Transcript Atomic mass - drseemaljelani

Atomic mass
12/04/2016
RELATIVE ATOMIC MASS, Ar
(“Mass number”) = number of
protons + number of neutrons
SYMBOL
PROTON NUMBER = number of
protons (obviously)
Relative formula mass, Mr
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The relative formula mass of a compound is the relative atomic
masses of all the elements in the compound added together.
E.g. water H2O:
Relative atomic mass of O = 16
Relative atomic mass of H = 1
Therefore Mr for water = 16 + (2x1) = 18
Work out Mr for the following compounds:
1) HCl
H=1, Cl=35 so Mr = 36
2) NaOH
Na=23, O=16, H=1 so Mr = 40
3) MgCl2
Mg=24, Cl=35 so Mr = 24+(2x35) = 94
4) H2SO4
H=1, S=32, O=16 so Mr = (2x1)+32+(4x16) = 98
5) K2CO3
K=39, C=12, O=16 so Mr = (2x39)+12+(3x16) = 138
CaCO3
More examples
40 + 12 + 3x16
HNO3
1 + 14 + 3x16
2MgO
2 x (24 + 16)
3H2O
3 x ((2x1) + 16)
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100
80
4NH3
2KMnO4
3C2H5OH
4Ca(OH)2
Moles – The relative formula mass of a substance, in grams, is known
as 1 mole of that substance.
E.g. 18g of H2O = 1 mole of H2O
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Calculating percentage mass
If you can work out Mr then this bit is easy…
Percentage mass (%) =
Mass of element Ar
Relative formula mass Mr
x100%
Calculate the percentage mass of magnesium in magnesium oxide, MgO:
Ar for magnesium = 24
Ar for oxygen = 16
Mr for magnesium oxide = 24 + 16 = 40
Therefore percentage mass = 24/40 x 100% = 60%
Calculate the percentage mass of the following:
1) Hydrogen in hydrochloric acid, HCl
2) Potassium in potassium chloride, KCl
3) Calcium in calcium chloride, CaCl2
4) Oxygen in water, H2O
Empirical formulae
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Empirical formulae is simply a way of showing how many atoms are in a
molecule (like a chemical formula). For example, CaO, CaCO3, H20 and
KMnO4 are all empirical formulae. Here’s how to work them out:
A classic exam question:
Find the simplest formula of 2.24g of iron
reacting with 0.96g of oxygen.
Step 1: Divide both masses by the relative atomic mass:
For iron 2.24/56 = 0.04
For oxygen 0.96/16 = 0.06
Step 2: Write this as a ratio and simplify:
0.04:0.06 is equivalent to 2:3
Step 3: Write the formula:
2 iron atoms for 3 oxygen atoms means the formula is Fe2O3
Example questions
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1) Find the empirical formula of magnesium oxide which
contains 48g of magnesium and 32g of oxygen.
2) Find the empirical formula of a compound that contains
42g of nitrogen and 9g of hydrogen.
3) Find the empirical formula of a compound containing 20g
of calcium, 6g of carbon and 24g of oxygen.
Calculating the mass of
a product
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E.g. what mass of magnesium oxide is produced when 60g of
magnesium is burned in air?
Step 1: READ the equation:
2Mg + O2
2MgO
IGNORE the
oxygen in step 2 –
the question
doesn’t ask for it
Step 2: WORK OUT the relative formula masses (Mr):
2Mg = 2 x 24 = 48
2MgO = 2 x (24+16) = 80
Step 3: LEARN and APPLY the following 3 points:
1) 48g of Mg makes 80g of MgO
2) 1g of Mg makes 80/48 = 1.66g of MgO
3) 60g of Mg makes 1.66 x 60 = 100g of MgO
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1) When water is electrolysed it breaks down into hydrogen and oxygen:
2H2O
2H2 + O2
What mass of hydrogen is produced by the electrolysis of 6g of water?
Work out Mr: 2H2O = 2 x ((2x1)+16) = 36
2H2 = 2x2 = 4
1.
