Atoms, Molecules, and Stoichiometry

Recapof Lecture 5

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Recap

Percentage Yield =Actual Yield

Theoretical YieldX 100%

Combustion Analysis

Values given in terms of mass

Values given in terms of volume

Find the mass of each element

Find mole ratio and hence empirical formula

Make use of Avogadro’s law to compare volumes

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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

Combustion Analysis

Mass of C

Mass of CO2

MC

MCO2

=

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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

Combustion Analysis

Mass of C Mass of CO2

MC

MCO2

= x

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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

Mass of oxygen in the compound = 1.000 – 0.409 – 0.045

= 0.546 g

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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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Avogadro’s Law: no of moles of gas volume of gas

Provided that temperature and pressure are kept constant

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 =

20

10= x = 2

Combustion Analysis

x

1=

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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= y = 6

Molecular formula is C2H6

Combustion Analysis

30

10=

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Points to note:

1. The volume of CO2 may be given as a decrease in volume when the residual gases are passed through NaOH or other alkali

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.

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.

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.

CXHY + (x + y/4) O2 x CO2 + y/2 H2O

y/2 molx mol(x + y/4) mol1 mol

0 cm310x cm310(x + y/4) cm310 cm3

H2OCO2O2CXHY

Mole ratio

Volume ratio

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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.

CXHY + (x + y/4) O2 x CO2 + y/2 H2O

y/2 molx mol(x + y/4) mol1 mol

0 cm340 cm310(x + y/4) cm310 cm3

H2OCO2O2CXHY

Mole ratio

Volume ratio

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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

20

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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 formula of a hydrocarbon given

the combustion analysis data In terms of mass In terms of volume

End of Lecture 6(End of part I)

Have a prosperous and happy lunar new year

(and do take care of your health!)