Transcript Mole calculations File

How do you know if it is balanced?
• The number of atoms of each element must balance
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
• LHS:
RHS:
1xC
1xC
4xH
2x2=4xH
2x2=4xO
2+2x1=4xO
Balanced or unbalanced?
Balanced or unbalanced?
2H2 + O2  2H2O

Balanced or unbalanced?
2 Mg + O2  2MgO
x
Balanced or unbalanced?
2 K + Cl2  2KCl
x
• When NaCl is added to water, the ions
are pulled off the structure, and they
become free to move in the water.
visible
+
invisible
• The ions can move independently, so a
solution of NaCl in water conducts electricity.
Activity
• mix copper (II) sulphate and barium chloride
solutions
• describe observations
• did you get them all
–
–
–
–
clear solutions
colours of solutions
colour of the precipitate
the word ‘precipitate’
• Can you explain them
Equations
copper (II) sulfate + barium chloride  copper (II) chloride + barium sulfate
CuSO4(aq) + BaCl2(aq)  CuCl2(aq) + BaSO4(s)
Why is the balanced equation more useful?
Why is the only chemical we can really see the
barium sulfate?
Cu2+(aq) + SO42-(aq) + Ba2+(aq) + 2 Cl-(aq) 
Cu2+(aq) + 2 Cl-(aq) + BaSO4(s)
Is this too much writing?
SO42-(aq) + Ba2+(aq)  BaSO4(s)
Learning objective:
06/04/2017
• To become familiar with the concept of the mole
• Sponge:
350g unsalted butter
350g caster sugar
½ doz medium eggs, beaten
350g self-raising flour
Filling:
6 tbsp strawberry jam, 1 l double cream
• Give the ingredients for: twice as much and half as much
sponge…comment on the units…especially for the eggs.
• What assumption is it making about the eggs? Why is it
useful?
Today
• Must:
Understand the need for the mole concept
• Should:
Link Avogadro’s number to the mole
• Could:
Understand the basis of molar mass
Masses and atoms
• What is relative atomic mass?
• What is relative isotopic mass?
• Why are they relative?
Some sums…
• Radius carbon atom = 73 pm; density (diamond) =
3.5 gcm-3
• Calculate number of carbon atoms in 12 g
• volume 12 g = mass/density
… unit
…convert to m3
• volume occupied by an atom
…assumption cubic or spherical?
• atoms in 12 g = vol 12g/vol atom
• Approx 1x1024…should be 6.02x1023 mol-1
Definition
• The Avogadro constant (symbols: L, NA), also
called the Avogadro number, is
the number of atoms in exactly 12 grams of 12C.
• A mole is defined as this number of "entities"
(usually, atoms or molecules) of any material.
• NA = 6.02x1023 mol-1
• Molar mass – mass of one mole of entities…using 3
d.p. masses…why are they 3 d.p.?
So how big is a mole…really…
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How many beaches like this would you
need to hold one mole of sand grains?
What assumptions will we make?
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Length
Width
Depth
Volume of
sand grain
So how big is a mole…really…
•
What assumptions will we make?
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Length ~ 3 km
Width ~ 200 m
Depth ~ 2 m
Volume of
sand grain ~
0.2x0.2x0.2
mm3
So how big is a mole…really…
•
A rough calculation
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Volume of sand = 3000 x 200 x 2 = 1200000 m3
Volume of sand grain = 0.2 x 0.2 x 0.2 = 0.008 mm3
= 8 x 10-12 m3
Grains =
1.2 x106/8x10-12
–
= 1.5x1017
For one mole
sand:
4 million
beaches
needed!
Part two…
Must:
Calculate amount of substance
Should:
Relate change in amount of
substance to balanced equation
Could:
Comment on method
Is this true?
magnesium+ oxygen  magnesium oxide
2 Mg + O2  2 MgO
• How could we test this experimentally?
Experiment
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Weigh a crucible and lid
Add between 1 and 5 cm strip of magnesium ribbon
Weigh the crucible, lid and magnesium
Place on a pipeclay triangle resting on a tripod.