36g of water produces 4g of hydrogen
2. So 1g of water produces 4/36 = 0.11g of hydrogen
3. 6g of water will produce (4/36) x 6 = 0.66g of hydrogen
2) What mass of calcium oxide is produced when 10g of calcium burns?
2Ca + O2
Mr: 2Ca = 2x40 = 80
2CaO
2CaO = 2 x (40+16) = 112
80g produces 112g so 10g produces (112/80) x 10 = 14g of CaO
3) What mass of aluminium is produced from 100g of aluminium oxide?
2Al2O3
4Al + 3O2
Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204
4Al = 4x27 = 108
204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al2O3
Actual Yield
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Even though no atoms are ever gained or lost in a chemical
reaction, it is not always possible to obtain the
calculated amount of product. Because:
• The reaction may not totally finish – it may be
reversible
• Some of the product may be lost when it is
separated from the reaction mixture – filtered
• Some of the reactants may react in different ways
to the expected reaction
Percentage yield
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The amount of product obtained is known as the yield. When
compared to the maximum theoretical (calculated) amount as a
percentage, it is called percentage yield.
Percentage yield (%) =
Actual yield made
Maximum yield possible
x100%
E.g. What mass of aluminium is produced from 100g of aluminium oxide?
2Al2O3
4Al + 3O2
Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204
4Al = 4x27 = 108
204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al
However, only 35.6g of Al was actually obtained during the experiment. What
is the percentage yield.
Percentage yield (%) =
= 67.3%
35.6
52.9
x100%
Atom economy
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This is simply a measure of the amount of starting
materials that end up as useful products.
It is important for sustainable development and economical
reasons that industrial reactions have High Atom
Economy.
Atom Economy (%) =
Total Mr of useful products
Total Mr of reactants
x100%
Reversible Reactions
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Some chemical reactions are reversible. In other words, they
can go in either direction:
A + B
e.g. Ammonium chloride
NH4Cl
C + D
Ammonia + hydrogen chloride
NH3 + HCl
When a reversible reaction occurs in a closed system (Where
nothing can escape), equilibrium is reached when both
reactions occur at exactly the same rate in each direction.
The relative amounts of all the reacting substances at
equilibrium depend on the conditions of the reaction.
Making Ammonia
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Guten Tag. My name is Fritz Haber and I won the Nobel
Prize for chemistry. I am going to tell you how to use a
reversible reaction to produce ammonia, a very important
chemical. This is called the Haber Process.
Nitrogen + hydrogen
Ammonia
N2 + 3H2
2NH3
Fritz Haber,
1868-1934
To produce ammonia from nitrogen and hydrogen you
have to use three conditions:
Nitrogen
Hydrogen
•High pressure
•450O C
•Iron catalyst
Mixture of NH3, H2 and
N2. This is cooled
causing NH3 to liquefy.
Recycled H2 and N2
Uses of Ammonia
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Ammonia is a very important chemical as it can be
used to make plant fertilisers and nitric acid:
Ammonia gas
Oxygen
Hot
platinum
catalyst
Nitrogen
monoxide
Cooled
Nitrogen
monoxide
Water and
oxygen
Nitric
acid
More ammonia can then be used to neutralise the nitric acid to
produce AMMONIUM NITRATE (a fertiliser rich in nitrogen).
Ammonia + nitric acid
NH3 + HNO3
Ammonium nitrate
NH4NO3
The trouble with nitrogen based fertilisers is that they can also create
problems – they could contaminate our drinking water.
Haber Process Summary
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A low temperature increases the yield of ammonia but is
too slow
A high temperature improves the rate of reaction but
decreases the yield too much
A high pressure increases the yield of ammonia but costs a lot
of money
To compromise all of these factors, these conditions are used
to make a reasonable Yield of ammonia, quickly:
Nitrogen
Hydrogen
•200 atm pressure
•450O C
•Iron catalyst
Mixture of NH3, H2
and N2. This is
cooled causing NH3
to liquefy.
Recycled H2 and N2