atoms, molecules, and stoichiometry
DESCRIPTION
Atoms, Molecules, and Stoichiometry. Recap. n Al =. n Cl2 =. m Al M Al. m Cl2 M Cl2. 27.0g 27.0gmol -1. 100g 71.0gmol -1. =. =. 14. Limiting Reagent. Example 1 What is the maximum amount of AlCl 3 that can produced from 27.0g of Al and 100g of Cl 2 by the following reaction? - PowerPoint PPT PresentationTRANSCRIPT
Atoms, Molecules, and Stoichiometry
Recap
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Example 1
What is the maximum amount of AlCl3 that can produced from 27.0g of Al and 100g of Cl2 by the following reaction?
2 Al + 3 Cl2 2 AlCl3
nAl =mAl
MAl
= 27.0g
27.0gmol-1nCl2 =
mCl2
MCl2
100g
71.0gmol-1=
= 1.0 mol = 1.41 mol
From the balanced equation: 2 mol Al 3 mol Cl2
Since 1.0 mol of Al would require 1.5 mol of Cl2
Hence Cl2 is the limiting reagent 14
Limiting Reagent
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3
nAlCl3
nCl2
=
nAlCl3 = 2/3 x nCl2 = 2/3 x 1.41 = 0.94 mol
The maximum amount of AlCl3 produced is 0.94 mol.
Limiting Reagent
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The mass of the product formed in a chemical reaction is called the yield.
If the yield is calculated based on the chemical equation and the amount of reactants present, it is called theoretical yield.
The amount of product actually obtained in a chemical reaction that is really carried out, is called the actual yield.
15Percentage Yield
Percentage Yield =Actual Yield
Theoretical YieldX 100%
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Example 1
15.0g of barium chloride was added to 10.0g of iron (III) sulphate and 15.6g of barium sulphate was precipitated. Calculate the percentage yield of barium sulphate.
3 BaCl2 + Fe2(SO4)3 3 BaSO4 + 2 FeCl3
15.0g
(137.3 + 2 x 35.5) gmol-1= 0.0720 mol=
mBaCl2
MBaCl2
nBaCl2 =
16
Percentage Yield
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Example 1
15.0g of barium chloride was added to 10.0g of iron (III) sulphate and 15.6g of barium sulphate was precipitated. Calculate the percentage yield of barium sulphate.
3 BaCl2 + Fe2(SO4)3 3 BaSO4 + 2 FeCl3
Percentage Yield
mFe2(SO4)2
MFe2(SO4)3
nFe2(SO4)3 =
10.0g
[2 x 55.8 + 3(32.1 + 4 x 16.0)] gmol-1= 0.0250 mol=
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Example 1
15.0g of barium chloride was added to 10.0g of iron (III) sulphate and 15.6g of barium sulphate was precipitated. Calculate the percentage yield of barium sulphate.
3 BaCl2 + Fe2(SO4)3 3 BaSO4 + 2 FeCl3
Percentage Yield
From the balanced equation: 3 mol BaCl2 : 1 mol Fe2(SO4)3
Hence 0.0720 mol of BaCl2 will require 0.0240 mol of Fe2(SO4)3
Since 0.0250 mol of Fe2(SO4)3 is provided, there is excess
And thus BaCl2 is the limiting reagent
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Example 1
15.0g of barium chloride was added to 10.0g of iron (III) sulphate and 15.6g of barium sulphate was precipitated. Calculate the percentage yield of barium sulphate.
3 BaCl2 + Fe2(SO4)3 3 BaSO4 + 2 FeCl3
Percentage Yield
nBaSO4 = nBaCl2 = 0.0720 mol
mBaSO4 = nBaSO4 x MBaSO4 = 0.0720mol x (137.3+32.1+64.0)gmol-1
(theoretical)= 16.8g
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Example 1
15.0g of barium chloride was added to 10.0g of iron (III) sulphate and 15.6g of barium sulphate was precipitated. Calculate the percentage yield of barium sulphate.
3 BaCl2 + Fe2(SO4)3 3 BaSO4 + 2 FeCl3
Percentage Yield
Percentage Yield =Actual Yield
Theoretical YieldX 100%
15.6g
16.8gx 100%=
= 92.8%
Atoms, Molecules, and Stoichiometry
Combustion Analysis
Combustion analysis apparatus
• To absorb water : anhydrous cobalt chloride / concentrated H2SO4
• To absorb CO2 : alkaline solutions e.g. aq NaOH
Note: CO2 is an acidic oxide
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Example 1:
Complete combustion of an unknown organic compound of mass 1.000g containing only carbon, hydrogen, and oxygen, gave 1.500g of carbon dioxide and 0.405g of water. What is the empirical formula of the organic compound?
Using mass proportion:
Mass of C in 1.500g CO2 =MC
MCO2 x mCO2
12.0
44.0 x 1.500= = 0.409 g
In 44g of CO2 , there is 12g of C ; In 1.5g of CO2 , there is ? of C
Combustion Analysis
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Example 1:
Complete combustion of an unknown organic compound of mass 1.000g containing only carbon, hydrogen, and oxygen, gave 1.500g of carbon dioxide and 0.405g of water. What is the empirical formula of the organic compound?
