DEMO

STOICHIOMETRY AND THE MOLE CONCEPT

By the end of the sub-topic, learners should be able to:
  1. State the symbols of elements and formulae of compounds.
  2. Determine the formula of an ionic compound from the charges on the ions present and vice-versa.
  3. Write balanced equations including ionic equations.
  4. Define and calculate relative molecular mass and formula mass.
  5. Calculate the stoichiometric reacting masses of reactants and products.
  6. Calculate volume of gases in chemical reactions and use the mole concept to calculate empirical and molecular formulae.

State symbols

  • The chemical substances that react are in certain states.
  • The state can be solid, gaseous, sand liquid or dissolved in water, that is, aqueous.
  • In a chemical equation these states are written as symbols and these symbols are referred to as state symbols.
  • For example, the chemical reaction below:
Aaq + Bl  Cs + Dg
  • A and B are reactants, while C and D are products. The state symbols are shown in brackets.
  • Table 2.1.1 is a summary of the state symbols and what they represent.

State State symbol
Solid (s)
Liquid (l)
Gas (g)
aqueous (dissolved in water) (aq)


Elements of the periodic table and their symbols

  • Elements in the periodic can be represented by symbols and it is these symbols that are used in writing the chemical equations.
  • The formulae of compounds are written using the symbols of the elements.
  • The symbols of the first twenty elements will be considered and given below:
  1. H – Hydrogen
  2. He – Helium
  3. Li – Lithium
  4. Be – Beryllium
  5. B – Boron
  6. C – Carbon
  7. N – Nitrogen
  8. O – Oxygen
  9. F – Fluorine
  10. Ne – Neon
  11. Na – Sodium
  12. Mg – Magnesium
  13. Al -  Aluminium
  14. Si – Silicon
  15. P – Phosphorus
  16. S – Sulphur
  17. Cl – Chlorine
  18. Ar – Argon
  19. K – Potassium
  20. Ca- Calcium

Formulae of compounds

  • As studied before, elements can form ions by either losing or gaining electrons.
  • The number of electrons that the element can gain or lose depends on the group the element is in.
  • Metals tend to lose electrons to gain a stable configuration. Non-metals tend to accept the electrons.
  • The compounds formed when metals lose electrons to non-metals are called ionic compounds.
  • Non-metals form compounds between themselves by sharing electrons depending on the group to which the element belongs. These are called molecular compounds.

Formation of compounds

  • A group 1 element will combine with a group 7 in the ratio of 1:1 since a group 7 element needs one electron and a group 1 metal can provide only one electron.
  • Simply put, since sodium forms Na+ and Chlorine forms Cl-, the neutral compound formed will have 1 sodium and 1 chlorine atom
  • Therefore, the formula for sodium chloride will be NaCl. Silver chloride will be AgCl, while potassium iodide will be KI.
  • Group 2 elements can also combine with two group 7 elements.
  • However, since calcium will form a Ca2+ and chlorine a Clion, it is natural that we will require 2 Chlorine atoms to balance out the charges.
  • The formula for calcium chloride will therefore be ions chloride will be CaCl2
  • Similarly, 2 group 1 metals will be needed to combine with one group 6 element for example sodium oxide will is represented as Na2O and water will be H2O.
  • Therefore knowing the valency of an element helps us know the ions that it is likely to form and this will help us to determine the formula of the compounds it will make.
  • By also having the formula of a compound, it becomes easy to determine the ions of the constituent elements.
  • It is important to note that any given, specific compound has a constant composition with the formula determined as above.
Table 2.1.2: Common cations 
Charge Formula Name Formula Name
1+ H Hydrogen ion NH4+ Ammonium ion
Li+ Lithium ion Cu+ Copper(I) or cuprous ion
Na+ Sodium ion    
K+ Potassium ion    
Cs+ Caesium ion    
Ag+ Silver ion    
2+ Mg2+ Magnesium ion Co2+ Cobalt(II) or cobaltous ion
Ca2+ Calcium ion Cu2+ Copper(II) or cupric ion
Sr2+ Strontium ion Fe2+ Iron(II) or ferrous ion
Ba2+ Barium ion  Mn2+ Manganese(II) or manganous ion
Zn2+ Zinc ion Hg22+ Mercury(I) or mercurous ion
Cd2+ Cadmium ion Hg2+ Mercury(II) or mercuric ion
Ni2+ Nickel(II) or nickelous ion
Pb2+ Lead(II) or plumbous ion
Sn2+ Tin(II) or stannous ion
3+ Al3+ Aluminium ion Cr3+ Chromium(III) or chromic ion
Fe3+ Iron(III) or ferric ion

