# Calculations in chemistry: stoichiometry

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```					    Calculations in
chemistry: stoichiometry
Chapter 15
Return of the mole
   The mole is a very useful unit
because it represents a ‘measurable’
number of particles – atoms, ions or
molecules.
   The mole concept effectively acts as
a scale up from individual particles to
measurable amounts of particles.
Stoichiometry
   In a chemical reaction, atoms are
neither created nor destroyed.
   Consequently, given the amount of a
reactant consumed or a product
formed we can calculated the
amounts of other reactants and
products involved.
   We are now applying the mole to
chemical reactions and not just
chemicals.
Balanced Chemical Equations

   All particles that exist at the start of a chemical
reaction must be accounted for after the reaction
is complete
Consider
2H2(g) + O2(g) → 2H2O(g)

   The equation gives the following information
• It identifies the reactants and products
• It identifies the states of the reactants and products
• Because it is balanced it gives the ratio in which the
substances react
Reacting Quantities

   Consider
2H2(g) + O2(g) → 2H2O(g)

   This reaction involves 2 moles of H2(g) reacting
with 1 mole of O2(g) to form 2 moles of H2O(g).

   In more general terms the amount of oxygen
used will always be equal to half the amount of
hydrogen used and half the amount of water
formed
What a chemical equation does not tell us

   An equation does not tell us about the rate of a
reaction

   It does not tell us whether heat is required or
given off

   It does not tell us what temperature or pressure
is required

   It gives no details as to how the individual atoms
or molecules are transformed from reacts to
products
Mass-Mass Stoichiometry

    We generally measure substances by mass but chemical
reactions depend on relative numbers of atoms, ions of
molecules as reflected in the chemical amount or number of
moles. So the relationships between mass and number of
mole is critical.
Chemical amount                                                        mass g
mol

OR
Chemical amount              m
m=nxM                                     mol           n=
M
mass g                     molar mass                                           molar mass
g mol-1                                              g mol-1

Conversion of metric units of mass:
÷103                         ÷103                 ÷103                        ÷103
microgram                 milligram                 gram                     kilogram                    tonne
µg                       mg                       g                          kg                         t
x103                         x103                x103                          x103
Mass-Mass Stoichiometry

    Given the amount of one substance involved in a
chemical reaction, the amounts of all other substances
involved can be calculated, provided a balance equation
for the reaction is known.

Mass-mass stoichiometry problems can be solved in four
steps:
1. Write a balanced equation for the reaction.

2.   Calculate the amount (in mol) of the substance with
the know mass.

3.   Use the mole ratio from the equation to calculate the
amount (in mol) of the required substance.

4.   Calculate the mass required.
Mass-Mass Stoichiometry
Worked Example
Determine the mass of oxygen used when 2.00 kg of water is
produced according to the equation 2H2(g) + O2(g) →
2H2O(l).

Suggested Solution
1.  Tag the equation with the data supplied and the quantity
you have to determine
2H2(g) + O2(g) → 2H2O(l)
?m      2.00kg

2.     Identify the starting point (the data you’ve got) and the
finishing point (the quantity you want) of the calculations.
m(H2O)                                               m(O2)
Mass-Mass Stoichiometry
3.     Link the data you’ve got and the quantity you want to
their number of mole.
÷ M(H2O)                            x M(O2)
m(H2O)              n(H2O)           n(O2)             m(O2)

4.     The link between n(O2) -what you want to find- and
n(H2O) –what you’ve got- is obtained from the equation.
The equation tells us that n(O2)/n(H2O) = ½, so n(O2) =
½ x n(H2O)
So the calculation flow chart can now be completed
÷ M(H2O)            X 1/2           x M(O2)
m(H2O)              n(H2O)           n(O2)             m(O2)
Mass-Mass Stoichiometry

   It is then relatively quick to follow through the
calculation sequence.
n(H2O) produced = m(H2O) / M(H2O)
= 2.00x103 g / 18.0 g mol-1
= 1.11x102 mol

n(O2) reacting   = ½ x n(H2O) produced
= ½ x 111 mol
= 55.6 mol

m(O2) required   =   n(O2) x M(O2)
=   55.6 mol x (2 x 16.0) g mol-1
=   1.78x103 g
=   1.78 kg
Mass-Mass Stoichiometry