Leave a small gap between the lid and the crucible
and heat strongly for five minutes (CARE: very hot)
• Allow the apparatus to cool
• Reweigh
• If time available:
Repeat the heating/cooling/weighing sequence until
constant mass is achieved
12T slap
NaCl
NaCl
Cu
Ar
SiO2
H2
A metal
An acidic gas
An element and a
A molecule
monatomic
element
A covalently
An ionic
solid
bonded
lattice
Cu
Ar
SiO2
H2
HCl
HCl
Calculation
• (By definition) one mole of a substance has a mass
equal to its relative mass in grams
• Amount (moles) = mass (g)/molar mass(gmol-1)
• For molecules and compounds:
Relative formula mass
= sum of atomic masses in the formula
Calculation
• Amount of solid (moles)
= mass (g)/relative mass(gmol-1)
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Calculate the mass of Mg in the crucible
Calculate the final mass of MgO in the crucible
Calculate the mass of O added
Calculate the amount of Mg in the crucible
Calculate the final amount of O in the crucible after the
reaction
• What is the ratio of Mg:O
Some sums…
• Calculate the mass of Mg in the crucible
• Calculate the final mass of MgO in the
crucible
• Calculate the amount of Mg in the crucible
• Calculate the final amount of MgOin the
crucible
Does the data fit?
What should the relationship be?
• Estimating error:
If X = A + B or X = A – B , then: DX = √{(DA)2 + (DB)2}
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•
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Find the % error with the masses
Use this to estimate the errors in the amounts
How confident are you in your conclusion?
What other sources of error are present?
Formula Mass
Formula mass (molar mass) = sum of atomic masses in formula
KCl
1 x 39.1 + 1 x 35.5 = 74.6
MgBr2
1 x 24.3 + 2 x 79.9 = 184.1
CuSO4
1 x 63.5 + 1 x 32.1 + 4 x 16.0 = 159.6
Mg(NO3)2
1 x 24.3 + 2 x 14.0 + 6 x 16.0 = 148.3
Empirical formula
• The simplest whole number ratio of the elements in
a compound
A hydrocarbon contains 2.625g carbon and 0.875g hydrogen.
1. Calculate amount of each element
amount (mol) = mass/molar mass
amount C = 2.625/12.0 = 0.219mol
amount H = 0.875/1.0 = 0.875mol
2. Calculate ratio by dividing through by smallest number
C:H = 0.219/0.219:0.875/0.219
C:H = 1.000:3.995 i.e. 1:4
3. Write formula
CH4
Molecular formula
• The formula of a molecule
A hydrocarbon contains 5.00g carbon and 1.25g hydrogen. One mole
of the hydrocarbon has a mass of 30.00g
1. Calculate amount of each element
amount (mol) = mass/molar mass
amount C = 5.00/12.0 = 0.417mol
amount H = 1.25/1.0 = 1.25mol
2. Calculate ratio by dividing through by smallest number
C:H = 0.417/0.417:1.25/0.417
C:H = 1.00:3.00 i.e. 1:3
3. Write formula
CH3
Molecular formula
• The formula of a molecule
A hydrocarbon contains 5.00g carbon and 1.25g hydrogen. One mole
of the hydrocarbon has a mass of 30.00g
4. Calculate molar mass of formula
= 12.0 + 3 x 1.0
= 15.0
5. Calculate ratio of empirical formula mass to molar mass of molecule
= 30.0/15.0
=2
6. Molecule contains two times the empirical formula
C2H6
Empirical formula from % by mass
• The simplest whole number ratio of the elements
in a compound
A compound contains the following percentages by mass:
Mg – 28.8%; C – 14.2%; O – 56.9%
1. Calculate amount of each element – assume % is mass in grams
amount (mol) = mass/molar mass
amount Mg = 28.8/24.3 = 1.19mol
amount C = 14.2/12.0 = 1.18mol
amount O = 56.9/16.0 = 3.56mol
2. Calculate ratio by dividing through by smallest number
Mg:C:O = 1.19/1.18:1.18/1.18:3.56/1.18
Mg:C:O = 1.01:1.00:3.02 i.e. 1:1:3
3. Write formula
MgCO3
Empirical formula from % by mass
• The simplest whole number ratio of the elements
in a compound
A compound contains the following percentages by mass:
Al – 52.9%; O – 47.1%
1. Calculate amount of each element – assume % is mass in grams
amount (mol) = mass/molar mass
amount Al = 52.9/27.0 = 1.96mol
amount O = 47.1/16.0 = 2.94mol
2. Calculate ratio by dividing through by smallest number
Al:O = 1.96/1.96:2.94/1.96
Al:O = 1.00:1.50
Al:O = 2:3
(If you get around 0.5 in the ratio double; around 0.33 or 0.67 triple)
3. Write formula
Al2O3
Amounts and balanced equations
1.