Table 2.1.3: Common anions
Charge Formula Name Formula Name
1- H- Hydride ion C2H3O2- Acetate ion
F- Fluoride ion ClO3- Chlorate ion
Cl- Chloride ion ClO4- Perchlorate ion
Br- Bromide ion NO3- Nitrate ion
I- Iodide ion MnO4- Permanganate ion
CN- Cyanide ion    
OH- Hydroxide ion    
2- O2- Oxide ion CO32- Carbonate ion
O22- Peroxide ion CrO42- Chromate ion
S2- Sulphide ion Cr2O72- Dichromate ion
    SO42- Sulphate ion
3- N3- Nitride ion PO43- Phosphate ion
   

* The most common ions are in bold

Radicals

  • A radical is an atom or molecule that has unpaired valence electrons and is unstable and highly reactive.
  • The table below shows some common radicals and their valency.
Table 2.1.4: Some radicals and their formulae
Radical Formula Valency
Hydroxyl OH- -1
Nitrate NO3- -1
Sulphate SO42- -2
Carbonate CO32- -2
Ammonium NH4+ +1

  • The radicals can also combine with elements based on their valence, for example, group 1 elements combine with one hydroxyl while two of them will combine with a carbonate.
  • That way the formula of the compound so formed is predicted.

Relative molecular mass

  • This refers to the ratio of the average mass of one molecule of an element or compound to one twelfth of the mass of an atom of 12C.
  • Mr=Average mass of one molecule of element or compound112of the mass of carbon 12
  • The relative molecular mass is generally found by adding the atomic masses of the constituent elements in the compound as they are given in the formula.
  • Relative molecular mass has no units and is given the symbol Mr.
Calculating the relative molecular mass.
Example 2.1.1
Calculate the relative molecular mass of the following compounds
(a) Sodium oxide (Na2O).
(b) Calcium carbonate (CaCO3).
Solution
(a) A sodium oxide molecule is made up of 2 sodium atoms and 1 oxygen atom.
Mr = (2 x 23) + 16
     = 62
(b) A calcium carbonate molecule has 1 calcium atom, 1 carbon atom and 3 oxygen atom.
Mr = 40 + 12 + (3 x 16)
     = 100
  • Relative formula mass is the term that is used if the compound contains both metals and nonmetals.
  • Molecular mass is normally used for substances that are covalent in nature, that contain nonmetals only.

Writing chemical equations

  • Chemical equations are written using chemical symbols and taking care to make sure that the compounds are properly written.
  • The compounds should have their state symbols written.
  • The reactants should be on the left side of the arrow while the products are at the right side.
  • The equation should be balanced, that is, the atoms on the right side should be equal to those on the right side.

Balancing equations.

  • We balance chemical equations based on the principal of conservation of mass; that is, the total mass of reactants = total mass of products (no mass is lost).
  • A balanced equation has an equal number of atoms of each kind on both sides of the chemical equation.
  • The coefficient of the compound multiplied by the subscript gives the number of atoms.
  • For example there are six oxygen atoms in 3O2.
  • The subscript of a molecule should not be changed for example, we cannot write oxygen as O in trying to balance the equation.
  • A coefficient cannot be put in the middle of a compound but at the beginning and it multiplies the whole compound.
Example 2.1.2: Balance the following equation
                            H2(g)+O2(g)H2O(l)
Solution
  • There are two hydrogen atoms on the left hand side and two of them on the right hand side.
  • There are two oxygen atoms on the left hand side but only one on the right making the equation unbalanced.
  • Putting a two on the water balances the oxygen but unbalances the hydrogen.
  • Putting a two on the hydrogen on the left will balance the whole equation such that the balanced equation becomes;

                          2H2(g)+O2(g)2H2O(l)
The mole
  • This refers to the amount of pure chemical substance that contains as many particles as there are in 12g of C12 atom.
  • The particles refer to atoms, molecules, ions or electrons.
  • The number of particles in the mole is 6,023 x 1023 and this is known as the Avogadro's constant (L).
  • Therefore one mole of a substance contains 6,023 x 1023 particles.
Worked examples.
Example 2.1.3
In the questions use L= 6,023 x 1023 mole-1
Calculate the number of atoms in 0,25 moles of sodium.
Solution
 1 mole contains 6,023 x 1023 mole-1
0,25moles contains 0,25 moles x 6,023 x 1023 mole-1
                        = 1,51 x 1023 atoms

Stoichiometric calculations

  • Stoichiometric calculations involve the amounts of substances in moles [n], mass of substance [m] (g) and molar mass [Mr] (g/mol).
  • Given any two of the above, it is possible to obtain the unknown quantity using the formula n = m/Mr.
  • If the amounts of the reacting species are known, the amount of the products can be calculated.
  • Stoichiometric coefficients (numbers used to balance to balance equations) are used to calculate ratios which tell us about the relative proportions of the chemicals in our reaction.
  • To be able to make correct stoichiometric calculations, the equation must be balanced.
Example 2.1.4
How many grams of NaOH will be required to react fully with 6.20g of H2SO4?
Solution:
The first step is the balanced chemical equation with state symbols;