Question
A phosphorous manufacturer is to extract 1.00 tonne of
phosphorous per day by the process given by the
equation:
2Ca3(PO4)2(s) + 6SiO2(s) + 10C(s) → P4(s) + 10CO(g) +
6CaSio3(s)

Calculate the mass required daily of:
a)  Calcium phosphate
b)  Silicon dioxide
   Page 263
   Question 1 - 3
Concentrating on Solutions
   Many chemical reactions take place between
substances that have been dissolved in a liquid,
most commonly water, to form a solution.

   The liquid in which the substances are dissolved
is called the solvent.

   The dissolved substance is called the solute.

   When working with solutions, the most easily
measured quantity is their volume. However, for
the volume of a solution to provide useful
information, the solution’s concentration must be
known.
Calculating concentrations of solutions
     Amounts of solutions are usually measured by volume.
     Common units for volume are litres (L) and millilitres (mL),
• where 1L = 103 mL
     Just as chemical amount (mol) and mass (g) are linked by
molar mass (g mol-1), then chemical amount (mol) and
volume (L) are linked by molar concentration (mol L-1)
     Molar concentration, symbol c, which has units mol per litre
(mol L-1), is often expressed as molarity (M).
     So 1.0 M HCl(aq), ie a 1 molar solution of hydrochloric acid
is the same as saying that c(HCl) = 1.0 mol L-1
Chemical amount                                   Chemical amount
mol                                               mol

Concentration
n = cV              mol L-1       c= n
V
Concentration            Volume                             Volume
mol L-1                   L                                  L
Preparation of a solution of known
concentration
    A solution of known concentration is known as a
standard solution.

    In order to prepare a particular volume of solution of
known concentration, the following five steps should be
followed
1. Calculate the number of moles of solute that are
needed to obtain correct concentration of solution for
the volume of solvent to be used, according to the
formula n = cV
2. Calculate the mass of the solute needed, using the
formula
m = nM
correct mass of solute
2. Dissolve the solute
3. Add water to the required volume
Preparation of a solution of known
concentration
Example
Calculate the number of moles of sodium chloride needed to
prepare 500 mL of a 0.0800 mol L-1 salt solution. What
mass of sodium chloride would be weighed out.
Solution

1.   List the known information
volume (V) = 500 mL = 0.5 L
concentration (c) = 0.0800 mol L-1

x V(NaCl)             x M(NaCl)
c(NaCl)               n(NaCl)               m(NaCl)
Preparation of a solution of known
concentration
2.   Calculate number of moles (n) of NaCl needed
n = cV
= 0.0800 x 0.500
= 0.0400 mol

3.   Calculate the mass represented by the number
of moles.
m=nxM
= 0.0400 x 58.5
= 2.34 g
Worked Example 15.1d
   What volume of 0.100 M sulfuric acid
reacts completely with 17.8 ml of
0.150 M potassium hydroxide?
   Page 263
   Questions 6 - 8
Limiting Reactant Calculations
   In many chemical reactions an excess of one reactant is
added to ensure complete reaction of another – generally
more valuable – reactant.

   The reaction stops when one reactant is used up (the
limiting reagent), even though some of the other substance
is unreacted. The other reactant is said to be in excess (the
excess reagent) – some of it remains when the reaction has
finished

   In associated problems it is often necessary to determine
which reactant is in excess before amounts of product can
be determined. Again there are logical calculation
techniques, which will lead to efficient solutions of such
problems

   Chemical reactions may be considered to involved three
stages: an initial stage where the reactants are added, a
reacting stage where the reactants combine in the mole
ration suggested by the equation, and a final stage where
the reactions appear to be complete
Limiting Reactant Calculations
Example 1
A shop advertises the sale of dining settings,
consisting of a table and four chairs. Using
chemical terminology, a dining setting would be
written as TaCh4. The arrangement can be set
out as an equation:
Ta + 4Ch → TaCh4
The shop does some stocktaking and discovers it
has 16 chairs and five tables, how many
complete settings can it sell?
Example 2

