Calculations in Chemistry

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					              Calculations in
•   You need to know how to carry out several calculations in Additional
    and Triple Chemistry

•   This booklet gives you a step by step guide to carrying out these
    equations and examples of each one. Follow the same rules for the
    questions in your exam and chemistry calculations are actually quite
    easy, honest!

•   These are:
     – Additional:
          • Relative formula mass
          • Mole masses
          • Percentage of element in a compound
          • Formula of a compound from its percentage composition
          • Masses of reactants and products
          • Percentage yield
          • Atom economy
     – Triple:
          • Titration calculations
          • Bond energy calculations
     Additional Chemistry
Relative atomic and Formula Masses
The mass of an atom is too small to deal with in real terms, so we use
‘relative’ masses – Carbon is given a mass of 12, and everything else is
compared with it and given a mass, e.g. Oxygen is ‘heavier’, so its
relative mass is 16.
 The relative atomic mass can be found by looking at
 the periodic table, It is always the larger of the two

Relative formula mass can be found by adding up the relative atomic
masses of each element in a compound.

E.g. Carbon Dioxide (CO2)
           Carbon has a relative atomic mass of 12
           Oxygen has a relative atomic mass of 16

The relative formula mass of Carbon Dioxide is therefore:
           12 + (16 x 2) = 44
 x 2 because it’s

Because saying ‘relative formula (or atomic!) mass in grams’ is a bit
clumsy, we simply say ‘moles’ instead. This means that 1 mole of Carbon
Dioxide is 44 grams, or 44g. Simple!
Percentage of an element in a compound
We can use the relative atomic mass (Ar) of elements and the relative
formula mass (Mr) of compounds to find out the percentage composition
of different elements.

E.g. What percentage mass of white Magnesium Oxide is actually
Magnesium, and how much is Oxygen?

                                              1. Work out the formula mass of
1.   Work out the mass of MgO                 the compound
     24 + 16 = 40                             2. Convert this into grams
2.   Convert to grams                         3. Work out the percentage by
                                              using this equation:
                                                       Mass of element
                                                                           x 100%
3.   Work out the percentage                        Total mass of compound
     24 X 100% = 60% is Magnesium, so 40% must be Oxygen!

Formula of a compound from its percentage composition
We can also do this backwards! If we know the percentage composition
of a compound we can work out the ratio of atoms. This is known as the
Empirical Formula. Sometimes this is the same as the molecular formula,
but not always (e.g. water has an empirical and molecular formula of H2O.
Hydrogen peroxide's empirical formula is HO, but it’s molecular formula
is H2O2.

E.g. If 9g of Aluminium react with 35.5g of Chlorine, what is the
empirical formula of the compound formed?
                                                1. Divide the mass of each
Aluminium                                          element by its relative
                                                   atomic mass to find out the
9                                                  number of moles reacted
    = 1/3 moles of Aluminium atoms
27                                              2. Create a ratio and simplify if
                                                3. Write a formula based on
Chlorine                                           the ratio
       = 1 mole of Chlorine atoms
35.5                                     Al : Cl
                                         1/3 : 1
                                         1 : 3
Masses of reactants and products

This is an important calculation when we want to know how much of each
reactant to react together. For example, sodium hydroxide reacts with
chlorine gas to make bleach. If we have too much Chlorine, some will be
wasted. Too little and not all of the sodium hydroxide will react.

                  2NaOH + Cl2  NaOCl + NaCl + H2O
                 Sodium Chlorine Bleach    Salt    Water

How much Chlorine gas should we bubble through 100g of Sodium
Hydroxide to make Bleach?
                                                  1. Work out the mass of one
1. NaOH                                              mole of Sodium Hydroxide
   23 + 16 + 1 = 40g is one mole of NaOH          2. Calculate how many moles
                                                     you have in your reaction
                                                  3. Work out how many moles
2. We have 100g in our reaction so…                  of Chlorine you need
                                                  4. Convert this into a mass
   100 = 2.5 moles                                   for Chlorine

3. The chemical equation tells us that we need 2 moles of Sodium
Hydroxide (2NaOH) for every mole of Chlorine (Cl2).
So we need: 2.5 = 1.25 moles of Chlorine
4. 35.5 x 2 = 71g is one mole of Cl2
   So we need 1.25 x 71 = 88.75g of Chlorine to react with 100g of
Sodium Hydroxide.
Percentage Yield
Rather than talk about the yield of a chemical reaction in terms of mass
(grams, tonnes etc.) we can talk about the percentage yield. This gives
us an idea of the amount of product that the reaction really makes,
compared to what it could possibly make under perfect conditions.
There are many reasons why we don’t make 100% every time, such as:
     –   The reaction may be reversible
     –   Some product could be left behind in the apparatus
     –   The reactants may not be pure
     –   It may be difficult to separate the products if more than one are made.

