CHEMISTRY

FORM THREE TERM ONE

CLASS NOTES UPDATED JANUARY 2014

BY ONYANGO NGOYEonyangoye@hotmail.com

For My Dear 2014 Form Three Students at Ruthimitu Mixed Secondary School

Called to Serve the Lord Jesus Christ eternally.

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THE GAS LAWSThe factors that affect the behavior of gases are-:

(i) Pressure(ii) Temperature &(iii) Volume

When we investigate the behavior of a gas, we normally keep one quantity constant while we investigate the other two.

BOYLE’S LAW

It states that the volume of a fixed mass of a gas is inversely proportional to its pressure if the temperature is kept constant.

Or P ∝ 1/v or

PV = a constant

From this law it follows that

P1 V1 = P2 V2

Where P1 is the initial pressure

V1 is the initial volume

P2 is the final pressure

V2 is the final volume.

P1V1 = P2 V2 is called Boyle’s Law equation.

We can use a syringe or a bicycle pump to show the relationship between pressure and volume

2

A decrease in volume causes an increase in pressure and vice versa. A graph of pressure against volume (v) gives a curve similar to the one shown below.

The graph shows that as the volume increases the pressure of the gas decreases.

NB: Both pressure and volume never reach zero, hence the circle does not touch both X and Y axis.

A graph of pressure (p) against volume

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The graph obtained is a straight line showing pressure is inversely proportional to volume.

When pressure is doubled the volume is halved and vice versa.

Example

1. If 60cm3 of oxygen is compressed from 20 atmospheres pressure to 40 atmospheres, what is the new volume of the gas at constant temperature?

P1 = 20 atm

V1 = 60 cm3

P2 = 40 atm

V2 = ?

P1 V1 = P2 V2

20atm x 60cm3 = 40 atm x V2

V2 = 20atm x 60 cm 3

40 atm

= 30 cm 3

2. A certain mass of a gas occupies 250 cm3 at 25oC and 750 mmHg. Calculate its volume at 25oC if pressure changes to 760 mmHg

P1 = 750mmHg

V1 = 250cm3

P2 = 760mmHg

V2 = ?

P1V1 = P2V2

750mmHg x 250cm3 =760mmHg x V2

750mmHg x 250cm 3 = V2

760mmHg

246.711cm3 = V2

4

CHARLES LAW

It shows the variation of volume and temperature at a constant pressure. It states that, the volume of a fixed mass of a gas at a constant pressure is directly proportional to its absolute temperature.

V ∝ T (Pressure constant)

V = a constant

T

V1 = V2

T1 T2

When the temperature of a given mass of a gas is increased and pressure kept constant, the volume of the gas increases. This is because kinetic energy of the gas molecules increases. The gas molecules move faster and hit the walls of the container harder, increasing the pressure.

When temperature of a given mass of a gas is decreased, the volume of the gas molecules decreases.

Experiment to verify Charles law

A pellet of mercury is drawn into a clean dry capillary tube which is then sealed after the mercury has been adjusted to be the centre of the tube. A fixed mass of a gas is trapped between mercury pellet and the tube.

The tube is held to a scale by a means of a small rubber band is placed in a water bath and stirred. Heat the water and read the length between mercury column and sealed end at temperatures of about 50oC, 60oC, 70, 80, 90 and 100oC.

5

When a graph of volume (length (h) of a gas is plotted against temperature a straight line graph is obtained showing that volume increase is proportional to temperature increase.

V ∝ T

A graph of volume against Temperature

Charles law

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The graph shows that the volume of a gas will be zero at -270oC or ok. This temperature when the volume of a gas is zero is called Absolute Temperature. It is the temperature at which the volume of a gas is assumed to be zero and kinetic energy of the gas molecules also becomes zero.

Absolute temperature is the temperature at which gas molecules loss all the kinetic energy and they stop moving.

Note: To obtain accurate readings

(i) The trapped gas must be dry.(ii) The water in the water bath has to continuously stirred to obtain uniform temperature.(iii) The experiment must be carried slowly to ensure that the gases inside the tube and

water bath are at the same temperature.(iv) The capillary tube is left open at one end so that the mercury pellet is free to move

without changing the pressure on the enclosed gas which is equal to atmospheric pressure plus pressure due to mercury column.

Worked out Examples;1. A gas has a volume of 100cm3 at 15oC

(a) Calculate the new volume if it is warmed at a constant pressure to 30oC(b) Calculate the new temperature of the gas if its volume decreases to 80cm3 at a

constant pressure.

(a) Convert temperature from oC to KelvinT1 = 15oC + 273 = 288KT2 = 30oC + 273 = 303KV1 = 100cm3

V2 = ?

100cm 3 = V2

288K 303K

V2 = 100cm3 x 303K 288K

=105cm3

(b) V 1 = V2

7

TI T2

V2 T1 = T2

V1

80cm3 x 288k = T2

100cm3

230.4k = T2

2. A sample of a gas occupies 225cm3 at 30oC. Find the new volume at 100oC if the pressure remains constant.

V1 = 225cm3

V2 = ?

T1 = 30oC + 273 = 303K

T2 = 100oC + 273 = 373K

V1 = V2

T1 T2

V1T1 = V2

T1

225cm 3 x 373k =V2

303k

276.98cm3 = V2

COMBINED GAS LAW (The General Gas Equation)

Boyle’s law and Charles law can be combined into a general gas equation.

Boyle’s law PV = a constant8

Charles law V/T = a constant

PV = a constant

T

General gas equation or ideal gas equation

P1 V1 = P2 V2

T1 T2

Note: standard temperature and pressure

stp temperature = 273k or 0oC

Pressure = 760 mmHg

Volume =22.4 dm3 =1L

Worked out Examples:

1. A fixed mass of a certain gas occupies 500cm3 at 28oc and 101325 pa. find its volume at 0oc and 100,000 pa.

P1 = 101325pa

P2 = 100,000pa

V1 = 500cm3

V2 = ?

T1 = 28 + 273K = 301KT2 = 0 + 273K = 273K

P 1 V 1 = P 2 V 2 T1 T2

P 1 V 1 T 2 = V2

T1P2

9

101325pa x 500 x 273 = V2

301k x 100,000pa

459.5cm3 = V2

2. A fixed mass of gas occupies 75cm3 at 36oc and 85kpa. Calculate the volume the gas will occupy at stp.V1 = 75cm3

T1 = 36 + 273 + = 309KP1 = 85KPa = 8500Pa

V2 = ?

T2 = 0oC + 273 = 273KP2 = hdg = 0.76m x 13600 x 10 = 101325Pa

P1V1 = P2V2

T1 T2

P1V1T1 = V2

T1P2

85000pa x 75cm 3 x 273K = V2

390k x 101325Pa

55.59cm3 = V2

GRAHAM’S LAW OF DIFFUSION

The volume of a fixed mass of gas can be varied by changing the pressure or temperature of the gas. The particles of a gas are always in a continuous random motion. The random motion increases.

Particles of gases and liquid moves from areas of high concentration to areas of low concentration through diffusion.

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Diffusion can be defined as the movement of particles from regions of high concentration to regions of low concentration.

Diffusion in terms of Kinetic Theory

This theory has two basic assumptions

(i) Matter is made of small particles.(ii) The particles are in constant random motion due to kinetic energy.

The Rate of Diffusion of Gases

The rate of diffusion of a gas is the measure of the quantity of the gas that passes through a point after a certain period of time.

To investigate the rate of diffusion of Hydrogen Chloride and Ammonia

Clamp a long glass tube horizontally with clamps. At the same time plug one end of the long tube with cotton wool soaked in concentrated Ammonia gas solution and the other end with another cotton wool soaked in concentrated hydrochloric acid.

Care should be taken when handling the two solutions because they are corrosive to the skin and eyes. They can cause some respiratory problems if inhaled in large quantities.

Ammonia is used in small salts since it acts on the heart and prevents fainting and dizziness.