2NaOH(l)+H2SO4(l)Na2SO4(aq) +2H2O(l)

  • The reactant masses should be converted to moles
  • Find the molecular mass of H2SO4 =(2 x 1)+32 +(4x16)=98g/mol
    • The number of moles of H2SO4 = m/Mr
      = 6,20g98gmol-1
      = 0,0633moles
  • The ratio of moles can be used to calculate the number of moles of the other reactant.
    • mole ratio NaOH : H2SO4
                        2    :  1
    • Number of moles of NaOH = number of moles of H2SO4 x2
                                      = 0, 0633mol x 2
                                       = 0, 1265 moles
                                      
     Mr of NaOH = 23 + 16 + 1 = 40g/mol
  • Having the number of moles of NaOH, ( the mass can be calculated from the relationship:   
                 mass = number of moles x Mr
                                               = 0,1265mol x 40g/mol
                                               = 5,056g
Example 2.1.5
How many moles of ZnCl2 will be formed in the reaction of zinc with HCl if 100g of zinc were used and the HCl was in excess? Solution
  • The equation is: Zn(s)+2HCl(l)ZnCl2(ag)+H2(g)
  • We can calculate the number of moles of zinc from the given mass.
        • Number of moles of zinc  = m/Mr
        • = 100g/65(g/mol)

          = 1.538 moles


          Ratio of Zn : ZnCl2

                       1  :  1

  • Number of moles of ZnCl2 produced = 1.538 moles
  • Mass of ZnCl2 produced   = number of moles x Mr
                                              = 1,538mol x 136g/mol
                                              = 209,17g

Molar gas volumes

  • This refers to the volume that is occupied by a gas at a given temperature and pressure.
  • A mole of any gas will occupy the same volume at the same temperature and pressure.
  • 1 mole of any gas occupies 28dmat room temperature and pressure under Zimbabwean conditions.
  • Knowing the volume of a gas, one can be able to calculate the number of moles of the gas and vice versa.
  • The following example illustrates that.
Example 2.1.6
What is the volume of 6g of carbon dioxide? (1 mole of gas occupies 28dm3). Solution
Given the mass of carbon dioxide, we can calculate the number of moles and then the volume. Number of moles = mass/Mr
    = 6g44g/mol
                  =0,136 moles
If 1mole occupies 28dm3, then 0,136moles occupy 0,136 x 28 = 3, 81dm3

Empirical formula

  • This is the simplest ratio in integers, of the atoms present in a compound.
  • It does not give the actual number or arrangement of the atoms.
  • The empirical formula of a compound can be calculated if the relative abundances of the atoms is given.
Example 2.1.7
Calculate the empirical formula of methyl acetate which has the following percentage composition:
carbon : 48,64%
hydrogen: 8,16%
oxygen :  43,20%
Solution
First we make the assumption that the mass of the methyl acetate is 100g such that the percentages become the mass of the atoms.
It therefore means that the ratio of atoms will be
   C         :          H            :      O
48,64g            8,16g                43,20g
These masses are then converted to moles.
   C         :         H              :     O
48,64/12          8,16/1            43,20/16
4,0533             8,16                2,7
Divide each by the smallest that is 2,7
  C      :              H       :         O
1,5                   3,0               1
To get whole numbers the values are multiplied by 2 such that we get C = 3,
H = 9 and O = 3.
Therefore the empirical formula is C3H3O3

Calculating the molecular formula from the empirical

  • The molecular formula can be obtained from the empirical by multiplying with an integer (whole number, n).

molecula formula= n (empirical formula)

  • The integer, n, is obtained by relating the actual molecular mass with the empirical mass.

n= nmolecular massemperical mass

  • For example, given that a substance ‘s empirical formula is CH2  (empirical mass= 12 + 2= 14) and the molecular mass is 28, we can obtain n:
    n= 2814
    = 2
  • Therefore the molecular mass is      = 2(CH2)
    =C2H4

Solution concentration

  • The concentration of solutions can be expressed as gdm-3 or moldm-3.
  • This means that the mass of the compound is divided by the volume to get the concentration in gdm-3.
  • A dm3 is the same as a litre or 1000cm3.
  • To get the concentration (c) in moldm-3 the number of moles (n) of the substance is divided by the volume (v)
  • The following equation can be used  c= nv
  • Concentration in moldm-3 is sometimes written as M.
  • To express concentration in gdm-3, simply divide the grams by 1000cm3.