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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