0.33 mol

Ratio of NaCl : HCl in the chemical equation is 1 : 1

amount of HCl needed = 0.33 mol
∴ mass of HCl needed = 0.33 mol × (1+35.5) g mol-1
= 12.16 g

This is rather a simplistic example; however, it shows the general use of amounts, molar masses and chemical equations.

The calculations get harder when the chemical equation involves more complex balancing numbers.

e.g. What mass of sodium sulphate would be formed when 5 g of sodium hydroxide is reacted with excess sulphuric acid?

Step 1 : chemical equation

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

Step 2 : calculate the amount of sodium hydroxide used.

molar mass of sodium hydroxide = (23+1+16) g mol-1
= 40 g mol-1
amount of sodium hydroxide = 5 g / 40 g mol-1
= 0.125 mol

Step 3 : use the ratio from the chemical equation

Ratio of NaOH : Na2SO4 in the chemical equation is 2 : 1

∴ amount of Na2SO4 produced = 0.125 mol / 2
= 0.0625 mol

Step 4 : calculate the mass of product

∴ mass of Na2SO4 produced = 0.0625 mol × ((2×23)+32+(4×16)) g mol-1
= 8.875 g

This is the standard calculation used when you have, or may easily work out, a chemical equation for a reaction.

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Atoms & Stoichiometry - Molar volume of a gas

Experimental sheet for determining the molar volume of a gas.

Experiments discovered that gases react together in volume ratios. That is to say that 1 mol of any gas occupies a particular volume at a particular temperature.

From physics, it is known that the volume of a gas is proportional to its temperature, according to the mathematical equation -

so the higher the temperature the greater the volume a gas occupies, keeping the pressure constant, at atmospheric pressure say.

Temperature Volume (dm3)
0 oC or 32 oF or 273.1 K (or 0 oR or 491.6 R) 22.4
20 oC or 68 oF or 293 K (or 16 oR or 527.6 R) 24.1
25 oC or 77 oF or 298 K (or 20 oR or 536.6 R) 24.5

The value normally used in calculations in examinations is 24 dm3 as this is closest to a value for room temperature.

This relationship between gas volume and amount forms another mathematical relationship -

or it can be expressed in a triangular relationship -

Using a similar method to that for the amount/mass/relative formula mass triangle, amounts of gas can be converted into volumes and volumes of gas converted into amounts

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Amount - Advanced Calculations

(1) Percentage composition in a compound :

There are two uses of this type of calculation,

(i) The calculation of the percentage of an element in a compound -

e.g. What is the percentage of oxygen in carbon dioxide gas?

mass of oxygen in one mole of carbon dioxide gas = 2×16 g
= 32 g
mass of one mole of carbon dioxide = 12 g + (2×16 g)
= 44 g
∴ percentage of oxygen = (32 g/44 g) × 100
= 72.7 %

(ii) The calculation of the percentage of water of crystallization in a compound -

e.g. What is the percentage of water of crystallization in hydrated copper(II) sulphate, CuSO4.5H2O?

mass of water present in one mole of hydrated copper(II) sulphate =(5×18 g)
= 90 g
mass of one mole of hydrated copper(II) sulphate = 64 g + 32 g + (4×16 g) + (5×18 g)
= 250 g
∴ percentage of water of crystallization present = (90 g/250 g) × 100
= 36 %

(2) Empirical formula of a compound :

This technique covers essentially the reverse of the above process. A typical question here involves the calculation of the lowest possible whole number ratio of atoms in an organic molecule, given the percentage elemental composition for that molecule.

e.g. An organic compound contains carbon and hydrogen only in the ratio of 85.7 % carbon to 14.3 % hydrogen. What is the empirical formula for that compound?

C H
% composition 85.7 % 14.3 %
mass of element in 100 g of compound 85.7 g 14.3 g
∴ amount of element in 100 g of compound 85.7 g/12 g mol-1 = 7.14 mol 14.3 g/1 g mol-1 = 14.3 mol
∴ lowest whole number ratio of element's amounts 7.14 mol/7.14 mol = 1 14.3 mol/7.14 mol = 2

The empirical formula for the compound is CH2.

(3) Molecular formula of a compound :

The molecular formula for a compound is simply the empirical formulae multiplied by some whole number ( 1, 2, 3, etc.. ).

