chemical calculations grade 11 (stoichiometry)
DESCRIPTION
PowerPoint to teach Chemical Calculations (Stoichiometry) including, mass mass, mass volume, volume of reacting gases, limiting reagent, percentage composition, empiracle formulaTRANSCRIPT
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Chemical Calculations (Stoichiometry)
K Warne
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Atomic WeightsYou must be able to…· Describe the mole as the SI unit for amount of substance
· Relate amount of substance to relative atomic mass
· Describe relationship between the mole and Avogadro’s
number
· Conceptualise the magnitude of Avogadro’s number
· Describe the relationship between molar mass and
relative molecular mass
· Calculate the molar mass of a substance given its
formula
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REACTIONSHydrogen and oxygen react to produce water.
(Synthesis reaction)
H2 + O2 H2O
Ar: H = 1, O = 16
Mr: H2 =2, O2 = 32
H2O = 16+2=18
Balanced Eqn:
2H2 + O2 2H2O
LHS: RHS:
2(2) + 32 = 36 2(2+16) = 36
LHS = RHS
MASS is CONSERVED!!
This is true for ALL REACTIONS.
This is the reaction used to
power the space shuttle.
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ASKEDGIVEN
Mole Calculations
MOLES MOLES
MASS MASS
VOLUME VOLUME
MOLAR
RATIO
Number
Of
particles
Number
Of
particles
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The MoleThe mole is defined as, “the amount of ………….. with the same number of
……………………… particles as ….. grams of carbon 12”. (n used as symbol for moles)
602 300 000 000 000 000 000 000Six hundred and two thousand, three hundred, billion billion !
6.023x1023 particles
12.00 g
CSymbol (….)
Number of particles = no of moles x no. particles in a mole
Particles = ……………..
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Moles & Particles
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...........................................particles
1 mol 1 mol 1 molmoles
(n)
X
L
6.023 x 1023
18.069 x 10236.023 x 1023
6.023 x 1023 6.023 x 1023
?3
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The Mole and MassThe mole is defined in such a way that the MASS NUMBER (A) of an element
is equal to the relative atomic mass mass of one mole of the substance. (in
grams) - THE MOLAR MASS
• Eg Na = 23g/mol, water(H2O)=18g/mol
Z
A
XAtomic Number
(smaller)
Mass Number
(bigger)
protons + neutrons
Periodic Table Symbol
Relative atomic
mass
or
mass(g) of one
mole
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Isotopes
• Isotopes - Atoms of the same element which have different
numbers of neutrons. Eg: 613C & 6
12C
Relative atomic mass is (actually) the average mass (of all the isotopes in a
random sample) of the atoms of an element relative to 1/12 of the mass of
a carbon-twelve atom.
613C
• 6 protons
• 6 electrons
• 13-6 = 7 neutrons
612C
• 6 protons
• 6 electrons
• 12-6 = 6 neutrons
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The Mole - equationsSodium reacts with water to form hydrogen and sodium hydroxide
according to the equation.
Na + H2O H2 + NaOH
If 46g of sodium are reacted with excess water what mass of hydrogen would be formed?
1. Balance the reaction
2Na + 2H2O H2 + 2NaOH
2 Work out moles of reactant (given).
n(Na)=m/Mr=46/23=2mol
3 Go through the equation to find out the number of moles reacting and being formed - the molar ratio:
Na : H2 2:1 => 1 mole H2 formed
4 Work out quantity asked for.
m(H2) = nxMr = 1 x 2 = 2 g
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Mole examples - B & J p119 21 & p120 22
1. Na + Cl2 NaCl
Calculate the mass of salt formed if 2.3g of sodium is reacted with XS chlorine.
2. Zn + HCl ZnCl2 + H2
What mass of HCl is needed to produce 100g of hydrogen?
3. KClO3 KCl + O2
What mass of oxygen is produced from 1kg of potassium chlorate?
4. Fe2O3 + H2 Fe + H2O
What mass of iron is produced if 3g of rust (Fe2O3) is reacted with XS(100g )of hydrogen?
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Mole examples - B & J p119 21 & p120 22
1. 2Na + Cl2 2NaCl
Calculate the mass of salt formed if 2.3g of sodium is reacted with XS chlorine.
1. n(Na) = m/Ar = 2.3/23 = 0.1mol
2. Molar Ratio
Na:NaCl 2:2 ie 1:1 => n(NaCl) = 0.1mol
3. m(NaCl) = nxMr = 0.1x(23+35.5) = 5.85g
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Limiting reagentIf 20 g of Na reacts with 20 g of water – what
mass of hydrogen will be formed?
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Percentage CompositionAnalysis of a compound by mass makes it
possible to work out the % mass of each element.
eg Table salt: NaCl mass analysis:One mole of NaCl would have a mass of
23 + 35.5 = 58.5g
• The % composition can be found using the formula:
Mass element X x100
Total Mass Compound
• %Na = […../ (…..) ]x100 = …………..% (by mass)
• %Cl = (…../ (…….) )x100 = …………%
% Mass Element X =
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Empirical and Molecular Formula.A compound consists of carbon, hydrogen and oxygen only. The
% by mass are Carbon 40.0% and 6.7% hydrogen. Calculate the
empirical and molecular formula of the compound if Mr = 60g·mol-1
%(O) = 100 – (40+6.7) = 53.3
C H O
In 100g: …….g ……..g ….…g
n=m/Mr:…/…
6.7/….53.3/……
…… …… ……..
……. …… …….
Simplest: … …… ….