2.
3.
1.
2.
This is easier than you are prepared to believe!
Find the ratio of substances in the equation (don’t forget
that no number = 1)
Use the ratio to convert the amount of known substance to
the amount of unknown
There isn’t a 3
CuO + H2  Cu + H2O
How many moles of copper from 0.1 mole copper oxide?
Ratio 1:1
0.1 gives 0.1  0.1 moles copper from 0.1 moles copper
oxide
Amounts and balanced equations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
How many moles of oxygen are required to
aerobically respire 0.2 moles glucose?
Ratio C6H12O6:O2 = 1:6
0.2 moles requires 6 x 0.2 = 1.2 moles O2
Mg + 2 HCl  MgCl2 + H2
How many moles of hydrogen can be made from 0.1
moles of hydrochloric acid reacted with excess
magnesium
Ratio HCl:H2 = 2:1
0.1 moles makes ½ x 0.1 = 0.05 moles H2
By the end of this
lesson you:
Must
be able to calculate
amounts (moles)
Activity:
06/04/2017
To use balanced
equations to perform
mole calculations
Should
be able to use balanced
equations to work out
number of moles
Could
mass (g)
amount
(moles)
formula
mass
(gmole-1)
relate gas volumes to
number of moles
How many moles of oxygen can be made from
0.5 moles hydrogen peroxide
2 H2O2  2 H2O + O2
Experiment
• Weigh 5 1 cm strips of magnesium ribbon
• Place 5 pieces of magnesium ribbon in the ignition tube
• EITHER: Fill a 100 cm3 measuring cylinder with water and
place upside down in a plastic bowl 2/3rds full of water.
Arrange an upward delivery tube to collect the hydrogen gas
• OR: Attach a gas syringe
• Add 40+ cm3 hydrochloric acid to the conical flask
• Push the bung into the boiling tube
• Tip the ignition tube so acid and magnesium can react
• Measure the volume of hydrogen gas collected
• Enter data on the spread sheet
Results Table
Mg (s) + 2 HCl (aq)  MgCl2 (aq) + H2 (g)
Experiment
• Weigh accurately between 0.9 and 1.0g sodium
carbonate decahydrate (Na2CO3.10H2O) and transfer to
ignition tube
• Place solid in an ignition tube
• Put about 40 cm3 in the flask
• Set up a gas syringe
• Push the bung into the flask
• Attach the gas syringe
• Topple ignition tube
• Measure the volume of carbon dioxide gas collected
Na2CO3 (s) + 2 HCl (aq)  2 NaCl (aq) + H2O (l) + CO2 (g)
Na2CO3.10H2O (s) + 2 HCl (aq)  2 NaCl (aq) + 11 H2O (l) + CO2 (g)
Gas Molar Volume
1 mole of any gas occupies 24000 cm3
(at room temperature and pressure)
1. Volume occupied by:
1 mole CO2 (g)
0.1 mole Cl2 (g)
1 x 24000 = 24000 cm3
0.1 x 24000 = 2400 cm3
volume (cm3)
2. Amount (moles) in:
12000 cm3 SO2 (g)
240 cm3 NH3 (g)
Gas molar
12000/24000
= 0.5
mole
amount
(moles)
240/24000
=
0.01
mole
volume
(24 000cm3)
Calculation
CaCO3 (s) –(heat) CaO (s) + CO2 (g)
Calculate the volume of carbon dioxide formed by the
thermal decomposition of 10 g calcium carbonate
1. Calculate amount (moles) CaCO3
amount = mass/formula mass [40Ca; 12C; 16O]
= 10/(40+12+3x16) = 0.1 moles
2. Use balanced equation ratio to state amount CO2 (g)
1 CaCO3 gives 1 CO2 so 0.1 moles CaCO3 gives 0.1 mole CO2
3. Calculate volume of gas
gas volume = amount x 24000
= 0.1 x 24000 = 2400 cm3
Bingo
Moles CO2 in 22g
NaCl (aq)
Formula mass NH3
Formula mass H2O
H2O, CO2 but not NaCl
% Mg (by mass) in MgO
Na+
Volume CO2 in 22g (at RT)
Moles BaSO4 in 466.8 g
CuO
6x1023
Formula mass Ca(OH)2
Moles SO2 in 36 dm3 at RT
Moles He in 1.2 dm3 at RT
Cl-
Cu2O
Learning objective:
• To
06/04/2017
• Which beaker contains most solute?
