Download The mole concept Since Compounds are formed

1/8/2016
Chemical calculations
Stoichiometry refers to the quantities of material which react according to a balanced
chemical equation.
Compounds are formed when atoms combine in fixed proportions.
E.g.
2Mg
+
2 atoms Mg
+
O2
2 atoms oxygen
(x4)
8 atoms Mg
+
8 atoms oxygen
(x6) 1 dozen atoms Mg + 1 dozen atoms oxygen

2MgO

2 molecules MgO


8 molecules MgO
1 dozen molecules MgO
BUT atoms and molecules are tiny, so counting them in ones and twos is silly.
Instead we count atoms and molecules in moles. A mole is 6.023 x 1023 units.
Count up the number of atoms in exactly 12g of pure 12C. That number is a mole.
Class Exercise
A mole (a) is 6 x 1023 units, and (b) weighs the atomic weight (for atoms) or molecular weight
(for molecules) in grams.
We can find the atomic weight (same as atomic mass or relative atomic mass) from the
periodic table, and molecular weights by adding atomic masses.
E.g. Atomic weights in g:
Na 23,
Cl 35.5,
C 12,
O 16
What is the weight of
(i) 3 moles of NaCl? (ii) 0.2 moles O2? (iii) 0.2 moles of O atoms? (iv) 11 moles CO2?
How many molecules are there in those quantities?
How many atoms?
1
1/8/2016
2Mg + O2  2MgO
How do we know what will happen when magnesium and oxygen react?
We know the reactants, so we write down their chemical symbols remembering that oxygen is a
diatomic molecule. Writing O for oxygen would imply that the reaction involves atomic oxygen
whose properties are very different from that of molecular O2 or ozone O3 and would thus result in
a serious error.
Finding the formula for the product is often more challenging. Experimentally, we could determine
the empirical formula by weighing the mass of magnesium that reacted with excess oxygen and
weighing the mass of the product formed.
2Mg (s) + O2 (g)  2MgO (s)
Equations must be balanced – an unbalanced equation implies creation or destruction of mass.
It is common to include the physical state of reactants and products in brackets by their symbols.
If an element exists as different allotropes (e.g. diamond and graphite, or white, red, violet or black
phosphorus) we also indicate this, usually as a subscript e.g. Cgraphite(s)
2
1/8/2016
Simple stoichiometry (works for most situations!)
1.
Write down the equation
2.
Balance it
3.
If necessary, convert the known amount(s) reactant/product to moles
4.
Use the balanced equation to set up the appropriate mole ratios
5.
Use the mole ratios to work out the number of moles of desired reactant/product
6.
If necessary, convert back to grams
How many grams of oxygen are required for the complete
combustion of 50 grams methane?
1. CH4 (g) + O2 (g)  CO2 (g) + H2O (g)
1. Write down the equation
2. Balance it
2. CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
3. If necessary, convert the known
amount(s) of reactant/product to
moles
3. 1 mole of methane weighs {12 + 4 x 1} = 16 g, so 50 g is 50/16 = 3.125
moles
4. Use the balanced equation to
set up the appropriate mole ratios
4. From the equation 1 mole of methane reacts with two moles of
oxygen
5. Use the mole ratios to work out
the number of moles of desired
reactant/product
6. If necessary, convert back to
grams
5. So the amount of oxygen required is 2 x 3.125 = 6.25 moles
6. To turn this back into grams we need the molecular weight of oxygen,
which is 2 x 16 = 32g (since the oxygen molecule consists of 2 atoms
joined together). Thus the amount of oxygen required is 6.25 moles x
32 g/mole = 200 g.
3
1/8/2016
Lithium hydroxide is used in space vehicles to absorb carbon
dioxide. How much CO2 can be absorbed by 1 kg of hydroxide?
1. Write down the equation
2. Balance it
3. If necessary, convert the known
amount(s) of reactant/product to
moles
4. Use the balanced equation to
set up the appropriate mole ratios
1. LiOH (s) + CO2 (g)  Li2CO3 (s) + H2O (l)
2. 2LiOH (s) + CO2 (g)  Li2CO3 (s) + H2O (l)
3. Molar mass of LiOH is {16.94 + 16 + 1} = 23.94 g/mol. Thus, # of moles
of hydroxide is 1000/23.94 = 41.8 mol
4. From the equation, 2 moles of LiOH react with 1 mole of CO2
5. Use the mole ratios to work out
the number of moles of desired
reactant/product
5. So the amount of CO2 that reacts is 41.8/2 = 20.9 moles
6. If necessary, convert back to
grams
6. The MW of CO2 is {12 + 2 x 16} = 44 g/mole. So the mass of CO2 that
reacts is 20.9 moles x 44 g/mole = 920g.