The white ring of Ammonium Chloride is formed far away from cotton wool soaked in Ammonia solution since ammonia gas has a higher rate of diffusion compared to Hydrogen chloride.

NH3 (g) + HCl (g) → NH4Cl( s)

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Example

In an experiment the distance covered by ammonia was 6cm while the distance covered by hydrogen chloride was 4cm with the same time interval of 2.5 minutes.

Rate of diffusion of Ammonia = 6cm

2.5min

= 2.4 cm/min

Rate of diffusion of HCl = 4cm

2.5min

=1.6 cm/min

Relative rate of diffusion of NH3 compared to HCl is

Rate of diffusion of NH3 = 2.4cm/min

Rate of diffusion of HCl = 1.6cm/min

= 1.5

This means Ammonia diffuses 1.5 times faster than hydrogen chloride.

The rate of diffusion of gas depends on:

(i) Molecular mass of the gas(ii) Density of the gas

The higher the molecular mass of the gas, the lower the rate of diffusion.

The higher the density, the lower the rate of diffusion.

GRAHAMS LAW OF DIFFUSION

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It states that at a constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density.

Rate of diffusion (r) ∝ √e e – density

r – rate

r ∝ √e

r = a constant

√e

r √e = a constant

When the rates of diffusion of two gases A and B are compared. The equations are

(i) Rate of diffusion of gas A = constant√Density of gas A

RA = constant

e A

rate of diffusion of gas B = constant

√density of gas B

RB = constant

√e B

= RA √e A = RB √e B

RA = √e B

RB e A

Since density is directly proportional to molecular mass, Grahams law can also be expressed as

Rate = constant

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√Molecular mass

The rate of diffusion of two gases A and B can be compared as

RA = √MB MA- relative molecular mass of gas A

RB MA MB – relative molecular mass of gas B

The rate of diffusion of a gas increases as its time (t) of diffusion decreases.

Rate of diffusion of a gas is inversely proportional to time.

RA = constant

Time of diffusion of gas A

RB = constant

Time of diffusion of gas B

RATA = constant

RBTB = constant

RATA = RBTB

RA = TB TA and TB are times of diffusion of gases A and B.

RB TA

But RA = √e B = √MB = √TB

RB e A MA TA

TB = √e B

TA e A

TB = √ MB

TA MA

RA = √MB

RB MA14

Worked out Examples

1. A gas x diffuses through a hole at a rate of 2cm3 / second. Oxygen diffuses through the same hole at a rate of 3cm3 / second. Calculate the molecular mass of gas x.

O = 16

Rx = √Mo

Ro Mx

2cm 3 /sec = √32

3cm/sec Mx

4 = 32

9 Mx

4Mx = 32 x 9

Mx = 32 x 9

4

= 72

2. A given volume of an unknown gas takes 6.3 seconds to pass through a small hole while oxygen takes 5.6 seconds to pass through the same hole under the same conditions of temperature and pressure. What is the molecular mass of unknown gas o = 16

M1 Molecular mass of unknown gasM2 Molecular mass of oxygent1 time taken by unknown gas = 6.3 sect2 time taken by oxygen = 5.6 sec

t1 unknown = √M1 unknownt2 O2 M2 of oxygen

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6.3 sec = √M1

5.6 32

(6.3sec) = M1

5.6 sec 32

39.69 = M1

31.36 321270.08 = M1

32

39.69 = M1

3. Calculate the time taken for given volume of methane (CH4) gas to diffuse through a small hole if the same volume of Sulphur (IV) oxide/ 802 under the same conditions take 80 seconds.H = 1, C = 12, O = 16, S = 32Molecular mass of CH4 = 12 + (4 X 1) = 16Molecular mass of SO2 = (r x 16) = 64t of CH4 = √MCH4

t of SO2 MS02

tCH4 = √1680sec 64t 2 = 166400 64

64t2 = 16 x 640

t2 = 16 x 6400 64

= 1600sec

t = 40sec

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4. The relative rates of diffusion of two gases x and y are in the ratio 3:2 respectively. Given that the relative formula mass of x is 489, calculate the relative formula mass of y.

r x = √ m y ry mx

2 = √m y 3 mx

4 = √ m y 9 mx4my = 48 x 4My = 48 x 4 9= 21.33

5. The setup below was used to study the movement of gas molecules. A white disc was formed where the two gases ammonia and hydrogen chloride mixed.

Determine the distance x where the white disc is formed from cotton wool soaked in HCl (aq)

(N = 14, H = 1, Cl = 35.5)

NH3 = 14 + (3 X 1) = 17

HCl = 1 + 35.5 = 36.5

r2 = √m1

r1 m2

x =√ 17

10 36.5

17

X 2 = 1

100 36.5

X2 = 17 x 100

36.5

= 6.82cm

6. Two gases x and y have relative density of 1.98 and 2.9 respectively. They diffuse under the same conditions.(i) Compare their rates of diffusion(ii) The relative molecular mass of y is 64, determine the molecular mass of x.

(i) r1 = e1r2 e2

= √1.98

2.9

=0.83

(ii) RA = √ex = √ MxRB ey My= ex = √Mx ey My1.98 = X2.9 64

1.98 X 64 = X 2.9

43.69 = X

7. A sample of an unknown gas diffuses in 11.1 minutes. An equal volume of hydrogen diffuses in 2.42 minutes. Calculate the R.M.M of the unknown gas.Let the volume be 100cm3

Rate of diffusion of unknown gas R.M.M = 100cm 3

11.1min

= 9.01cm/min18

Rate of diffusion of hydrogen= 100cm 3 = 41.32 cm3/min 2.42min

Rate of x = MH2

Rate of H2 Mx

9.01cm 3 /min = 2 41.32cm3/min mx

Mx = 3415.0781.1801

Mx = 42

\

THE MOLEA mole of a substance is the amount of substance which contains as many particles as there are carbon atoms in 12g of carbon - 12 isotopes. The number of particles is one mole of any substance is 6.023 x 1023. This number is called the Avagadro number or Avagadro’s constant. The amount of any substance that contains Avagadro’s number of particles is called a mole.A mole is the standard counting unit or SI unit of the amount of a substance.The mass in grams of one mole of a substance is called molar mass. For elements and compounds the molar mass of a substance is equal to its formula mass in grammes.

Molar masses and Avagadro’s constant

Element Symbol R.A.M Molar Mass No. of ParticlesHydrogen H 1 1 6.023 x 1023

Carbon C 12 12 6.023 x 1023

Magnesium Mg 24 24 6.023 x 1023

Sulphur S 32 32 6.023 x 1023

Iron Fe 56 56 6.023 x 1023

Calcium Ca 40 40 6.023 x 1023

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Nitrogen N 14 14 6.023 x 1023

Calculating the relative molecular massRelative molecular mass of a compound is the sum of all the relative atomic masses of the atoms in a molecule in the compound.

Examples.

1. R.M.M of water = H2O= (2X1) + (1 X 16)=18

2. R.M.M of carbon (iv) oxide = CO2

= (1 X 12) + (2 X 16)

12 + 32

=44

3. R.MM of Sodium Carbonate = Na2CO3

R.A.M Na = 23, C = 12, O = 16

Na C O

(2 X 23) + (1 x 12) + (3 x 16)

46 12 48 = 106

4. R.M.M of Ammonium Sulphate = (NH4)2SO4

R.A.M N = 14, H = 1, S = 32, O = 16

N H S O20

(2X14)+ (8X1) + (1X32) + (4X16)

28 + 8 + 32 + 64 = 132

CALCULATIONS INVOLVING MOLES

The number of moles = Mass given

Relative atomic mass or relative molar mass

Mass of substance = Number of moles x molar mass

Molar mass = mass given

No. of moles

No. of particles in a substance = No. of moles x Avagadro constant

o. of moles of particles = Number of particles

Avagadros constant

1. Calculate the number of moles in 2.2g of carbon(iv)oxide CO2

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C = 12, O = 16

R.M.M of CO2 = 12 + (16X2) =44g

No. of moles = mass R.M.M = 2.2

44

= 0.05 moles

2. Calculate the number of moles in 112g of Iron. (Fe = 56)

No. of moles = 112 56 =2Moles.

