L.S.T. Leung Chik Wai Memorial School

Ch2: p1

L.S.T. Leung Chik Wai Memorial School

F.6 Chemistry

Mole Concept

Chapter 2:The Mole Concept

2.1The mole

I.The mole

The amount of substance is a physical quantity. It has the symbol (mol.) and its base unit is the mole. The mole is a unit of counting.

Note:

<1>The elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

<2>The number of atoms of 12 g carbon-12 is 6.022x1023 . This is called Avogadro constant, L.

That is one mole of substance should contain 6.022 x 1023 elementary entities.

Example:

One mole of hydrogen atoms = 6.022 x 1023 hydrogen atoms

One mole of oxygen molecules = 6.022 x 1023 oxygen molecules

One mole of electrons, e = 6.022 x 1023 electrons

<3>The number of moles of a substance can be calculated as follow:

II.Molar mass

The relative molecular mass (Mr) of a compound is defined as:

Relative molecular mass (Mr) = Mass of one molecule of the cpd. / (1/12 mass of one atom of 12C)

For simplicity, the relative molecular mass of a compound can also be calculated as the sum of relative atomic masses of the atoms existing in that compound.

Example: The relative molecular mass of CO2 = 12 + 16 x 2 = 44 (no unit)

Note: In the case of ionic compounds, which do not consist of molecules, formula mass is used instead.

Example: The formula mass of NaCl = 23 + 35.5 = 58.5 (no unit)

Molar mass of an element or compound is the mass of one mole of that element or compound.

Example: Molar masses of

(1)Oxygen atoms

(2)Oxygen molecules

(3)Carbon dioxide

(4)Potassium nitrate

Note: The mass of several moles of substance can be calculated using the following relationship:

Where m denotes the mass of substance (g)

n denotes the amount of substance (mol)

M denotes the molar mass of the substance (g mol-1)

Similarly, the mass of a substance can be converted to the amount of substance of using the following relationship:

2.2.Ideal Gas Equation

I.Moles in Gases

(A)Gay-Lussac’s Law and Avogadro’s Hypothesis

Gay-Lussac studied chemical reactions between gases. Among the observations he made were the facts that one volume of oxygen reacts with exactly twice its volume of hydrogen to form two volume of steam.

Later, Avogadro’s Hypothesis was suggested to explain the simple relationship, which Gay-Lussac had found. Avogadro’s hypothesis states that

Notes:

<1>There are two general condition for temperature and pressure:

(a)R.T.P.= room temperature (250C) and pressure ( 1 atm)

(b)S.T.P.= Standard temperature (00C) and pressure (1 atm)

<2>The number of molecules in one mole of gas is always 6.022 x 1023 .

(B)Molar volume of gases

Since one mole of any gas always contains the same number of molecules, it follows from Avogadro’s law that one mole of any gas always occupies the same volume. This volume is called molar volume.

By studying the densities of most gases, it was found that

This information is used in calculations on the volumes of gases formed in chemical reactions.

Also,

(C)The Ideal Gas Equation

Review: <1>Boyle’s Law: (PV = constant for a given mass of gas under a fixed temperature.)

<2>Charle’s Law:(V/T = constant for a given mass of gas under a fixed pressure)

Boyle’s Law

For ideal gases which obey Boyle’s Law and Charle’s Law the dependence of volume upon external conditions is given by the equation

PxV / T = constant ( for a given mass of gas)

It follows from Avogardo’s Hypothesis that , if one mole of gas is considered, the constant will be the same for all gases. It is called the Universal Gas constant , and given by the symbol R, so that the equation becomes, when V = Vm , the gas molar volume

P Vm = R T

For n moles of gases, the equation becomes

The value of the constant R can be calculated. Consider one mole of an ideal gas at stp. Its volume V is 22.414 dm3 . Inserting values of P, V and T in SI units into the ideal gas equation gives

P = 1.0132 x 105 N m-2

T = 273 K

V = 22.414 x 10-3 m3

n = 1 mole

R = PV / nT= 8.314 N m mol-1 K-1

= 8.314 J mol-1 K-1

Note:The ideal gas equation is an equation of state. For a gas , an equation of state relates pressure, volume and pressure.

A gas that obey ideal gas equation exactly is called ideal gas or perfect gas. An ideal gas does not exist in reality. Real gases obey the ideal gas equation closely at low pressure and high temperature.

(D)The Density and Molar Mass of gas

The density of a gas is found by weighing a know volume of gas. Since

( where m = mass and M = molar mass ) and density  ( in g dm-3 )

 = m /V

Therefore,

Name: ______Class No.:______Date:______Marks:______

Exercises

1. A 512 cm3 sample of gas weighed 1.236 g at 200C and a pressure of one atmosphere. Calculate the molar mass and relative molecular mass of the gas. ( R = 8.314 J K-1 mol-1 ) ( 5 marks )

1. A sample of radioactive element is sealed in a thin walled tube inside an evacuated vessel. The -particles emitted pass into the outer vessel, leaving other decay products inside the tube. After 300 days the volume of Helium gas collected in the outer vessel by this way is 0.040 cm3 at 298 K and 1.00 x 105 N m-2 ; Helium gas being formed from 1.00g of the element.