Empirical Formulae:
……. (12+2+16 = …..)
Molecular Formula: 2(CH2O)
……… (Mr = …. X 30)
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Empirical and Molecular Formula.A compound consists of carbon, hydrogen and oxygen only. The
% by mass are Carbon 40.0% and 6.7% hydrogen. Calculate the
empirical and molecular formula of the compound if Mr = 60g·mol-1
%(O) = 100 – (40+6.7) = 53.3
C H O
In 100g: 40.0g 6.7g 53.3g
n=m/Mr:40/12
6.7/153.3/16
3.33 6.7 3.33
3.33 3.33 3.33
Simplest: 1 2.01 1Empirical Formulae:
CH2O (12+2+16 = 30)
Molecular Formula: 2(CH2O)
C2H4O2 (Mr = 2x30)
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Concentration - MolarityThe concentration of a solution is defined as the ………………. of
……………………… per ………………. (dm3) of ………………….
solute
solute
Final volume of
……………..
500cm3
=+
Concentration =Amount of ……… (……….)
Volume of ………………
30g of
NaCl
C =n
v
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ASKEDGIVEN
Mole Calculations
MOLES MOLES
MASS MASS
VOLUMEVOLUME
CONCENTRATIONCONCENTRATION
MOLAR
RATIO
Number
Of
particles
Number
Of
particles
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Volume Conversions
1 dm = ….. cm
1 dm3 = ………… cm3
1 m3 = …………….. dm3 = ………………….cm3 (10….)
1cm3
1 dm3 (1 litre)
10 cm3
10 cm3
10 cm3
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Decimal Conversions
King Henry Died a miserable death called measles
Kilo Hecta Decca m(unit) deci centi milli
1000 100 10 1 1/101/100
1/1000
1m
?km
0. 0 0 1km
1 m = 0.001 km
1 km = 1000 m
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Conversions
1 cm = 0.1 dm
1 cm2 = (0.1)2 dm2
1 cm2 = 0.01 dm2
1 cm3 = (0.1)3 dm3
1 cm3 = 0.001 dm3
1 dm3= 1000 cm3
25 cm3 = 0.025 dm3
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Molar VolumesOne mole of an ideal (ANY) gas occupies a volume of ………….3 at
………………………… temperature and pressure. (STP)
STP: T= ….ºC, ……K P =1 atmosphere (……...kPa)
Fe2O3 + 3H2 2Fe + 3H2OWhat volume of hydrogen reacts with 50g of Fe2O3
Fe2O3 : H2
… : …..
n(H2) =…..n(Fe2O3) = ……………………
v(H2) = …………………………… dm3
n(Fe2O3) = m/Mr = ………………….= ………………mol
moles = volume/molar volume ==> n = …./…..
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Concentration CalculationsHydrochloric acid reacts with calcium carbonate according to the following
equation:
HCl + CaCO3 CaCl2 + H2O + CO2
What mass of calcium carbonate would be needed to react completely with 25 cm3 of a 0.1 mol.dm-3 hydrochloric acid solution?
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Volume - Volume Calculations1. Balance the equation
2. Calculate the moles of the substance given.
3. Work through the molar ratio to find out the moles of the substance asked.
4. Calculate the quantity asked for. (Volume V = n x Mv)
Mv = 22.4dm3 At STP
EG: H2 + N2 --> NH3If 3.00 dm3 of nitrogen are reacted to produce ammonia, what
volume of hydrogen will be required? (At STP)
H2 + N2 --> NH3
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Mass Volume Calculations1. KClO3 KCl + O2
What volume of oxygen is produced by the decomposition of 1kg of potassium chlorate?
1. Balance the equation - 2KClO3 2KCl + 3O2 (1)
2. Calculate the moles of the substance given.
n(KClO3) = m/Mr = 1000/(39+35.5+3(16)) = 8.16mol (1)
3. Work through the molar ratio to find out the moles of the substance asked.
KClO3 : O2 2 : 3
n(O2) = 3/2n(KClO3) = 3/2(8.16) = 12.24 mol (1)
4. Calculate the quantity asked for. (Volume V = n x Mv)
Mv = 22.4dm3 At STP
v(O2) = n(O2)Mv = 12.24(22.4) = 275 dm3 (2)
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Standard Solution
A standard solution is one for
which the concentration is precisely known.
Since
c = n(solute)/v(solvent)
= m/Mr V
• The number of moles of solute(Mass)
• The volume of solution.
These values must be accurately determined.
2.45g
Mass is determined accurately
using an electronic balance.
• Possible accuracies of 0.1 -
0.0001g
KMnO4
Volume is measured
using a volumetric
flask.
• 250 cm3
• 100 cm3, 200 cm3,
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Volumetric FlaskMaking a standard solution.
1. Rinse a clean & dry 100 cm3 beaker with a little distilled water.
2. Transfer the correctly weighed amount of salt to the beaker. Ensure NO SALT IS LOST.
3. Add 50 - 80 cm3 water the salt and stir gently with a glass rod until all salt is dissolved. DO NOT REMOVE THE ROD FROM THE SOLUTION NOR ALLOW ANY DROPS OF SOLUTION TO ESCAPE.
4. Add ALL the solution to volumetric flask via funnel. Ensure glass rod and beaker are thoroughly rinsed. (Include rinsings.)
5. Add enough solvent to bring the level up to the mark.
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Hi -
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My FULL presentations, which contain a lot more more slides and other resources, are freely
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Have a look and enjoy!
WarneScience