1. Calculate the number of moles in 5.3g of Sodium Carbonate.Na = 23, C = 12, O = 16R.M.M Na2CO3

(23 X2) + (12 X1) + (16 X3) = 106

No. of moles = mass R.M.M

= 5.3 106

=0.05moles

2. What is the mass of 0.2 moles of magnesium. (Mg = 24)

1 mole of Mg = 24

0.2 moles of Mg = 24 x 0.2

=4.8g

3. What is the mass of 0.1 moles of sodium carbonate.22

(Na = 23, C = 12, O = 16R.M.M of Na2CO3

23X2 + 12 + 16X3 =1061mole of Na2CO3 = 1060.1 mole of Na2CO3 = 106 X 0.1

=10.6g

4. Calculate the number of atoms in (i) 4.2g Nitrogen (N = 14)(ii) 6.9g of sodium (Na = 23)

(i) No. of moles of nitrogen = 4.2 14 =0.3 mole

1 mole of Nitrogen contains 6.023 x 1023 atoms

0.3 moles = 6.023 x 1023 x 0.3

= 1.8069 x 1023 atoms

(ii) No. of moles of Na = 6.9

23

=0.3 moles

0.3 moles = 6.023 x 1023 x 0.3

= 1.8069 x 1023 atoms

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EMPIRICAL AND MOLECULAR FORMULA

Empirical formula is the simplest formula of a compound. It shows the simplest ratio of the different atoms present in a compound.

To determine the empirical formula of magnesium oxide

Weigh a clean dry crucible and lid.

Clean about 15cm length of magnesium ribbon.

Coil the magnesium ribbon and place it in the crucible.

Weigh the crucible and magnesium ribbon and record the mass.

Heat the contents strongly for a few minutes, occasionally lifting the lid slightly by tilting it using a pair of tongs.

When there are no more flame ups, remove the lid and heat the crucible strongly. Allow the crucible to cool and weigh again.

Repeat the heating and cooling until a constant mass is obtained. Record your results in a table.

Mass of empty crucible + lid =19.52g

Mass of crucible + lid + magnesium =20.36g

Mass of crucible + lid + magnesium oxide =20.92g

Mass of magnesium = 20.36 – 19.52g =0.84g

Mass of magnesium oxide = 20.92 – 19.52g =1.40g

Mass of oxygen =1.40 – 0.84g =0.56g

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Magnesium Oxygen

Mass (g) 0.84 0.56

Atomic mass 24 16

No. of moles 0.84 0.56

24 16

Mole ratio 1 1

Empirical formula of magnesium oxide MgO

NOTE: It is necessary to clean the magnesium ribbon to remove any oxide film on it.The products weigh more due to the oxygen combining with magnesium.

- It is important to keep the lid in place to prevent any solid from escaping.- It is necessary to lift the lid from time to time to allow air in. The reason for heating until

a constant mass is obtained is to ensure that all the magnesium has reacted.

To determine the empirical formula of copper(ii)oxide.

Weigh an empty porcelain boat containing copper (ii). Place the contents in a combustion tube as shown below. Pass a stream of hydrogen gas through the tube for a few minutes to ensure all air is driven out.

When the gas collected in ignition tube burns quietly, it is safe to light the jet.

Heat the black copper (ii) oxide and record your observation. Continue heating the tube until there is no further change.

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Observation

The black copper (ii) oxide is reduced to reddish brown copper metal. The oxygen combines with hydrogen to form water. Before heating, hydrogen gas was passed through the tube to remove air which may result into an explosion.

Copper was cooled in a stream of hydrogen to prevent reoxidation of heat metal by air.

Sample results of copper (ii) Oxide reduced.

Mass of empty porcelain boat 15.6 g

Mass of empty porcelain boat + copper (ii) oxide 19.1 g

Mass of empty porcelain boat + copper 18.4 g

Mass of copper (ii) oxide 3.5 g

Mass of copper 2.8 g

Mass of oxygen 0.7 g

Element Cu O

Mass 2.8 0.7 g

Atomic mass 63.5 16

No. of moles 2.8 = 0.044 0.7 = 0.0044

63.5 16

Mole ratio 1 : 1

CuO

Note the Empirical formula can be calculated from percentage composition of the element.

The percentages are taken to present the actual masses of the elements in the compound.

E.g. the percentage composition by mass of copper oxide is 80% copper and 20% oxygen. Determine its empirical formula (Cu=63.5, O=16)

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Element Cu O

% Composition by mass 80 20

Atomic mass 63.5 16

No. of moles = 80 = 20

63.5 16

= 1.25 1.25

Mole ratio = 1 : 1

The empirical formula = CuO.

2) The percentage composition by mass of an oxide of iron is 70%, iron and 30% oxygen. Determine its empirical formula (Fe=56, O=16)

Element Fe O

% Composition by mass 70 30

Atomic mass 56 16

No. of moles 70 = 1.25 30 = 1.875

86 16

Divide by smallest 1.25 1.875

Number 1.25 1.25

Mole ratio 1 1.5

To make whole number multiply the mole ratio by 2

2 : 3

Empirical formula of iron oxide is Fe2O3

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MOLECULAR FORMULA

For ionic compounds, the empirical formula is the same as the chemical formula.

Molecular formula shows the actual number of each element present in a molecule of the compound.

If the empirical formula is known, the molecular formula can be determined by the relationship.

(Mass of empirical formula) n = molecular mass

n is always a whole number.

Examples:

1. A hydrocarbon has a percentage composition by mass of 87.8% Carbon the rest is hydrogen. If its relative molecular mass is 84, determine the molecular formula of the compound.

Element Carbon (C) Hydrogen (H)

% composition by mass 87.8 12.2

R.A.M 12 1

No. of moles 87.8 12.2

12 1

7.312 12.2

Mole ratio 1 : 2

Empirical formula CH2

(CH2) n = 82

(12+2) n = 82

14 n = 82

N = 82

= 6

C6H12

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2. A compound has the following percentage composition by mass 40% Carbon 6.7% hydrogen and 53.3% oxygen. The relative formula mass of the compound is 180.

(i) Determine the empirical formula of the compound.(ii) Determine the molecular formula of the compound.

Element C H O

% composition by mass 40 6.7 53.3

R.A.M 12 1 16

No. of moles 40 6.7 53.3

12 1 16

= 3.33 6.7 3.33

Mole ratio = 1 2 1

Empirical formula = CH2 O

(CH2O) n = 180

(12=2=16) n = 180

30 n = 180

n =6

C6H12O6

3. Samples of 3.57g of hydrated salt MSO4.XH2O were heated to a constant mass of 3.21 g of the anhydrous salt. Calculate the value of X (M=63.5, S=32, O=16, H=1)

Element M S O H2O

Mass given 3.21 0.36

Molecular mass 3.21 = 0.02 0.36 = 0.02

63.5 +32 + 64 18

Divided by smallest number 0.02 = 1

0.02

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X = 1

4. A hydrated salt of Barium chloride contain Barium 56.1%, chloride 29.1% and water of crystallization 14.8%. Determine the formula of the compound.

(Barium = 137, Cl = 35.5, O = 16, H = 1)

Element Ba Cl H2O

Mass given 56.1 29.1 14.8

Mole ratio 13.7 35.5 18

0.41 0.82 0.82

0.41 = 1 0.82 = 2 0.82 = 2

0.41 0.41 0.41

BaCl2.2H20

5. The Oxide of iron weighs 6.40g and it contains 4.48g of iron. Determine the formula of this oxide. (Fe = 56, O =16)

Element Fe O

Mass 4.48 1.92

R.A.M 56 16

No. of moles 4.48 = 0.08 1.92 = 12

56 16

Mole ratio 0.08 =1 0.12 =1.5

0.08 0.08

1 1.5

Multiply by 2 to make a whole number.