Calculate the Avogadro’s number from these data, given that 1.00 g of the element emits 3.80 x 1010 particles per second. (Universal gas constant, R = 8.314 J mol-1 K-1 )

( 5 marks )

2.3 Experimental Determination of Relative Molecular Mass

The relative molecular masses of gases and volatile substances can be determined by methods dependent on the ideal gas equation.

(Note : The most accurate method is by Mass spectrometry)

Determination of Relative molecular mass of a volatile liquid e.g. propanone

(see fig. 2-11)

Procedure:

<1>A few cm3 of air is drawn into the gas syringe.

<2>Steam is passed continuously to maintain a steady temperature (T1) . The initial volume of air is also recorded.

<3>The hypodermic syringe is filled with the volatile liquid, e.g. propanone.

<4>About 0.2 cm3 propanone is injected into the air space of the gas syringe through the rubber tube. The mass of the hypodermic before the (m1) and after (m2) injection are recorded.

<5>The final volume (V2) of air plus the vapor in the graduated syringe is recorded.

Calculation :

Assuming the vapor obeys ideal gas behaviour, the ideal gas equation can be applied.

Note:

<1>The temperature should be steady but not necessary 1000C.

<2>The method is limited to determine the molecular mass of substance with boiling point lower than 800C .

(______) may occur for the substances with boiling points higher than 800C.

<3>Possible errors:

(i)The assumption that the vapor behave as ideal gas.

1. The measurement of m1–m2 and the temperature.

<4>The measurement of temperature and pressure can be avoided by using a comparative method whereby the gas density of the gas of unknown molar mass is compared with the gas density of a gas of known molar mass under identical conditions of temperature and pressure. Trichloromethane and propanone are often used for this purpose.

2.4Dalton’s Law of Partial Pressure

The Partial pressure of a component gas is the pressure of the gas would exert if it alone occupied the total volume at the same temperature.

Dalton’s Law can be expressed mathematically

Ptotal = Pa + Pb+ Pc + ………(1)

Where Pa is the partial pressure of component a and so on.

Since Pa = naRT / V,

Ptotal = (na + nb+ nc +……) RT / V`(2)

By combining equation (1) and (2),

Pa = ( na /(na + nb+ nc +……)) Ptotal

The quantity na / (na + nb+ nc +……) is known as the mole fraction of component a. It is the ratio of the number of moles of gas a to the total number of moles of gas present.

Note: From Avogadro’s Hypothesis, it follows that

na / (na + nb+ nc +……)= Va / ( Va + Vb + Vc + ….)

=Volume of gas a / total volume

Name: ______Class No.:______Date:______Marks:______

Exercises

1. 4.00 dm3 of oxygen at a pressure of 400 kPa and 1.00 dm3 of nitrogen at a pressure of 200 kPa are introduced into a 2.00 dm3 vessel. What is the total pressure in the vessel?

( 4 marks )

1. An enclosed vessel contains 4.2 g of nitrogen and 21.3 g of chlorine at one atmosphere and 300 K. What will be the partial pressure of the nitrogen if the temperature is raised to 400 K?

( 4 marks )

2.5The Faraday and The Mole

I.Electrolysis

When a direct current of electricity is passed through an electrolyte, chemical reaction take place at the electrodes. This process is called electrolysis. It literally means breaking up or decomposition by electricity.

Terms involved in electrolysis

Electrolyte : A liquid which reacts chemically when electricity is pass through.

e.g. A molten salt such as molten Lead (II) bromide, an aqueous solution of dilute acid, alkalis, or salts.

Electrode:They are wires, rods or plates which make electrical contact with the electrolyte.

e.g.metals such as platinum, copper and zinc etc.
non-metals : graphite

Cathode:the negative electrode at which reduction takes place.

Anode:the positive electrode at which oxidation takes place.

Inert electrode:electrode which do not react chemically when in contact with electrolytes and when electricity is passed through them.

e.g. Graphite, Platinum

Cation:positively charged ion which moves to cathode and discharge (by gaining electrons) during electrolysis.

Anion:negatively charged ion which moves to anode and discharge (by losing electrons) during electrolysis.

II.Ionic Theory of Electrolysis

In this theory, the direct current of electricity is carried through the electrolyte by ions.

At the electrodes, electrons are transferred to or form the ions. The processes which take place at the electrodes are thus either reduction or oxidation half-reactions. Electrolysis is thus a reduction-oxidation (redox) process.

Example:The electrolysis of molten lead (II) bromide PbBr2 consist of two half reactions:

At the anode: Oxidation take place, bromide ions are discharged. The half equation is

2Br-(l) Br2(g) + 2e

The anode acts as a sink for the electrons from the anions.