14g methane and 14g water react to give hydrogen and carbon
monoxide. Which reactant is in excess? How much hydrogen is formed?
1. CH4 (g) + H2O (g)  H2 (g) + CO (g)
1. Write down the equation
2. Balance it
3. If necessary, convert the known
amount(s) of reactant/product to
moles
4. Use the balanced equation to
set up the appropriate mole ratios
5. Use the mole ratios to work out
the number of moles of desired
reactant/product
6. If necessary, convert back to
grams
2. CH4 (g) + H2O (g)  3H2 (g) + CO (g)
3. MW of methane is {12 + 4 x 1} = 16 g/mole, so there are 14/16 =
0.875 moles methane. MW of water is {16 + 2 x 1} = 18, so there are
0.778 moles of water
[3a. The equation shows that methane and water are used up at the
same rate, so water is the limiting reactant - it will be consumed
completely]
4. 1 mole water reacts to give 3 moles of hydrogen
5. So the # of moles of hydrogen formed is 3 x 0.778 = 2.334 moles
6. MW of H2 is 2 x 1 = 2, so the weight of hydrogen formed is 2 x 2.334 =
4.668 g
4
1/8/2016
Finding the empirical formula
• Suppose we started with 211.5 mg magnesium and obtained 350.1 mg product. What is the
empirical formula?
As the product only contains oxygen and magnesium any gain in mass must be due to oxygen
chemically bound in the compound.
350.1 mg – 211.5 mg = 138.6 mg of oxygen in the product.
To get to the stoichiometric ratio of the two elements and express an empirical formula we need to
convert mass to moles:
1 mole of Mg is 24.31 g, so 0.2115 g Mg / 24.31 = 8.70 x 10 -3 mol
1 mole of oxygen (atoms) is 16 g, so 0.1386 g O is 0.1386/16 = 8.663 x 10 -3 mol
Dividing both amounts by the smaller number of moles yields the mole ratio of the two elements:
8.702 x 10 -3 mol / 8.663 x 10 -3 mol = 1.004
8.663 x 10 -3 mol / 8.663 x 10 -3 mol = 1.000
Within experimental error the Mg:O ratio is 1:1 so we write the formula MgO (not Mg1O1). The
empirical formula expresses the smallest whole number ratio of atoms in a compound.
There are two major kinds of stoichiometry problems you will encounter:
a) You know the amount of one reactant
b) You know the amount of at least two reactants
E.g. a) How much magnesium oxide will form if 5.73 g magnesium burns in excess oxygen?
0) Think moles!
1) Convert mass of Mg to moles of Mg
5.73 g /24.31 g mol-1 = 2.36 x 10 -1 mol Mg
2) Convert moles of Mg to moles of MgO using the coefficients in the balanced equation:
2.36 x 10 -1 mol Mg  2.36 x 10 -1 mol MgO
3) Convert moles of MgO to mass of MgO To find the molar mass (or formula weight) of MgO add
the mass of all atoms in the formula (24.304 g mol-1 + 16.000 g mol-1 = 40.304 g mol-1 ).
2.36 x 10 -1 mol MgO = 9.50 g MgO
Keep more significant figures in intermediate calculations to avoid rounding errors.
5
1/8/2016
b) A sample of 1.51 g magnesium is ignited in a sealed 1L flask that contains 1.42 g oxygen gas. How
much magnesium oxide can be formed and how much of which reactant is left behind?
First determine which reactant limits the amount of product:
0) Think moles!
1) Convert the masses of the reactants to moles of reactant
1.51 g Mg = 6.21 x 10 -2 mol Mg
1.42 g O2 = 4.44 x 10 -2 mol O2
2) Identify the limiting reactant
The limiting reactant is the chemical that will be used up first.
There are more moles of magnesium than oxygen, but the two reactants may not be consumed at
the same rate. It’s simplest to calculate how much product could be formed from the given amount
of each reactant.
6.21 x 10 -2 mol Mg  6.21 x 10 -2 mol MgO (if enough O2 is available)
4.44 x 10 -2 mol O2  8.88 x 10 -2 mol MgO (if enough Mg is available)
Though there is more magnesium by mass and mols it is the limiting reactant, so all the magnesium
will be consumed and some oxygen will be left over.