1 x2 1.5 x 2

2 : 3

30

Fe2O3

6. The following is a composition of a hydrated compound zinc = 22.7% sulphur = 11.1% oxygen 22.3% water of crystallization = 43.9%.

Determine its molecular formula (Zn = 65, S=32, O =16, H2O =18)

Element Zn S O H2O

Mass 22.7 11.1 22.3 43.9

R.A.M 65 32 16 18

No. moles 22.7 = 0.35 11.1 = 0.35 22.3 = 1.4 43.9 = 2.44

65 32 16 18

Divided by the smallest 0.35 = 1 0.35 = 1 1.4 = 4 2.44 = 7

Number. 0.35 0.35 0.3 0.35

CONCENTRATION OF A SOLUTION

Concentration of a solution is the amount of a solute contained in a given volume.

Concentration of a solution may be expressed in terms of mass of solute in grams per given volume or number of moles of the solute per given volume

Example.

If 4g of sodium hydroxide were dissolved in200cm3 of distilled water then made up to 500cm3 of solution. Determine

(i) Concentration in grams(ii) Concentration in moles

i) The concentration is 4g/ 500cm3 of Solution.NaOH

ii) No. of moles = 440

0.1 Moles.

The concentration is 0.1 moles per 500cm3 of solution.

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Normally concentration of solutions usually expressed in grams per litre solution (g/dm3) or in moles per litre of solution (md/dm3)

Expressing the concentration per g/litre or md/litre

(i) Mass in 500cm3 = 4g

Mass in 1cm3 = 4g

500cm3

Mass in 100cm3 = 4 x 100cm3

500cm3

= 8g /dm3

1 moles

500c3m → 0.1 mole

1cm3 → 0.1 mole

500cm3

1000cm3 → 0.1 moles x 1000cm3

500cm3

→ 0.2 moles

Molar solution

It is a solution which contains one mole substance dissolved in water and made one litre (dm3) of solution. Molarity of a solution is the number of moles of the solute per litre of solution.

Example

Suppose 8g of sodium hydroxide are dissolved in enough water then made to one litre of solution. Determine the Molarity of the solution formed. (Na = 23, O = 16, H = 1)

Mass of NaOH dissolved = 8g

32

No. of moles = 8 = 1

40 5

= 0.2 moles

Molarity is 0.2 moles NaOH (aq)

Determine the Molarity of a solution containing 10.6g of sodium carbonate in 250cm3 of distilled water. (Na = 23. C = 12, O = 16)

Molar mass Na2CO3

(2 X23) (1 X12)+ (3 X16) = 106

No. of moles = 10.6g

106

= 0.1 mole.

250cm3 → 0.1 moles

1cm3 → 0.1 moles

250cm3

1000cm3 → 0.1 moles x 1000cm3

250cm3

= 0.4 moles NaOH

Molarity of a solution = No. of moles of solute

Volume of solution in litre

Molarity of solution is expressed in mol/litre

Molarity or concentration = mass per litre

Molar mass

No. of moles of solute = volume in cm3 x Molarity

1000cm333

= volume (dm3) x Molarity

Mass per litre (dm3) = Molarity x molar mass

Worked out examples

1. Calculate the concentration in mole/litre /Molarity of a solution containing 53g of anhydrous sodium carbonate ,Na2CO3 in one litre

Method

R.M.M of Na2CO3 = 106g

Mass per litre = molarity x molar mass

Mass per litre = molarity

Molar mass

53 = molarity

106

0.5 mole = molarity

2. Calculate the number of moles of solute in 400cm3 of 0.25m HCl. Calculate also mass per litre of HCl (aq).

(i) No. of moles = volume in cm3 x Molarity 1000cm3

= 400cm 3 x 0.25 1000cm3

= 0.1 moles

(ii) Mass per litre = Molarity x molar mass = 0.25 x 36.5= 9.125g

Calculate the molarity of a solution of sodium carbonate containing 1.06g in 50 cm3 solution.

R.F.M of Na2CO3 = 46+12 + 48 = 106g

Mass of Na2CO3 per dm3 = 1.06 x 1000cm3

50cm3

34

= 21.2g

Molarity = mass per litre

Mass

= 21.2g

106g

= 0.2 m Na2CO3 (aq)

STOICHIOMETRY OF CHEMICAL REACTIONS

In a chemical reaction the mole ratio in which the reactions combine and the mole ratio in which the products are formed is known as the stoichiometry; it also means balancing of chemical equations.

This relationship can be established through a simple experiment.

Example.

Reaction between copper (ii) sulphate and zinc metal.

Take about 6g of finely powdered copper (ii) sulphate and dissolve in a beaker of water.

Weigh accurately about 1g of zinc powder.

Add the zinc powder to the copper (ii) sulphate solution. Warm the copper (ii) sulphate

Observation: The blue color fades and brown copper metal is deposited.

CuSO4 (aq) + Zn (s) →ZnSO4 (aq) + Cu (s)

Filter off the copper using a filter paper and wash the small particles of copper sticking on the sides of the beaker by a jet of water.

Wash the copper three times with hot distilled water

Wash again with methylated spirit.

35

Allow the copper to dry and then reweigh it.

Calculations

Mass of zinc used = 0.810g

Mass of copper displaced = 0.780g

(Zn = 65, Cu = 63.5)

Element Zn Cu

Mass 0.81 0.78

R.A.M 65 63.5

Mole ratio 0.81 0.78

65 63.5

= 0.012 0.012

= 1 : 1

CuSO4 (aq) + Zn(s) →ZnSO4 (aq) + Cu (s)

1

IONIC EQUATIONS

When a chemical reaction takes place between ionic compounds, it may happen that only certain ions undergo changes while others do not.

Example:

Na2CO3 (aq) + BaCl2 (aq) → BaCO3 (s) + 2NaCl (aq)

36

Ions present in solutions before mixing, are Na+, CO32-, Ba2+ and Cl- ions.

When the solutions are mixed Ba2+ ions unite with CO32- ions to form insoluble Barium

Carbonate, BaCO3 (aq) which leaves the system as a precipitate.

The Na + ions do not undergo any changes. They were free ions at the start of the reaction and they are free at the end of the reaction such ions are known as spectactor ions, they just watch from the sides. They are omitted from the ionic equation.An ionic equation includes only those ions that changes in some way during a reaction.

Examples: (i) Formation of a precipitate

(ii) Formation of water

(iii) Formation of a Gas

The ionic equation of the reaction above

Ionic equation CO32- (aq) + Ba2+ (aq) → BaCO3 (s)

When silver Nitrate solution is added to sodium chloride solution, a white precipitate of silver chloride is formed.

AgNO3 (aq) + NaCl (aq) →NaNO3 (aq) + AgCl (s)

Ions present:

Ag ++(aq) NO3-(aq) + Na+

(aq) + Cl –(aq) → Na+

(aq) + NO3-(aq) + AgCl(s)

Omit those ions which line present on both sides of the equation

Ag + (aq) Cl- (aq) → AgCl (s)

When lead nitrate solution is added to the solution of potassium iodide, a yellow precipitate of lead (ii) iodide is formed.

Pb (NO3)2 (aq) + 2KI (aq) → PbI2 (aq) + 2KNO3 (aq)

Yellow

Pb2+ + 2NO3-+ 2K+ + 2I- → PbI2 + 2K+ + 2NO3

-

37

Ionic equation Pb2+ (aq) + 2 I –(aq) → PbI2 (aq)

IONIC EQUATIONS INVOLVING SOLIDS

When dilute hydrochloric acid reacts with zinc carbonate, aqueous, zinc chloride, carbon (iv) oxide and water are formed.

Zn CO3 (s) + 2HCl (aq) → ZnCl2 (aq) + CO2 (g) + H2Cl (l)

Solid zinc carbonate contains ions but they are not free to move.

