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
    about the reaction it represents
    • 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
               Your Turn
   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
    1. Partially fill a volumetric flask with water, and add the
         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?
               Your Turn
   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?

Suggested answer
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?
               Your Turn
   Page 267
   Question 9, 10 and 11
          Volumetric analysis
   This is what your first outcome is all
    about.
   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
   A volumetric flask –
    is used to prepare the
    standard solution. An
    accurately weighed
    sample is placed in the
    flask and dissolved in
    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
    into a conical flask ready
    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.
              Your Turn
   Page 268
   Question 12

				
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