3) Calculate the mass of product that can form
6.21 x 10 -2 mol MgO = 2.50 g MgO
(Use MW of MgO, which is the AW of Mg and O combined)
Some oxygen remains. Since we know the masses of reactants and products we can use the law of
conservation of mass to calculate the mass of oxygen left over.
The mass of the reactants is 1.51 g Mg + 1.42 g O2 = 2.93 g reactants.
Since only 2.50 g of product is formed the difference must be O 2 left over:
2.93 g – 2.50 g = 0.43 g oxygen left un-reacted.
6
1/8/2016
Had we mistakenly assumed that oxygen was the limiting reactant, we’d calculate that
8.88 x 10 -2 mol of MgO should have been produced. This amount of MgO would weigh:
8.88 x 10 -2 mol MgO = 3.58 g MgO
which is more than the mass of reactants available (2.93 g). An error is not always found so readily
but it pays to check if your result makes sense.
This example has practical relevance. Magnesium is sometimes used as a getter in tungsten light
bulbs where it removes trace amounts of oxygen and nitrogen; this greatly enhances the lifetime of
the light bulb (as long as you ensured that oxygen is the limiting reactant).
Solution Stoichiometry
In aqueous solution the number of moles of a component dissolved per litre of solution is molar
concentration, or molarity of that component.
This is simpler to use than other possible measures of concentration, such as g/L, %, or ppm.
7
1/8/2016
• A solution contains 2.4 g NaOH in 50 mL solution; what is the molarity?
• Molarity is moles/litre, so we need the number of moles and the volume in litres.
• # moles = wt/MW = 2.4 g / 40 g mol-1 = 0.06 moles
• Volume = 0.050 L, so molarity = 0.06 mol / 0.05 L = 1.2M
23.4 mL of a 1.200 M solution of NaOH was used to neutralize 10.00 mL of a sulfuric acid sample of
unknown concentration.
(i)
Write down and balance the equation for the reaction.
(ii) How many moles of NaOH react?
(iii) What is the molarity of the sulfuric acid?
2NaOH + H2SO4  Na2SO4 + 2H2O
The number of moles of sodium hydroxide is:
M × V = 1.200 mol/L × 0.0234 L = 0.02808 mol NaOH
8
1/8/2016
2NaOH + H2SO4  Na2SO4 + 2H2O
At the equivalence point (neutral) the number of moles of sulfuric acid can be calculated using the
coefficients in the balanced chemical equation
Remember, we need two moles of base per mole of acid:
0.02808 mol NaOH = 0.01404 mol H2SO4
As this is the number of moles in 0.01L, the molarity of the sulfuric acid is thus:
M = 0.01404 / 0.01 = 1.404 M
3g of ethanoic (acetic) acid is dissolved in 50 mL water. The
resulting solution is neutralised by 41 mL of sodium hydroxide
solution. Determine the molarity of both solutions
1.
Molarity = moles/litre
2.
MW of ethanoic acid, CH3COOH, = {12 + 3 x 1 + 12 + 2 x 16 + 1} = 60 g/mol, so 3g = 3/60
=0.05 moles
3.
Total volume of solution = 53 mL = 0.053L, so molarity of the acid is 0.05/0.053 = 0.9434M
4.
CH3COOH + NaOH  CH3COONa + H2O, so 1 mole of acid reacts with 1 mole of base.
5.
Therefore number of moles of NaOH = 0.05
6.
Volume of base = 41 mL, so molarity of the base = 0.05/0.041 = 1.22M
9
1/8/2016
% Yield
Often we need to calculate the % yield of a reaction. This is the amount of product isolated from
the reaction mixture divided by the maximum theoretical amount of product, expressed as a
percentage.
Example: 4.75 g metallic copper is precipitated from a solution containing 25.23 g CuSO4.5H2O by
adding excess iron metal. Calculate the % Yield.
Cu2+ (aq) + Fe(s)  Cu(s) + Fe2+
25.23 g (CuSO4.5H2O) = 0.1010 mol CuSO4.5H2O, which contains 0.1010 mol Cu2+
This is the maximum amount of copper that could be formed
0.1010 mol Cu2+ = 6.42 g Cu
% yield = 4.75/6.42 x 100 % = 74.0 %
10