ZnCO3 (s) + 2H+ (aq) + 2Cl- (aq) → Zn2+ (aq) + 2Cl-

(aq) + CO2 + H2O

Ionic equation

ZnCO3 (s) + 2H+ (aq) →Zn2+ + CO2 (g) + H2O (l)

IONIC EQUATION INVOLVING DISPLACEMENT

These are redox reactions which involve the displacement of one element from a solution of one of its salts by another element.

e.g. when zinc powder is added to warm copper (ii) sulphate solution turns colourless from blue colour and red- brown precipitate of copper is formed.

38

Zn (s) + Cu2+ (aq) + SO4 (aq) → Zn2+ (aq) + SO42- (aq) + Cu (s)

Ionic equipment Zn (s) + Cu2+ (aq) → Zn 2+ (aq) + Cu(s)

Ionic equation involving the evolution of Gases When zinc metal reacts dilute hydrochloric acid, hydrogen gas is liberated.

Zn (s) + 2HCl (aq) →ZnCl2 (aq) + H2 (g)

Zn (s) + 2H+ (aq) + 2Cl-

(aq) → Zn 2+ (aq) + 2Cl- (aq) + H2(g)

Zn (s) + 2H+ (aq) → Zn2+ (aq) + H2 (g)

IONIC EQUATION IVOLVING WATER (NEUTRALISATION)

Alkali + Acid → salt + water

1. NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)

Na2+ + OH- (aq) + H+

(aq) + Cl- → Na+ + Cl- (aq) + H2O

OH- (aq) + H+(aq) → H2O(l)

2. H2SO4 (aq) + CuO(s) → CuSO4 (aq) + H2O (l)

2H+ (aq) + SO42- (aq) + CuO(s) → Cu 2+ (aq) + SO4

2- + H2O (l)

2H +CuO (s) → Cu2+ (aq) + H2O

IONIC EQUATIONS INVOLVING WATER AND GAS

Na2CO3 (aq) + 2HCl (aq) → 2 NaCl (aq) + CO2 (g) + H2O (l)

39

2Na+ + CO32- (aq) + 2H (aq) + 2Cl-

(aq) → 2Na+ + 2Cl- (aq) + CO2 (g) + H2O

CO32- (aq) + 2H+(aq) → CO2 (g) + H20 (l)

TITRATION (VOLUMETRIC ANALYSIS)

Volumetric analysis is a method where solution of unknown concentration is used to determine the unknown concentration of another solution.

The solution of known concentration is called Standard solution.

A good standardizing agent should be

(i) Highly pure

(ii) Not react with air

(iii) Not be efflorescent, deliquescent or hygroscopic

e.g. sodium carbonate and solid ethane-dioic acid.

Titration is the running of one solution from a burette into a known volume of the other solution in a conical flask, in which an appropriate indicator has been added.

The volume of the known concentration is measured by a pipette.

40

The running of the unknown is stopped just when the end point is reached. The end point is noticed when the colour of the indicator just changes.

The volume of the solution used from the burette (titre) is noted down as accurately as possible to the nearest = 0.2.

The titre should be recorded in one or two decimal places. When in two decimal places, the last digit should be zero or five.

e.g. 20.10 or 20.15

Note: when you near the end point, add the solution from burette drop by drop until the colour changes.

Properties of indicators used in titration

Indicator Colour in acid solution

Colour in alkali solution

Colour in neutral solution

Phenolphthalein Colourless Pink Colourless

Litmus solution Red Blue Purple

Screened methyl orange

Red Green Grey

FORMAT OF THE ARRANGEMENT OF THE READINGS

Titration Trial 1 2 3

Final burette readings (cm3)

22.05 41.40 19.25 38.50

Initial burette 0.00 22.05 0.00 19.2541

readings (cm3)

Volume of HCl used (cm3)

22.05 19.25 19.25 19.25

Average volume of HCl used = 19.25 + 19.25 + 19.25

3

= 19.25 cm3

Note: Average titre is determined by considering the values which are consistent to within 0.1 cm3 range only.

Conclusion

19.25cm3 of 0.1m HCl solution neutralized

25.0cm3 of sodium hydroxide solution

Equation

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O

1mole 1 mole 1mole 1 mole

Calculations of the result

Moles of HCl used = V X M

1000cm3

= 19.25cm 3 x 0.1M

1000cm3

= 0.001925 moles

No. of moles of NaOH used = 0.001925moles

42

Molarity of NaOH

25cm3 of NaOH contain → 0.001925 moles

1cm3 of NaOH contain → 0.001925 moles

25cm3

1000cm3 of NaOH → 0.001925 X 1000cm 3

25cm3

= 0.08

Molarity of NaOH = 0.08m

Example

1. 20.0cm3 of hydrochloric acid required 25.0cm3 of 0.2m sodium hydroxide for complete neutralization. Calculate the molarity of the acid and its concentration in grams per litre.

NaOH(aq) + HCl(aq) → NaCl (aq) + H2O

1 mole 1 mole 1 mole 1 mole

V1 = 25 cm3 V2 = 20cm3

M1 = 0.02m M2 = ?

V1M1 = V2M2

V1M1 = M2

V2

25cm 3 x 0.2M = M2

20cm3

43

0.25M = M2

Concentration in g/litre = molarity x molar mass

= 0.25M x 36.5

= 9.125g/litres

2. 20cm3 of hydrochloric acid required 25cm3 of 0.1m sodium carbonate for complete neutralization. Calculate the molarity of the acid.

Na2CO3(aq) + 2HCl(aq) → 2NaCl (aq) + CO2 + H2O

1mole 2 moles

V1 = 25cm3 V2 = 20cm3

M1 = 0.1M M2 = ?

Moles of Na2CO3 = V1M1 = 1

Moles of HCl V2M2 2

25cm 3 x 0.1M =1

20cm3 x M2 2

2 X 25cm 3 x 0.1M = M2

20cm3

44

0.25M = M2

3. A solution of sodium hydroxide contains 8gdm-3. 25.0cm3 of this solution neutralized 10cm3 of a solution of sulphuric acid.Calculate (i) the molarity of sodium hydroxide

(ii) the concentration of sulphuric acid in md/dm33 and also in g/litres.

(i) R.M.M of NaOH = 23 + 16 +1=40

Molarity of NaOH = Mass per litre

Molar mass

= 8

40

= 0.2M

(ii) 2NaOH(aq) + H2SO4 = NaSO4(aq) + H2O

2moles 1 mole

V1 = 25cm3 V2 = 10cm3

M1 = 0.2M M2 = ?

No. of moles of NaOH = V1M1 = 2

No. of moles of H2SO4 = V2M2 1

25cm 3 x 0.2 M = 2

10cm3 x M2 1

25cm3 x 0.2M = M2

10cm3 x 2

0.25m = M2

45

Concentration in g/litre of H2SO4 = Molarity x Molar mass

= 0.25M x 98

= 24.5g/litres

4. 25.0cm3 of an alkali, MOH, completely reacted with 20.0cm3 of Nitric acid. The concentration of the alkali was 3.2g/dm3.Calculate (i) the molar mass of the alkali

(ii) the relative atomic mass of M.

MOH(aq) + HNO3(aq) → MNO3(aq) + H2O

1mole 1 mole

V1 = 25cm3 V2 = 20cm3

M1 = ? M2 = 0.1m

Moles of MOH = V1M1 = 1

Moles of HNO3 = V2M2 = 1

25cm3 x M1 = 1

20cm3 x 0.01M 1

M = 20cm 3 x 0.1M 46

25cm

= 0.08m

Concentration in g/dm3 = molarity x molar mass

3.2g/dm3 = 0.08m x molar mass

3.2g/dm 3 = molar mass

0.08

40 = molar massThe relative atomic mass of m = molar mass - (m + 16 + 1)

= 40 – 17

= 23

REVISION QUESTIONS

REVIEW QUESTIONS 1.1

Gas Laws

I. State:

(a) (i) Boyle’s Law

(ii) Charles’ Law

(b) Write mathematical expressions for each of the laws stated in (a) above and combine them to give the combined gas equation.

