 # 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:
• 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 $N{H}_{{4}^{+}}$ Ammonium ion
$L{i}^{+}$ Lithium ion $C{u}^{+}$ Copper(I) or cuprous ion
$N{a}^{+}$ Sodium ion
${K}^{+}$ Potassium ion
$C{s}^{+}$ Caesium ion
$A{g}^{+}$ Silver ion
2+ $M{g}^{2+}$ Magnesium ion $C{o}^{2+}$ Cobalt(II) or cobaltous ion
$C{a}^{2+}$ Calcium ion $C{u}^{2+}$ Copper(II) or cupric ion
$S{r}^{2+}$ Strontium ion $F{e}^{2+}$ Iron(II) or ferrous ion
$B{a}^{2+}$ Barium ion Manganese(II) or manganous ion
$Z{n}^{2+}$ Zinc ion $H{{g}_{2}}^{2+}$ Mercury(I) or mercurous ion
$C{d}^{2+}$ Cadmium ion $H{g}^{2+}$ Mercury(II) or mercuric ion
$N{i}^{2+}$ Nickel(II) or nickelous ion
$P{b}^{2+}$ Lead(II) or plumbous ion
$S{n}^{2+}$ Tin(II) or stannous ion
3+ $A{l}^{3+}$ Aluminium ion $C{r}^{3+}$ Chromium(III) or chromic ion
$F{e}^{3+}$ Iron(III) or ferric ion

Table 2.1.3: Common anions
Charge Formula Name Formula Name
1- ${H}^{-}$ Hydride ion ${C}_{2}{H}_{3}{{O}_{2}}^{-}$ Acetate ion
${F}^{-}$ Fluoride ion $Cl{{O}_{3}}^{-}$ Chlorate ion
$C{l}^{-}$ Chloride ion $Cl{{O}_{4}}^{-}$ Perchlorate ion
$B{r}^{-}$ Bromide ion $N{{O}_{3}}^{-}$ Nitrate ion
${I}^{-}$ Iodide ion $Mn{{O}_{4}}^{-}$ Permanganate ion
$C{N}^{-}$ Cyanide ion
$O{H}^{-}$ Hydroxide ion
2- ${O}^{2-}$ Oxide ion $C{{O}_{3}}^{2-}$ Carbonate ion
${{O}_{2}}^{2-}$ Peroxide ion $Cr{{O}_{4}}^{2-}$ Chromate ion
${S}^{2-}$ Sulphide ion $C{r}_{2}{{O}_{7}}^{2-}$ Dichromate ion
$S{{O}_{4}}^{2-}$ Sulphate ion
3- ${N}^{3-}$ Nitride ion $P{{O}_{4}}^{3-}$ Phosphate ion

* The most common ions are in bold

• 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
Hydroxyl $O{H}^{-}$ -1
Nitrate $N{{O}_{3}}^{-}$ -1
Sulphate $S{{O}_{4}}^{2-}$ -2
Carbonate $C{{O}_{3}}^{2-}$ -2
Ammonium $N{{H}_{4}}^{+}$ +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.
• 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 ${M}_{r}$.
Calculating the relative molecular mass.
Example 2.1.1
Calculate the relative molecular mass of the following compounds
(a) Sodium oxide ($N{a}_{2}O$).
(b) Calcium carbonate ($CaC{O}_{3}$).
Solution
(a) A sodium oxide molecule is made up of 2 sodium atoms and 1 oxygen atom.
${M}_{r}$ = (2 x 23) + 16
= 62
(b) A calcium carbonate molecule has 1 calcium atom, 1 carbon atom and 3 oxygen atom.
${M}_{r}$ = 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

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;

$2{H}_{2\left(g\right)}+{O}_{2\left(g\right)}\to 2{H}_{2}{O}_{\left(l\right)}$
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 ${H}_{2}S{O}_{4}$?
Solution:
The first step is the balanced chemical equation with state symbols;

• The reactant masses should be converted to moles
• Find the molecular mass of ${H}_{2}S{O}_{4}$ =(2 x 1)+32 +(4x16)=98g/mol
• The number of moles of ${H}_{2}S{O}_{4}$ = $m/{M}_{r}$
= $\frac{6,20g}{98gmo{l}^{-1}}$
= 0,0633moles
• The ratio of moles can be used to calculate the number of moles of the other reactant.
• mole ratio NaOH : ${H}_{2}S{O}_{4}$
2    :  1
• Number of moles of NaOH = number of moles of ${H}_{2}S{O}_{4}$ x2
= 0, 0633mol x 2
= 0, 1265 moles

${M}_{r}$ 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 ${M}_{r}$
= 0,1265mol x 40g/mol
= 5,056g
Example 2.1.5
How many moles of $ZnC{l}_{2}$ 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: $Z{n}_{\left(s\right)}+2HC{l}_{\left(l\right)}\to ZnC{l}_{2\left(ag\right)}+{H}_{2\left(g\right)}$
• We can calculate the number of moles of zinc from the given mass.
• Number of moles of zinc  = m/${M}_{r}$
• = 100g/65(g/mol)

= 1.538 moles

Ratio of Zn : $ZnC{l}_{2}$

1  :  1

• Number of moles of $ZnC{l}_{2}$ produced = 1.538 moles
• Mass of $ZnC{l}_{2}$ produced   = number of moles x ${M}_{r}$
= 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/${M}_{r}$
= $\frac{6g}{44g/mol}$
=0,136 moles
If 1mole occupies 28dm3, then 0,136moles occupy 0,136 x 28 = 3, 81$d{m}^{3}$

#### 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 ${C}_{3}{H}_{3}{O}_{3}$

#### 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.

• For example, given that a substance ‘s empirical formula is $C{H}_{2}$  (empirical mass= 12 + 2= 14) and the molecular mass is 28, we can obtain n:
n= $\frac{28}{14}$
= 2
• Therefore the molecular mass is      = 2(CH2)
=${C}_{2}{H}_{4}$

#### 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
• Concentration in moldm-3 is sometimes written as M.
• To express concentration in gdm-3, simply divide the grams by 1000cm3.