At the cathode:Reduction takes place, lead (II) ions are discharged. The half-equation is

Pb2+(l) + 2ePb(s)

The cathode acts as a source of electrons for the cations.

III.Faraday’s Law of Electrolysis

The relationship between the mass of product formed at an electrode and the quantity of electricity passed through an electrolyte is given by Faraday’s Laws of electrolysis.

Note: <1>The quantity of electricity is expressed in terms of the Faraday. The Faraday is the charge carried by one mole of electrons or one mole of singly charged ions. It has a constant (Faraday constant) value of 96500 C mole-1 . Thus

Where 6.02 x 1023 mol-1 is the Avogadro’s constant.

<2>The quantity of electricity (Q) passed through can be calculated from the current (I) and time (t)

Where Q = quantity of electricity (coulomb, C)

I = Current (ampere , A)

t= time (second, s)

Name: ______Class No.:______Date:______Marks:______

Exercises

1.Calculate the mass of lead produced at a cathode by a current of 2A flowing through molten
lead(II) bromide for 30 minutes.

R.A.M. of Pb = 207

( 4 marks )

2.Find how long will it take to deposit 1.00 g of chromium when a current of 0.120 A flows through a solution of chromium (III) sulphate, Cr2(SO4)3 solution.

(R.A.M. of Cr = 52, S=32, O=16 )

(4 marks )

Experiment for demonstrating the Faraday’s First Law of Electrolysis:

The electrolysis of copper (II) sulphate solution using copper electrodes

Note:

<1>The ammeter is used to measure the current passing through the circuit.

<2>The rheostat is used to vary the resistance of the circuit, thereby regulating the current passed.

<3>The rheostat is adjusted to give a constant current. This constant current is allowed to pass through the electrolytic cell for a certain period of time. The electrodes are then washed and dried to determine their change in mass.

<4>Specimen results:

The mass of copper is deposited is potted againstthe quantity of electricity.

Interpretation:

<1>A straight line passes through the origin of the graph. This shows that the mass of copper is deposited is directly proportional to the quantity of electricity . This agrees with Faraday’s First Law ofelectrolysis.

<2>The result can be explained by considering the following equation:

Cu2+(aq)Cu(s)

<3>The passage of 2 moles of electrons (2 Faradays) liberate 1 mole of copper. If the quantity of electricity ( no. of mole of electrons) is doubled , the number of mole and hence the mass of copper deposited is doubled.

Example:

Pb2++2ePb

1 mol.2 mol.1 mol.

One mole of Lead (II) ions required 2 Faradays ( 2 moles of electrons) to be discharged at the cathode.

Note:

The second law of electrolysis can also be stated as:

Suppose there are three electrolytic cells connecting in series for discharging lead, aluminium and bromine. When the same quantity of electricity is passed through these 3 celss, the number of moles of the element deposited can be expressed as follows:

Pb2+ / Al3+ / Br-
Ratio of ionic charges / 2 / 3 / 1
Ratio of quantity of electricity ( in Faraday) to deposit one mole of element / 2 / 3 / 1
Ratio of no. of moles of element formed when 1 faraday is passed through. / 1/2 / 1/3 / 1/1

Experiment for demonstrating the Faraday’s Second Law of Electrolysis:

The electrolysis of copper (II) sulphate solution and silver nitrate solution.

Note: <1>The two voltameters are connected in series so that the same current and hence the same quantity of electricity passes through the two electrolytic cell.

<2>Specimen results:

Copper / Silver
Mass of metal liberated / 0.318 g / 1.080 g
No. of moles of metal liberated / 0.318 / 63.5 = 0.005 / 1.080 / 108 = 0.01
Mole ratio / 1 / 2

Interpretation:

-The results can be explained by the following equations:
Cu2+(aq) + 2e  Cu(s)
Ag+(aq) + e Ag(s)

-The passage of 2 moles of electrons (2 Faradays) will liberated 1 mole of copper and 2 moles of silver.

Conclusion:

Therefore, when the same quantity of electricity is passed, the number of moles of copper and silver liberated is in simple ration 1: 2 whichis the inverse proportional of their ion charges.

Name: ______Class No.:______Date:______Marks:______

Exercises

1.10 g of zinc is found to be deposited at the cathode when electricity is passed through for some time. What will be the mass of silver obtained when same quantity of electricity is used for electrolysis?

( 4 marks )

2.Silver nitrate solution is electrolyzed, using graphite electrode, until 1 g of silver is liberated at the cathode. Calculate the volume of gas liberated at the anode at room temperature and pressure.

(Given: the molar volume of gas in r.t.p. = 24000 cm3 , R.A.M of silver = 107.9)

( 4 marks )

Name: ______Class No.:______Date:______Marks:______

1. After a passage of current of 0.5A for 2 hours 42 minutes and 51 seconds through a molten solution of MBrn . 5.24 g of metal M deposited on the cathode.

(a)Calculate the value of n.

(b)Write an equation for the liberation of M at the cathode. Indicate the state symbols in the equation.

(Give: The R.A.M. of M = 207 )

( 4 marks )