Using this reaction “A + B  C”, it was found that in perfect conditions,
scientists could make 2.5g of C. However, when they tried it out, they
only made 1.5. What is the percentage yield of this reaction?

         Amount of product produced
                                                                        The higher the
                                         x 100%                      percentage yield and
      Maximum amount of product possible
                                                                      atom economy, the
                                                                   better the reactions are
1.5 x 100% = 60% percentage yield                                 for the Earth’s resources,
2.5                                                                 as there’s less waste!

Percentage Atom Economy
Sometimes reactions will make more than one product. Working out the
atom economy is a way of finding out how much of the reactant ends up
in the useful product.
      Relative formula mass of useful product x 100%
       Relative formula mass of all products

Ethanol (C2H5OH) can be converted into ethene (C2H2) to make
polyethene. However, the reaction also produces water.

                              C2H5OH  C2H2 + H2O

Mass of useful product = (12 x 2) + (1 x 2) = 28
Mass of all products = 28 + [(1x2) + 16] = 28 + 18 = 46

28   x 100% = 61% atom economy
          Triple Chemistry
Titration Calculations
Titrations are used to find out the volumes of solutions that react
exactly. If you know the concentration of one of the solutions, and the
volumes that react together are known as well, you can use this to work
out the concentration of the other solution.

It is important that you know that mol/dm3 is a term for concentration of a solution.
It means that X moles of substance are dissolved in 1000cm 3 (1dm3) of a solvent.

1mol/dm3 of Sodium Hydroxide means that 40g (the mass of one mole) is dissolved
in 1000cm3 of the solvent. 20g dissolved in the same volume would be 0.5mol/dm 3.

E.g. a student put 25.0cm3 of sodium hydroxide solution with an
unknown concentration into a conical flask using a pipette. The sodium
hydroxide reacted with exactly 20.0cm3 of 0.50mol/dm3 hydrochloric
acid added from a burette. What was the concentration of the sodium
hydroxide solution?

                                           1.     Calculate the number of moles
                                                  reacted in the known substance:
The equation for this reaction is:                (Concentration/1000) x volume used

                          NaOH + HCl  NaCl + H2O

0.50 moles of Hydrochloric acid are dissolved in 1000cm3 of acid. We
used 20cm3 in this reaction. So…

1000 X 20 = 0.10 moles of HCl in 20cm3 of acid.
Titration Calculations continued…

We know from the equation that 1 mole of HCl reacts with 1 mole of
NaOH. So, if there is 0.10 moles of HCl in the reaction, there must be
0.10 moles of NaOH as well.

Now we can work out the concentration of NaOH.

                                              2.    Use the volume of the
If 0.10 moles of NaOH are in 25cm3                  second unknown solution
of solution, we can use the equation in the         to work out the
                                                    concentration in 1dm3
box on the right to work out how many are           (number of moles/volume
in 1000cm3, therefore the concentration!             of solution) x 1000

0.10 X 1000 = 0.40 moles of NaOH in 1000cm3 of solution = 0.4 mol/dm3
                                                   This is the concentration!
  And Finally…

  Bond Energy Calculations
  The energy needed to break a bond between two atoms is called the
  bond energy. It is measured in kJ/mol – kilojoules per mole. We can use
  this to work out the energy change (ΔH) for chemical reactions.
  However, we need a list of the bond energies to do this:

                        Bond energy                                  Bond energy
      Bond                                         Bond
                          (kJ/mol)                                     (kJ/mol)
      C-C                    347                   H-Cl                   432
      C-O                    358                   H-O                    464
      C-H                     413                  H-N                    391
      C-N                    286                   H-H                    436
      C-Cl                   346                   O=O                    498
      Cl-Cl                  243                   NΞN                    945
                                                                                    You don’t need to
                                                                                     remember this!
  To calculate bond energy reactions we need to know two things:
        –     How much energy is needed to break the chemical bonds in the reactants
        –     How much energy is released when the new bonds are formed in the products
  e.g. Hydrogen and Nitrogen are used in the Haber process to make
  ammonia. What is the overall energy change in this reaction?
            3H2 + N2 = 2NH3
                                                                1. Draw out the structure of each
                                                                     atom so you can work out what
                                                                          bonds are involved
                                                                2. Work out the energy needed to
                                                                    break the reactants into atoms.
2. Energy needed to break one mole of                            3. Work out the energy released
nitrogen, and three moles of hydrogen                               when the bonds in the products
945 + (3 x 436) = 2253 kJ/mol                                                 are formed
                                                                  4. Work out the energy change
3. Energy released when 3 N-H bonds are
formed in one mole of ammonia = 3 x -391 = -
Energy released in two moles of ammonia = 2                         The bond energies are:
x -1173 = -2346 kJ
                                                                      NΞN = 945 kJ/mol
 Overall energy change = +2253 – 2346 = -93kJ                         H-H = 391 kJ/mol

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