• Concentration (moldm-3) = amount (mol)/volume (dm3)
• The unit (moles per decimetre cubed) gives you the
formula:
amount per (divided by) volume
What is a standard solution?
• A standard solution has an accurately known
concentration
• HCl (aq) and NaOH (aq) cannot be used as standard
solutions…
…why not?
• Anhydrous sodium carbonate (Na2CO3) is a good
primary standard
• You are going to make a solution of accurately
known concentration around 0.05 moldm-3
Preparing a standard solution
1. Calculate the mass of sodium carbonate required to
produce 250 cm3 of 0.05 moldm-3 Na2CO3:
–
–
Amount (mol) = volume (dm3) x concentration (moldm-3)
1 dm3 = 1000 cm3 and 1 cm3 = 1 x10-3 dm3
2. Calculate the mass of this amount of Na2CO3
3. Rinse a 250 ml beaker and a 250 cm3 volumetric flask
with distilled water
Why is this necessary?
Why have I used ml & cm3?
4. Weigh a clean dry weighing boat (do not zero balance)
5. Add the mass Na2CO3 calculated in 2 (+/- 0.1 g )
6. Tip into the beaker and reweigh the weighing boat
Preparing a standard solution
1. Dissolve in about 100 ml distilled water – use a
stirring rod to crush lumps
2. When it has all dissolved transfer to the rinsed
volumetric flask using a rinsed funnel
3. Rinse the beaker, rod and funnel into the
volumetric flask
4. Make up the solution in the flask until the bottom
of the meniscus touched the line
5. Mix by inversion
6. Check that the volume is the same after inversion
Preparing a standard solution
1. Calculate the actual [Na2CO3 (aq)]
–
Concentration (moldm-3) = amount (mol)/volume (dm3)
2. Label the volumetric flask with: your intitials;
[Na2CO3 (aq)] 0.0XXX moldm-3
3. We will use the solution to standardise some HCl
(aq) next week
4. The HCl (aq) will then be used to test the
concentration of the bench NaOH (aq) that is
approximately 1 moldm-3
Homework
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Show working
Look up symbols/formula e.g. type barium sulphate into google and
the first link will give you the correct formula
Ask for help before the work is due in – in person or via email
Attempt all of the questions
%Mg in Mg3N2
Mr (Mg3N2) = 24 x 3 + 14 x 2 = 100;
Mass Mg = 3 x 24 = 72
% Mg = 72 x 100/100 = 72 %
Formula of: 29.1% sodium, 40.5% sulfur & 30.4% oxygen
Amount Na = 29.1/23 = 1.27;
Amount S = 40.5/32 = 1.27;
Amount O = 30.4/16 = 1.9;
Ratio (Na:S:O) = 1.9/1.27 = 1: 1: 1.5;
Formula = Na2S2O3
Homework
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•
Formula chloride of calcium. Calcium chloride: 9.435g of
which calcium 3.400g
Amount Ca = 3.4/40 = 0.085;
Amount Cl = (9.435-3.4);/35.5 = 0.17;
Ratio (Cl:Ca) = 0.17/0.085 = 2:1;
Formula = CaCl2
Mass toluene and nitric acid required to make 10 tonnes
TNT
Amount TNT = 10 x 106/(7 x 12 + 5 x 1 + 3 x 14 + 6 x
16); = 44.1 x 103 mol;
mass toluene = 44.1 x 103 x (7 x 12 + 8 x 1); = 4.06
tonnes;
mass nitric acid = 44.1 x 103 x 3 x (1 + 14 + 3 x 16); =
8.33 tonnes
Homework
•
Volume chlorine gas made by electrolysing 60g
NaCl
Amount NaCl = 60/(23 + 35.5); = 1.03 mol;
Amount Cl2 = 1.03/2; = 0.51 mol;
Volume = 0.51 x 22.4; = 11.5 dm3
Titrations
Standardising HCl (aq)
1. Rinse burette with distilled water then your solution of
sodium carbonate.