2. A gas occupies 4dm3 at -23°C and 152 mmHg pressure. At what pressure will its volume be halved if the temperature is 27°C.

3. At 27°C and 74OrnmHg, a sample of oxygen occupied 30cm3. What will be its volume at s.t.p.

47

4. A given mass of gas occupies 60cm3 at 0°C and 670mmHg. Find out the volume it will occupy at:

(a) -23°C and 76OmmHg

(b) 0°C and 390mmHg

5. A gas occupies 40 dm3 at 0°C and 10atmospheric pressure. What volume will it occupy at 27°C and 1 atmospheric pressure.

6. With respect to the kinetic theory and with special reference to the particles, state two differences between solids and liquids.

7. 10 litres of nitrogen gas at a pressure of 7.5 atmospheres and a temperature of -23°C was heated to a temperature of 27°C. If the pressure was lowered by 2.5 atmospheres, determine the final volume of the gas.

8. It takes oxygen gas 45 seconds to diffuse through a pin-hole. If the molecular mass of a gas, X is 44g. Calculate the time taken by gas X to diffuse through the pin-hole under the same conditions.

9. Determine the relative molecular mass of W given that the relative molecular mass of Z is 64 and the density of W and Z is 1.98g/cm3 and 2.9 g/cm3 respectively.

10. In an experiment, it was found that a mole of a gas occupies 24 dm3 at 20°C and 1 atmospheric pressure. What volume would it occupy at s.tp?

11. A fixed mass of a gas occupies 100cm3 at -15°C and 600mm.Hg. At what temperature will it have a volume of 200cm3 if the pressure is adjusted to 700mmHg?

12. Study the set-up below and answer the questions that follow.

48

(a) State and explain the observation made in the glass tube.

(b) Indicate with a cross (x) the likely point where the above observation is made.

13. If it takes 44 seconds from the time a bottle of concentrated ammonia is opened on the demonstration desk until the students in the front row can smell ammonia, how long would it take from the time a little hydrogen sulphide gas escapes from the same place until the students in the front row can smell hydrogen sulphide?

14. A given volume of methane (CH4) diffuses from a certain apparatus in 96 seconds. Calculate the time taken by an equal volume of sulphur (IV) oxide (SO2) to diffuse under the same conditions.

15. A volume of 102dm3 of a gas was collected at s.t.p. Determine the volume occupied by this gas at 25°C and 760mmHg.

16. A fixed mass of a gas has a volume of 250cm3 at a room temperature of 27°C and 750mm Hg pressure. Calculate the volume the gas would occupy at 42°C and 750mm Hg pressure.

17. The densities of sulphur (IV) oxide (SO2) and carbon dioxide are 2.9g /dm3 and 1.98g/dm3 respectively. How much faster does carbon (IV) oxide diffuse through a porous pot than sulphur (IV) oxide?

18. A certain mass of a gas occupies 250cm3 at 25°C and 750mmHg pressure. Calculate its volume at 10°C and 760mmHg.

49

19. A certain mass of a gas occupies 500cm3 at 27°C and 720mmHg. Calculate the pressure of the same gas at 90°C if the volume remains the same.

20. Calculate the relative formula mass of a gas X, given that the time taken for equal volumes of oxygen and gas X to diffuse through the same hole is 40 seconds and 48 seconds respectively.

REVIEW QUESTIONS 2.1

Take Avogadro constant, NA =6 x 1023 wherever necessary.

1. Define each of the following:

(a) A mole

(b) Relative atomic mass

2. What is the mass of 1 mole of :

(a) Sodium (b) Copper (c) Carbon (d) Zinc?

50

3. One atom of a metal Q has a mass of 9.833 x 10-23g.

What is the relative atomic mass of metal Q?

4. Calculate the number of atoms present in:

(a) 1115 g of sodium

(b) 0.04g of magnesium

(c) 2.3g of tungsten

(d) 23.4 g of potassium

(e) 3.2 g of oxygen gas

(f) 0.71 g chlorine gas

5. Calculate the number of electrons lost when;

(a) 4.6g of sodium forms sodium ions

(b) I .35g of aluminum forms aluminum ions

(c) 2.lg of lead forms lead (II) ions.

(d) 4g of hydrogen gas forms hydrogen ions

6. Which contains more atoms?

(a) 23g of sodium or 24g of carbon?

(b) 5.6g of silicon or 9.3g of phosphorus?

(c) 28g of nitrogen or 8g of hydrogen?

(d) 207g of lead or 48g of magnesium?

7. Which one weighs more?

(a) 20 atoms of sodium or 10 atoms of zinc?

(b) 100 atoms of hydrogen or 25 atoms of helium?

51

(c) 10 molecules of oxygen or 10 atoms of sulphur?

(d) 207 atoms of hydrogen or 1 atom of lead?

8. Atomic mass of silver is 107.87. Calculate the mass of silver in grammes which are contained in;

(a) 2 moles of silver atoms

(b) 0.5 moles of silver atoms

(c) 10 moles of silver atoms

(d) 0.1 moles of silver.

9. The relative atomic mass of carbon is 12. Calculate the mass of;

(a) 4 moles of carbon atoms

(b) 10 moles of carbon atoms

(c) 6 moles of carbon atoms

(d) 5 moles of carbon atoms.

10. Calculate the number of atoms present in:

(a) 112g of iron

(b) 128g of sulphur

(c) 40g of helium

(d) 22.3g of francium

11. Calculate the number of moles of molecules present in:

(a) 128 g of sulphur (IV) oxide

(b) 98 g of oxygen gas

(c) 170 g of ammonia

52

(d) 30 g of ethane gas.

12. Calculate the mass of substance present in:

(a) 18 x 1023 atoms of zinc

(b) 60 x 1023 atoms of helium

(c) 1.8 x 1023 molecules of carbon (IV) oxide

13. Calculate the number of electrons required to convert;

(a) 32 g of oxygen gas into oxide ions

(b) 5.6 g of nitrogen gas into nitrite ions

(c) 14.2 g of chlorine gas into chloride ions.

14. How many moles of;

(a) Al3+ are there in 2 moles A12(SO4)3?

(b) SO42- are there in 0.7 mole CuSO4?

(c) PO43- are ti, .ere in 0.2 mole Ca3(P04)2?

(d) NH3 are there in 0.05 mole Ag(NH3)2?

15. How many moles of ions (i.e. total of all the ions) are there in:

(a) 0.2 mole FeSO4 .7H2O

(b) 20gNaOH

(c) l0g CaCO3.

53

16. Calculate the mass of the following:

(a) 3 moles of H2

(b) 0.1 mole of HC1

(c) 0.04 mole of CO2

17. Find the total number of atoms in each of the following substances:

(a) 1.65 x 10-4 moles of magnesium

(b) 2.60 kg sulphur.

REVIEW QUESTIONS 2.2

1. Write down the chemical formula of;

(i) Sodium hydroxide

(ii) Sodium carbonate decahydrate

(iii) Zinc sulphate heptahydrate

(iv) Copper (II) sulphate pentahydrate

(v) Potassium dichromate.

2. Calculate the formula mass or molecular mass of the following:

(a) CuSO4.5H2O (b) NH3 (c) H2SO4

(d) NiSO4 (e) PbO2 (f) Fe2O3

(g) Phd4 (h) A12O6 (i) FeCl3

54

3. Determine the percentage composition by mass of each of the elements present in;

(a) HgCl2 (f) N2H4

(b) CaO (g) CO2

(c) NH3 (h) H2O

(d) Al2(SO4)3 (i) H2O2

(e) NH4HCO3 (j) KMnO4

4. For each of the compounds in question 3, give an appropriate name.

5. Given below are mass compositions of a number of compounds.

Calculate the empirical formula for each of them.

(a) P = 22.5% ,Cl = 77.5%

(b) Ca = 40.0% ,C = 12.0% ,O = 48.0%

(c) Pb = 86.5% ,0 = 13.5%

(d) C = 86.5% ,H = 7.6%

(c) N = 26.2% ,H = 7.5%, Cl = 66.3%

(f) Na = 29.1% ,S = 40.5% ,O = 30.4%

(g) C = 82.8g%,H = 17.2%

6. 12.8 g of copper completely reacted with 1.6 g of oxygen atoms.