Knowing the molar mass of the compound enables the molecular formula of it to be found.

e.g. The actual molar mass of the compound in the previous example is 42 g mol-1. What is the molecular formula for this compound?

The molar mass of CH2 = (12+(2×1)) g mol-1= 14 g mol-1

The molecular formula of this compound is (CH2)n, where n is a positive whole number.

∴ The value of n = 42 g mol-1 / 14 g mol-1
= 3

the molecular formula is (CH2)3 or more properly C3H6.

(4) Percentage yield for a reaction :

e.g. In the reaction between iron metal and an excess of copper(II) sulphate solution, 0.56g of iron is reacted and 0.48 g of copper metal is formed. What is the yield for this reaction?

the equation - Fe(s) + CuSO4(aq) → Cu(s) + FeSO4(aq)

the amount of iron used = 0.56 g/ 56 g mol-1
= 0.01 mol

From the equation, the maximum amount of copper metal that could be formed = the amount of iron used, as the ratio of balancing numbers is 1 : 1.

∴ the maximum possible amount of copper metal = 0.01 mol
∴ the maximum possible mass of copper metal formed = 0.01 mol × 64 g mol-1
= 0.64 g
∴ the percentage yield of copper =(mass produced / maximum possible mass)× 100
=(0.48 g/ 0.64 g) × 100
= 75 %

(5) Percentage purity of a compound :

e.g. Q/ In the reaction between aluminium metal and an excess of copper(II) sulphate solution, 4.0 g of aluminium is reacted and 0.48 g of copper metal is formed.  Assuming that all the aluminium present reacted to give copper, what is the purity of the aluminium?

The equation -

2Al(s) + 3CuSO4(aq) → 3Cu(s) + Al2(SO4)3(aq)

the amount of copper made = 4.8 g/ 64 g mol-1
= 0.075 mol

From the equation, the maximum amount of aluminium metal that could be present = 2/3 × the amount of copper made, as the ratio of Al : Cu is 2 : 3.

∴ the maximum possible amount of aluminium metal =2/3 × 0.075 mol
=0.05 mol
∴ the maximum possible mass of aluminium metal present = 0.05 mol × 27 g mol-1
=1.35 g
∴ the percentage purity of aluminium used =(mass present / total mass of compound) × 100
= (1.35 g / 4.0 g) × 100
= 33.75 %
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Amount - Practice Calculations

(1) 3.5 g of lithium is heated and reacted with an excess of oxygen to make lithium oxide. What mass of lithium oxide is produced?

Take the atomic mass of lithium to be 7 g mol-1 and oxygen to be 16 gmol-1

First balance the equation for the reaction -


Li(s) + O2(g)Li2O(s)

Then fill in the boxes below with your answers and press check to see if you are correct -

amount of lithium used = mol
amount of lithium oxide formed = mol
mass of lithium oxide formed = g

(2) 18 g of pentane is combusted in excess oxygen to make carbon dioxide and water only. What is the minimum volume of oxygen needed?

Take the atomic mass of carbon to be 12 g mol-1 and hydrogen to be 1 gmol-1, and the molar volume of a gas to be 24 dm3 mol-1

First balance the equation for the reaction -


C5H12(l) + O2(g)CO2(g) + H2O(l)

Then fill in the boxes below with your answers and press check to see if you are correct -

amount of pentane used = mol
amount of oxygen needed = mol
volume of oxygen needed = dm3

(3) When sodium nitrate is heated strongly it decomposes to give sodium nitrite and oxygen gas. If 60 cm3 of oxygen is formed from a sample of sodium nitrate, what mass of sodium nitrate must have been used?

Take the atomic mass of sodium to be 23 g mol-1, nitrogen to be 14 g mol-1 and oxygen to be 16 gmol-1, and the molar volume of a gas to be 24 dm3 mol-1

First balance the equation for the reaction -


NaNO3(s)NaNO2(s) + O2(s)

Then fill in the boxes below with your answers and press check to see if you are correct -

amount of oxygen formed = mol
amount of sodium nitrate used = mol
mass of sodium nitrate used = g
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written by Dr Richard Clarkson : © Saturday, 1 November 1997

Updated : Sunday 15th July, 2012

mail to: chemistryrules

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