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Stoichiometry
Chapter 15
What is it?
• It is the term used when calculations are used in
chemistry to predict the amount of product or
reactant needed or produced using a balanced
equation.
• Stoichiometry is important as it allows scientists to
predict the amount of chemical that will be needed
or will be produced.
• Stoich is vitally important for industries that
manufacture chemicals, drugs, foods, or process
potentially explosive substances.
Balanced Equations
• Balanced equations are the basis for all stoich
calculations.
• It is vital that your chemical formula is correct for
each reactant and product and the equation is
correctly balanced.
• Once balanced the formula is telling us the molar
ratio for that chemical reaction.
1 mole
3 mole
2 mole
Don’t forget that one mole is 6.02 x 1023 particles.
Balance the following equations
• Balance the following equations.
1. Hydrogen gas reacts with Oxygen gas to produce
water.
2. Hexane combusts in oxygen gas.
3. Nitric acid is neutralised by Calcium Hydroxide.
4. NH3(g) + O2(g)  NO(g) + H2O(g)
Steps in solving a stoich problem
1.
2.
3.
4.
Write out a correct and balanced equation.
Identify the known and unknown chemicals.
Convert amounts to correct units. (grams and Litres)
Calculate the known amounts in mole (double check the
Molar Mass as this is a common error)
5. If needed, work out which is excess / limiting
6. Use the mole ratio of the balanced equation to
determine the number of mole of the unknown.
7. Calculate the amount / concentration of the unknown.
8. Ensure your answer has correct significant figures and
units (grams, kilograms, ml)
9. Check that you have answered the question.
Mass – Mass Stoichiometry
• Involves solving a problem in which the mass of one
product or reactant is known.
• By using a balanced equation you are then required
to calculate the mass of an unknown chemical.
• It requires you to use the following formula:
• n = m
m = n x M
M
• Don’t forget all mass calculations must be in grams.
Stoich problem using the steps.
• To extract pure aluminium in a smelter
alumina (Al2O3) is reacted with solid carbon to
produce pure aluminium and carbon dioxide
gas.
a) How much alumina (kg) is needed to produce
1000kg of pure aluminium?
b) How much carbon dioxide (kg) would be
produced?
Limiting Reactant
• It is rare that we will have the exact amount of each
reactant for a reaction.
• One or more reactants are in excess – ie there is
more than what is needed.
• The reactant that is completely used up is said to be
the limiting reactant.
• It is the limiting reactant that must be used in
stoichiometry.
• Eg. 2Mg(s)
0.5mol
+
O2(g)
0.3mol

2MgO(s)
Concentration of solutions
• Most reactions, especially organic, occur in aqueous
environments.
• The most common liquid is water. The liquid is the solvent
and the dissolved substance the solute.
• Solutions are measured by their volume and the
concentration of substance dissolved in them, known as
molarity (M). This has the unit mol/L.
• A solution of concentration 2.5 M means it has 2.5 mol of the
solute dissolved in 1 litre of solvent.
• The formula used for solution stoichiometry is:
n=CxV
C= n
V
V = n_
C
Concentration of water
• All pure liquids have a concentration.
• The concentration of water can be calculated to
be 55.5M. Refer to page 344 of text.
• Concentration of ionically bonded chemicals in solution can also
be calculated. It must be remembered that the ions dissociate
in a solution.
• Dissolving sodium carbonate will
results in twice the concentration
of sodium ions compared with
carbonate ions.
• If aluminium oxide forms a solution with a concentration of
1.5M aluminium ions, what concentration of oxide ions will be
present?
Dilutions
• Solutions may be diluted by adding more solvent
while the number of moles of solute remains the
same.
• By diluting a solution its volume will increase and the
concentration of the solute will decrease.
• Dilution can be calculated using the formula.
c1 V1 = c2 V2
Q. What volume of water is needed to make 500ml of
0.5M sulphuric acid solution from 15M stock
solution?
Mass – Concentration Stoich
• Same process as mass-mass stoichiometry except
one of the chemicals is in solution.
• Can include reactions such as precipitation, acids and
redox* where solids and liquids or aqueous solutions
are both involved in the chemical reaction.
• Requires the use of two formula:
n = _m_
and
n = cV (volume in Litres)
M
Q. What volume of 1.3M HCl will neutralise 2.5g of
Ca(OH)2?
Solution Stoichiometry
• Same process as for mass-mass and mass-conc.
except the entire reaction occurs in solution.
• Only need to use the formula n = cV.
• Mainly used for acid base reactions and analysis such
as titrations.
Q. An average titre of 22.56ml of a standardised 1.5M HCl
solution was used to neutralise 20ml of an unknown cleaning
product containing NaOH. What is the concentration of this
product?
Calculating pH
• The pH of an acid or a base is a measure of the
concentration of H3O+ ions present.
• pH = -log10[H3O+]
• [H3O+] = 10-pH
To find pH when acid conc is known
To find the acid conc when given the pH.
• In year 11 we assume all acids completely dissociate
(ionise) thus the concentration of H3O+ will be the
same as the concentration of the acid.
The ionic product of water
• Pure water is neutral and has a pH of 7 at 25oC.
• The concentration of hydronium and hydroxide ions
are equal with a combined product of 10-14M
• [H3O+] x [OH-] = 10-14 M at 25oC.
• The pH is a measure of [H3O+] ions, NOT [OH-] ions.
• How do we calculate the pH of a base that produces
hydoxide ions?
Calculating the pH of a base:
1. Determine the concentration of Hydroxide ions
produced referring to the balanced equation.
2. Calculate the concentration of hydronium ions using
the formula:
[H3O+] x [OH-] = 10-14 M
3. Now use the formula:
pH = -log10 [H3O+]
If you are given the pH of a alkali and want to work out
the concentration of the base, do this in reverse
except use the formula [H3O+] = 10-pH in step 3.
Percentage Yield
• Stoichiometry can be used to predict the theoretical
amount of product produced by a reaction.
• The theoretical amount is never achieved in real life.
• This is because:
 Reactions do not always go the completion
 Impurities and side reactions interfere
 During purification product is usually lost.
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