–
Why is this necessary?
2. Fill the burette with sodium carbonate solution using a
funnel – remove the funnel
3. Rinse a pipette with distilled water and dilute hydrochloric
acid (approx. 0.1 moldm-3)
4. Rinse a wide necked conical flask with distilled water
only.
5. Use a safety filler to transfer exactly 25.0 cm3 of acid into
the conical flask and add 3-4 drops of methyl orange
indicator
Titrations
1. Place the conical flask on a white tile and add
sodium carbonate, with swirling, until the indicator
changes colour
2. Record the volume to the nearest 0.05 cm3
3. Record to 2 d.p. even if you get 24.00 cm3
4. Rinse the flask with distilled water and repeat the
process until you have two concordant titres (two
results within 0.20 cm3).
5. Calculate the mean of the two concordant titres
Calculation
Na2CO3 (aq) + 2 HCl (aq)  2 NaCl (aq) + H2O (l) + CO2 (g)
1. Calculate the amount of Na2CO3 in your mean titre
Amount (mol) = volume (dm3) x concentration (moldm-3)
2. Use the balanced equation to calculate the amount
of HCl in the conical flask – this will be a simple
sum
–
For example: the same, half as much or twice as much
as your answer to 1
3. Calculate the concentration of the hydrochloric
acid
Concentration (moldm-3) = amount (mol)/volume (dm3)
Titrations
Standardising bench NaOH (aq)
1. Rinse a pipette with distilled water and bench
sodium hydroxide (approx.1 moldm-3)
2. Rinse a volumetric flask with distilled water only.
3. Use a safety filler to transfer exactly 25.0 cm3 of
hydroxide into the volumetric flask and make up to
the line – mix by inversion
4. Rinse a burette with distilled water and then your
dilute sodium hydroxide solution
5. Fill the burette using a funnel
BELOW EYE LEVEL
Titrations
1. Rinse the pipette with distilled water and then your
standardised hydrochloric acid
2. Pipette 25.0 cm3 hydrochloric acid into a wide necked
conical flask and add 2-3 drops phenolphthalein indicator
3. Place the conical flask on a white tile and add sodium
hydroxide, with swirling, until the indicator changes
colour permanently
4. Record the volume to the nearest 0.05 cm3
5. Record to 2 d.p. even if you get 24.00 cm3
6. Rinse the flask with distilled water and repeat the process
until you have two concordant titres (two results within
0.20 cm3).
7. Calculate the mean of the two concordant titres
Calculation
NaOH (aq) + HCl (aq)  NaCl (aq) + H2O (l)
1. Calculate the amount of HCl in your conical flask
Amount (mol) = volume (dm3) x concentration (moldm-3)
2. Use the balanced equation to calculate the amount
of NaOH in the mean titre – this will be a very
simple sum
3. Calculate the concentration of the sodium
hydroxid
Concentration (moldm-3) = amount (mol)/volume (dm3)
4. Calculate the concentration of bench sodium
hydroxide
–
How much did you dilute the original solution by?