A container has a mixture of eight molecules of
hydrogen gas and six molecules of oxygen gas.
When ignited, hydrogen gas and oxygen gas
react to form water according to the equation:
2H2(g) + O2(g) → 2H2O(g)
How many water molecules can be formed?
Limiting Reactant Calculations

A real example
A gaseous mixture of 25.0g of hydrogen gas and 100.0g of
oxygen gas are mixed and ignited. The water produced is
collected and weighed. What is the expected mass of
water produced?

1.  Tag the equation with the data supplied and the quantity
you have to determine
2H2(g) + O2(g) → 2H2O(g)
25.0 g 100.0 g   ?g
Limiting Reactant Calculations
2.  Calculate the amount of each reactant.
n(H2) = m(H2)/M(H2)        and    n(O2) = m(O2)/M(O2)
= 25.0 g / 2 x 1.01 g mol-1   = 100.0 g / 2 x 16.0 g mol-1
= 12.4 mol                     = 3.13 mol

3.   i) Work out which reactant is fully used (the other must be
in excess). This requires comparison of initial amounts of
reactants with the mole ratios suggested in the equation.
Assume all the H2 reacts, then
2H2(g) + O2(g) → 2H2O(g)
Initially     12.4 mol    3.13 mol
Reacting      12.4 mol    6.2 mol

This is not possible because n(O2) = 2 x n(H2) – according to
the equation – and there is simply not enough O2 present.
So H2 is in excess and O2 will be fully used.
Limiting Reactant Calculations
3.  ii) Show that all the O2 does react and find
amounts present at end of reaction
2H2(g) + O2(g)          → 2H2O(g)
Initially     12.4 mol       3.13 mol
Reacting     6.26 mol       3.13 mol → 6.26mol
Finally       6.14 mol        -            6.26 mol

4. Complete the calculations
m(H2O) = n(H2O) x M(H2O)
= 6.26 mol x 18.02 g mol-1
= 113 g
Therefore, 113 g of water is produced from this
hydrogen – oxygen mixture
Worked Example 15.2b
  2.50 g of aluminium is mixed with
5.00 g of iodine and allowed to
react according to the equation:
2Al(s) + 3I2(s) ―› 2AlI3(s)

(a)   What mass of aluminium iodide
would be produced.
(b)   What is the mass of the reactant in
excess?
   Page 267
   Question 9, 10 and 11
Volumetric analysis
   This is what your first outcome is all
   There are many situations when it is
essential to know the exact amount
of acid or base in a substance.
   The concentration of solutions of
acids and bases can be determined
accurately by a technique called
volumetric analysis.
Volumetric Analysis.
   This involves reacting the solution of
an unknown concentration with a
solution of accurately known
concentration.
   A solution of accurately known
concentration is a standard
solution.
Volumetric Analysis
   If we want to find the concentration
of a solution of hydrochloric acid, the
hydrochloric acid can be reacted with
a basic standard solution.
   This is more of an exact procedure
than anything you have done to date
in the lab so we need to use
precisely calibrated equipment.
Precise Equipment
is used to prepare the
standard solution. An
accurately weighed
sample is placed in the
de-ionised water to
form a specific volume
of solution.
A pipette
   Is used to measure
accurately a specific
volume of the solution.
This known volume, or
aliquot, is then poured
for analysis.
A Burette
   A burette delivers accurately known,
but variable volumes of solution
(titres).
   A volume of acid or base is slowly
titrated to the solution of base or acid
in the conical flask until the reactants
are present in stoichiometrically
equivalent amounts. This is called the
equivalent point.
The equivalent point
   For the reaction:
Na2CO3(aq) + 2HCl(aq) ―› 2NaCl (aq) + H2O(l) + CO2(g)

   The equivalence point is reached when
exactly 2 mol HCl has been added for each 1
mol Na2CO3
   This whole process is called a titration.
   When the equivalence point is reached the
reaction is complete.