(a) Calculate the empirical formula of the copper oxide

(b) Determine the formula mass of the oxide.

7. 10 g of calcium reacted completely with 9.5 g of fluorine atoms.

55

Work out the empirical formula of the compound.

8. When 8 g of a metal oxide were reduced using hydrogen, 6.4g of metal M, were obtained.

(a) Given that the relative atomic mass of the metal is 64, determine the simplest formula of the oxide.

(b) Write an equation for the reaction which occurred between the metal M and hydrogen gas.

9. A hydrate of zinc sulphate, ZnSO4 x H2O contained 22.8% of zinc sulphate by mass. Find the value Of X.

10. An anhydrous salt, Q has a relative formula mass of 152 and forms a hydrated salt of formula, Q.nH2O. If 3.8 g of the salt was found to combine with 3.15 g of water, calculate the value of n.

11. In an experiment determine the formula of hydrated sodium sulphate, Na2SO4. x H20, the following results were obtained:

Weight of dish = 15.14 g

Weight of dish + hydrated salt = 18.36 g

Weight of dish + anhydrous salt

After heating to constant mass = 16.56 g.

What is the formula of hydrated sodium sulphate?

12. Which of the following compounds contain the greatest mass of copper Cu2O, CuS, Cu2(NO3)2 and CuCO3, if you consider 10 g of each compound.

13. Find the mass of chlorine in 1.0 g of each of the following compounds

(a) Cl2O

(b) CH2Cl2

56

(c) C14H9Cl5

(d) C12H4Cl402

14. The simplest formula of a compound of nitrogen is CNC1. Work out its molecular formula if its relative molecular mass is 184.

15. The atomic mass of a divalent metal is 9. Calculate the percentage of the metal in its chloride.

16. A hydrocarbon contains 82.8% carbon by mass.

(a) Determine its empirical formula.

(b) If its relative molecular mass is 58, determine its molecular formula.

17. A compound of carbon, hydrogen and oxygen contains 40% carbon 6.67% hydrogen and the rest oxygen. What is its simplest formula? Its relative molecular mass is 180, what is its molecular formula?

18. Ethylene glycol, the common automobile anti-freeze, contains 38.79 carbon, 9.70% hydrogen and 51.6% oxygen. What is the empirical formula of the compound?

57

REVIEW QUESTIONS 2.3

1. What mass of ethanol in grams would you weigh out in order to a solution of ethanol and water which contains;

(a) 50 per cent ethanol

(b) 0.2 per cent ethanol

(c) 200 per cent ethanol.

2. What mass of sodium hydroxide would you weigh out to make a solution which is;

(a) 70 % sodium hydroxide

(b) 10 % sodium hydroxide

(c) 0.4 % sodium hydroxide.

3 What is the molarity of a solution containing;

(a) 0.05 moles in 150 cm3

(b) 0.75 moles in 250 cm3

58

(c) 0.025 moles in 3 dm3

(d) 0.05 moles iii 500 cm3

(e) 0.04 moles in 200 cm3

4. Find the molarity of each of the following solutions:

(a) 2.35 g of KC1 in 35.0 cm3 of, solutions

(b) 10.6 g of Ca(NO3)2 in 125cm3 of solution.

(c) 3.5 g of BaCI2 in 250 cm3 of solution.

(d) 4.75 g of KOH in 25.0cm3 of solution.

(e) 5.3 g of anhydrous sodium carbonate in 100cm3 of solution.

(f) 8.0 g of potassium hydroxide in 200 cm3 of solution.

5. What volume of 0.125 M AgNO3 solutions is needed to prepare 25.0cm3 of 0.01M solution

6. What mass of solute is needed to prepare 100cm3 of each of the following solutions:

(a) 3.0 M KI solution.

(b) 1.50 M Na2SO4 solution

(c) 1.1 M K3PO4 solution

(d) 1.25 M BCl2 Solution?

7. What mass of solute is needed to prepare each of the following solutions?

(a) 100 cm3 of 1.2M (Ca(N03)2 solution

(b) 250cm3 of 2.0 M NaOH solution

(c) 500 cm3 of 0.250 M NaCl solution

(d) 1.50 litres of 3.0 M Na2SO3 solution?

8. What mass of the substance would you weigh out to make the molarity

of the substances stated below:

(a) 0.05 M KMnO4

59

(b) 0.02 M (NH4)2SO4.FeSO4.6H2O

(c) 0.06 M Al2 (SO4) 24H20

(d) 0.5 M H2S04

(e) 0.2 M Na2CO3.10H2O

(f) 0.05 M CuSO4.5H2O?

9. What is the molarity of a solution prepared by dissolving 75.0 g of glucose (C6H12O6) in water to make 1 .2 litre of solution?

REVIEW QUESTIONS 2.4

1. Write balanced stoichiometric equations and then the ionic for the reaction between;

(a) aluminium and dilute nitric acid

(b) iron and dilute hydrochloric acid (to form iron (II) chloride)

(c) zinc and dilute sulphuric acid

(d) calcium and sulphorous acid (to form calcium sulphite)

(e) copper (II) oxide and dilute hydrochloric acid.

(f) lead (II) oxide and dilute nitric acid

2. Calculate the maximum mass of sodium chloride that could be obtained when 11.5g of sodium are burnt in excess chlorine.

3. (a) Calculate the minimum volume of oxygen measured at r.t.p required to react

with 24.8g of phosphorous.

(Assume that only phosphorus (V) oxide is formed)

60

(b) Determine the mass of phosphorus (V) oxide obtained in (a) above

4. What mass of zinc oxide is obtained when 25 g of zinc carbonate heated to a constant mass?

5. Magnesium ribbon was burned in a gas jar of nitrogen. If the the ribbon Was 16 g,

(a) Calculate the mass of magnesium nitride formed.

(b) State any assumption(s) you made in calculating the value

6. Determine the mass of the residue formed when 33.6 g of sodium hydrogen carbonate is heated to a constant mass.

7. Iron (III) hydroxide decomposes when strongly heated as represented by the following word equation.

iron (III) hydroxide → iron (III) oxide + steam

(a) Write the stoichiometric equation for this reaction.

(b) What mass of iron (III) oxide should be heated in order to produce 32 g of iron (III) hydroxide?

(c) What mass of steam would be produced in (b)?

8. When dry chlorine is passed over heated aluminium, aluminium chloride, A12C16 is obtained.

(a) Write the equation for this reaction.

(b) Calculate;

9. Burning magnesium reduces carbon (IV) oxide to carbon (IV) oxide to carbon.

(a) Write the equation for the reaction between magnesium and carbon (IV) oxide.

(b) Calculate:

(i) The mass of aluminium that would react exactly with 400cm3 of dry chlorine gas measured at r.t.p.

(ii) The mass of aluminium chloride that would result in (b) (i).

10. Oxygen can be prepared in the laboratory by strongly heating potassium chlorate which decomposes according to the equation below:2KClO3(s) → 2 KCl(s) + 3O2(g)

Determine the maximum volume of oxygen measured at r.t.p. that can be obtained by strongly heating 24.5g of potassium chlorate.

11. When sodium hydroxide pellets are exposed to air, they react with carbon (IV) oxide and water vapour, and are converted to crystalline sodium carbonate according to the equation below:

61

2NaOH(s) + CO2 (g) + 9H2O (g) → Na2CO3.10H2O(s)

(a) Calculate the maximum mass of sodium Carbonate decahydrate that could result from 120 g of sodium hydroxide pellets.

(b) When this solid is exposed to air, it effloresces to produce sodium carbonate monohydrate.

(i) Write an equation showing this reaction

(ii) Calculate the mass of crystals of the monohydrate that will result from the mass of the decahydrate calculated in (a).