Homework
• Bestchoice section 1.3 Balancing equations
to concentrations
• AS labskills – work through titration
exercise as far as acid/base titration
exercises
Amount (moles) – concentrations
For solutions:
amount (moles) = volume (dm3) x concentration (moldm-3)
concentration (moldm-3) = amount (moles) / volume (dm3)
volume (dm3) = amount (moles) / concentration (moldm-3)
Find the amount of solute in:
25.0 cm3 0.1 moldm-3 KCl
25.0 x 10-3 x 0.1 = 2.50 x 10-3 mol
12.5 cm3 0.5 moldm-3 MgBr2
12.5 x 10-3 x 0.5 = 6.25 x 10-3 mol
50.0 cm3 0.01 moldm-3 CuSO4
50.0 x 10-3 x 0.01 = 5.0 x 10-4 mol
Amount (moles) – concentrations
For solutions:
amount (moles) = volume (dm3) x concentration (moldm-3)
concentration (moldm-3) = amount (moles) / volume (dm3)
volume (dm3) = amount (moles) / concentration (moldm-3)
Find the concentration of solute in:
2.50 x 10-3 moles in 25.0 cm3 KCl
2.50 x 10-3 / 25.0 x 10-3 = 0.1 moldm-3
2.50 x 10-3 moles in 50.0 cm3 MgBr2
2.50 x 10-3 / 50.0 x 10-3 = 0.05 moldm-3
1.0 x 10-3 moles in 100.0 cm3 CuSO4
1.0 x 10-3 / 100.0 x 10-3 = 0.01 moldm-3
Amount (moles) – concentrations
For solutions:
amount (moles) = volume (dm3) x concentration (moldm-3)
concentration (moldm-3) = amount (moles) / volume (dm3)
volume (dm3) = amount (moles) / concentration (moldm-3)
Find the volume of solution containing:
0.01 mol of 0.2 moldm-3 KCl
0.01/0.2 = 0.05 dm3
2.50 x 10-3 mol of 0.1 moldm-3 MgBr2
2.50 x 10-3/0.1 = 25.0 x 10-3 dm3 = 25.0 cm3
1.25 x 10-3 mol of 0.5 moldm-3 CuSO4
1.25 x 10-3/0.5 = 2.50 x 10-3 dm3 = 2.50 cm3
Combining ideas
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lithium + water  lithium hydroxide + hydrogen
2 Li (s) + 2 H2O (l)  2 LiOH (aq) + H2 (g)
How could we demonstrate this is true?
What can we measure?
How would we measure it?
Which reactant should be in excess?
Instructions
1. Pour about 50 cm3 distilled water into a rinsed narrow
mouth conical flask
2. Set up a gas jar (capacity c. 400 ml) to collect gas over
water
3. Cut and clean a small piece of lithium
(GLOVES and GOGGLES – corrosive and flammable)
4. You are aiming for a mass of about 0.13g – measured
accurately
5. Add the lithium to the water and connect the delivery tube
6. When the reaction has finished mark the volume of gas
collected on the gas jar
…and then
1. Use water and a 100 ml measuring cylinder to
estimate the volume of gas collected
2. Transfer the lithium hydroxide solution into a
rinsed 250 cm3 volumetric flask, rinse the conical
flask and add to the volumetric flask and then
make up to the mark with distilled water.
3. Titrate against standardised hydrochloric acid
using phenolphthalein indicator
4. Repeat until you have recorded two concordant
titres
Calculation
1. Calculate the amount of lithium you started with
–
Amount (mol) = mass (g)/formula mass (gmol-1)
2. Calculate the amount of hydrogen gas you
collected
–
Amount (mol) = volume (dm3)/gas molar volume (dm3mol-1) –
assume it is 24.0 dm3mol-1
3. Calculate the concentration of lithium hydroxide in
the volumetric flask
–
Concentration (moldm-3) = amount (mol)/volume (dm3)
4. Calculate the amount of lithium hydroxide made in
the reaction
–
Amount (mol) = concentration (moldm-3) x volume (dm3)
Combining ideas
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lithium + water  lithium hydroxide + hydrogen
2 Li (s) + 2 H2O (l)  2 LiOH (aq) + H2 (g)
Do your results match the equation?
Comment on the accuracy of the data
Which do you think was the most accurate value?
Explain your choice
I’d expect it to be the titration – explain my choice!