Combustion Analysis
Mass of H in 0.405g H2O =2 x MH
MH2O
x mH2O
= 2 x 1.0
18.0x 0.405 = 0.045 g
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Mass of oxygen in the compound = 1.000 – 0.409 – 0.045
= 0.546 g
CXHYOZ + ? O2 ? CO2 + ? H2O
By conservation of atoms/mass,
C in CO2 comes from C in compound
H in H2O comes from H in compound
Mass of C + mass of H + mass of O = mass of compound
0.409 g + 0.045 g + mass of O = 1.00g
Combustion Analysis
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3 4 3
C H O
0.409 0.045 0.546Mass ratio
0.40912.0 0.0451.0 0.54616.0
= 0.0341 = 0.045 = 0.0341
Mole ratio
0.03410.0341 0.0450.0341 0.03410.0341
= 1 = 1.320 = 1
Simplest whole number ratio
Compound’s empirical formula is C3H4O3
Combustion Analysis
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When the unknown organic compound is a gaseous hydrocarbon, the quantities are commonly expressed in terms of gaseous volume under specified conditions.
A hydrocarbon is an organic compound containing only C and H as constituent elements. Expressed as CXHY.
CXHY + (x + y/4) O2 x CO2 + y/2 H2O
Useful combustion equation for hydrocarbons only:
Combustion Analysis18
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Avogadro’s Law: no of moles of gas volume of gas
Provided that temperature and pressure are kept constant
Applicable for all gaseous substances
This means that, Mole ratio is equivalent to the Volume ratio
10 (y/2) cm310x cm310(x + y/4) cm310 cm3
y/2 cm3x cm3(x + y/4) cm31 cm3
y/2 molx mol(x + y/4) mol1 mol
H2OCO2O2CXHY
Mole ratio
Volume ratio
CXHY + (x + y/4) O2 x CO2 + y/2 H2O
Combustion Analysis
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Example 1
When 10 cm3 of a gaseous hydrocarbon was completely burnt in excess oxygen, 20 cm3 of carbon dioxide and 30 cm3 of steam was produced (all gas volumes measured under the same conditions).
Calculate the molecular formula of the hydrocarbon.
CXHY + (x + y/4) O2 x CO2 + y/2 H2O
By Avogadro’s law,
nCO2
nCxHy
VCO2
VCxHy =
x
1
20
10= x = 2
Combustion Analysis
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Example 1
When 10 cm3 of a gaseous hydrocarbon was completely burnt in excess oxygen, 20 cm3 of carbon dioxide and 30 cm3 of steam was produced (all gas volumes measured under the same conditions).
Calculate the molecular formula of the hydrocarbon.
CXHY + (x + y/4) O2 x CO2 + y/2 H2O
By Avogadro’s law,
nH2O
nCxHy
VH2O
VCxHy =
y/2
1
30
10= y = 6
Molecular formula is C2H6
Combustion Analysis
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Points to note:
1. The volume of CO2 may not be given directly
2. The volume of H2O may be given as
a. Decrease in volume when the residual gases are passed through anhydrous CaCl2 or conc. H2SO4
b. Decrease in volume when the residual gases are cooled to below 100oC at atm pressure
3. Oxygen is usually added in excess and not in stoichiometric amount, so there will be excess oxygen as residual gas
4. Contraction = volume of reactants – volume of products
5. Expansion = volume of products – volume of reactants
19Combustion Analysis
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Example 2
In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon.
Volume of CO2 gas = 40.0 cm3 [absorbed by aq NaOH]
By Avogadro’s Law,nCO2
nCxHy
VCO2
VCxHy =
x
1
40
10= x = 4
CXHY + (x + y/4) O2 x CO2 + y/2 H2Oliquid state
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Example 2
In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with an excess of oxygen, and a contraction of 35.0 cm3 occurred. The resultant volume was passed through aqueous sodium hydroxide, and the volume was reduced by another 40.0 cm3. Assuming that all volumes were measured at room temperature and pressure, determine the empirical formula of the hydrocarbon.
35.0 = Initial volume of gases – final volume of gases
35.0 = { 10 + 10( x + y/4 ) + excess O2 } – { 40 + excess O2 }
65 = 10(x+y/4)
Since x = 4, solving y = 10
Hence the empirical formula of the hydrocarbon is C4H10
CXHY + (x + y/4) O2 x CO2 + y/2 H2O
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Example 3
In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with 100.0 cm3 of oxygen. The volume of the resulting gas mixture was 80.0 cm3, which decreased to 40.0 cm3 when passed through aqueous sodium hydroxide. All volumes were measured at room temperature and pressure. What is the formula of the hydrocarbon?
Volume of CO2 produced = 80.0 – 40.0 = 40.0 cm3
CXHY + (x + y/4) O2 x CO2 + y/2 H2O
nCO2
nCxHy
VCO2
VCxHy =
x
1
40
10= x = 4
Volume of resulting gas mixture = 80.0 cm3
= volume of CO2 + volume of unreacted O2
Hence volume of unreacted O2 = 80.0 – 40.0 = 40.0 cm3
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Example 3
In an experiment, 10.0 cm3 of a gaseous hydrocarbon was exploded with 100.0 cm3 of oxygen. The volume of the resulting gas mixture was 80.0 cm3, which decreased to 40.0 cm3 when passed through aqueous sodium hydroxide. All volumes were measured at room temperature and pressure. What is the formula of the hydrocarbon?
CXHY + (x + y/4) O2 x CO2 + y/2 H2O
Therefore volume of O2 used up in the reaction
= 100.0 – 40.0 = 60.0 cm3
nO2
nCxHy
VO2
VCxHy =
x + y/4
1
60
10=
y = 8
The formula for the hydrocarbon is C4H8
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What have I learnt? Determine the percentage yield of a product Determine the formula of a hydrocarbon given
the combustion analysis data In terms of mass In terms of volume
End of Lecture 5
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