12. Carbon attacks steam at white red heat forming carbon monoxide and hydrogen.

(a) Write the balanced equation for this reaction

(b) (i) Determine the mass of steam required to react exactly with 60g of carbon

(ii) What volume of carbon monoxide, measured at r.t.p. will be produced in (b) (i)?

(c) The mixture of carbon (II) oxide and hydrogen is an important fuel.

(i) Explain why it is used as a fuel.

(ii) Write an equation to support your answer to (c) (i).

13. Calculate the mass of copper (II) oxide that would be obtained by heating 16 g of copper in air.

14. Calculate the mass of calcium oxide that would result if 15 g of calcium carbonate were strongly heated to a constant mass.

15. Calculate the mass of sodium chloride obtained when 9.2 g of sodium are completely burnt in chlorine. Use the equation below.

2Na(s) + Cl2 (g) 2NaCl(s)

16. Determine the maximum mass of zinc oxide that could be obtained by reacting 13 g of zinc with excess steam.

17. Dolomite is a naturally occurring compound consisting of 50% magnesium carbonate and 50% calcium carbonate by mass.100 g of dolomite is strongly heated in air.

Determine the mass of;

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(a) Calcium Oxide (b) Magnesium Oxide obtained.

18. When magnesium is burnt in nitrogen, magnesium nitride ( a white solid) is obtained:

(a) Write an equation for the reaction which occurs between magnesium and nitrogen to form magnesium nitride, Mg3N2.

(b) Calculate the mass of magnesium nitride that would be obtained from 1.2 g of magnesium.

19. When lead (IV) oxide is strongly heated in air, it decomposes to produce lead (II) oxide and oxygen.

What mass of zinc will react exactly with 250 cm3 of 2M nitric acid?

REVIEW QUESTIONS 2.5

1. Calculate the volume of 2M NaOH that is required to neutralize:

(a) 20 cm3 of 1M HC1

(b) 50cm3 of 0.5M H2SO4

(c) 35 cm3 of 0.3M HNO3

(d) 15 cm3 of 0.25 M H3PO4

2. Calculate the volume of 1M HCI required to react exactly with:

(a) 400 cm3 of 0.25M NaOH

(b) 150cm3 of O.2MNH4OH

(c) 5.8 g of zinc oxide

(d) 20cm3 of 1M lead (II) nitrate solution

(e) 15 g of potassium hydrogen carbonate

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(f) 14.4 g of magnesium

(g) 26.5 g of anhydrous sodium carbonate

3. Calculate the molarity of;

(a) Sulphuric acid, 30 cm3 of which neutralized 50 cm3 of 0.25 M KOH

(b) Hydrochloric acid, 15 cm3 of which neutralized 30 cm3 of 0.5 NaOH

(c) Nitric acid, 20 cm3 of which neutralized 25 cm3 of 2M KOH.

(d) Sodium hydroxide, 30 cm3 of which neutralized 10 cm3 of IM HNO3

(e) Potassium carbonate solution, 40 cm3 of which neutralized 15 cm3 of 2M hydrochloric acid.

(f) Potassium hydroxide solution, 80 cm3 of which neutralized 40 cm3 of 0.2 M sulphuric acid.

4. 40 cm3 of 2M sodium chloride solution was required to precipitate all the lead ions in 200cm3 of a solution of lead (II) nitrate. Calculate the concentration of the lead (II) nitrate solution in g / I.

(Pb=207, N=14, O=16)

REVIEW QUESTIONS 2.6

1. 0.162 g of a certain metal displaced 200cm3 of dry hydrogen at s.t.p. Calculate the relative atomic mass of the metal.

2. The relative atomic mass of magnesium is 24. How much magnesium will react with a dilute acid to liberate 112cm3 of hydrogen at s.t.p.?

3. 10cm3 of hydrogen at s.t.p. are burnt in excess chlorine. How much Chlorine is used and what is the volume of the gaseous product?

4. (i) What volume of hydrogen will combine with 3 litres of chlorine and what is the volume of the product, all volumes measured s.t.p?

(ii) What mass of silver chloride is precipitated if the product in treated with an excess

of silver nitrate.

5. (i) What volume of oxygen will be required to oxidize 560cm3 of carbon monoxide

and what is the volume of the products? (Volumes measured at s.t.p.)

(ii) If the gaseous product formed in (i) is bubbled through excess water, calculate the 64

Mass of calcium carbonate precipitated.

6. The weight of 1 litre of a gas at s.t.p. is 1.5. g. What is its Molecular Mass?

7. Given the equation:

2Fe(s) + 3Cl2(g) 2FeCl 3(s)

Calculate:

(i) Volume of chlorine (at s.t.p) required to react with 8 g of iron

(ii) Mass of iron (II) chloride formed

REVISION QUESTIONS 2.7

1. (a) State Gay-Lussac’s Law of gases.

(b) 200cm3 of a gaseous element A2 reacted with 650cm3 of B2 to form 450cm3 of a mixture of AB3

and B2. 50cm3 of B2 did not react.

(all volumes were measured at the same temperature and pressure)

(i) Work out the volume of AB3 formed.

(ii) Write a statement showing the relationship between volumes of A2 and B2 used and the volume

ofAB2 formed.

(iii) Write a balanced formula equation for the reaction between A2 and B2 to form AB3.

2 (a) State Avogadro’s law.

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(b) Calculate the mass occupied by 2400 litres of nitrogen gas at room temperature and pressure. (Molar gas volume 24 litres, N=14)

3. Calculate the volume at s.t.p. occupied by carbon dioxide gas produced when 860g of sodium hydrogen carbonate is heated very strongly.

4. Calculate the volume occupied by the following at standard temperature and pressure (s.t.p):

(a) 142g of chlorine gas

(b) 128Og of sulphur (IV) oxide

(c) 280g of nitrogen gas

(d) 320g of oxygen gas

(e) 4.4g of carbon dioxide gas

(f) 7.lg of chlorine gas

5. Calculate the mass of:

(a) 4 moles of neon gas

(b) 40 moles of oxygen gas

(c) 5650 moles of hydrogen gas

6. Work out the volume of hydrogen gas produced when 6.Sg of zinc is reacted with excess sulphuric acid at room temperature and pressure.

7. Calculate the volume of gas liberated when 20g of calcium is reacted with excess dilute nitric acid at room temperature and pressure

8. When a gas from an oil refinery burns completely, it forms Carbon (IV) oxide and water only. When 250cm3 of the refinery gas burns, 500cm3 of oxygen required to form 250 cm3 of carbon dioxide and 500cm3 of water vapour

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(all volumes measured under the same conditions of temperature pressure). Deduce the equation for the reaction and the formula of the refinery gas.

9. When 50 cm3 of a gaseous hydrocarbon was burnt in excess oxygen 200 cm3 of carbon dioxide and 500 cm3 of steam were obtained, all volumes measured at 150°C and 760mm Hg. Use this information. To deduce the formula of the hydrocarbon and hence the equation for this - combustion.

10. 10cm3 of a gaseous hydrocarbon requires 50cm3 of oxygen for complete combustion. 30cm3 of carbon dioxide is produced, all volumes measured at the same temperature and pressure. Deduce the formula of the hydrocarbon.

REVIEW QUESTIONS 2.8

1. 25cm3 of 0.1 M sodium hydroxide solution required 10 cm3 of sulphuric acid for complete neutralization what was the molarity of the acid?

2. 24 cm3 of 0.1 M sulphuric acid required 32 cm3 of potassium hydroxide solution for complete neutralization. Calculate the concentration of the alkali in g/1.

3. An excess of powdered calcium carbonate was added to 20 cm3 of 2 hydrochloric acid.

(a) Give the equation for the reaction.

(b) Find the mass of powder which reacts.

(c) Find the volume at s.t.p. of the gas evolved.

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4. 25 cm3 of 0.12M sodium hydroxide was neutralized by 30 cm3 of a solution of a dibasic acid H2X, containing 6.3 g of acid in a litre. Calculate:

(i) the Molarity of the acid used.

(ii) the Relative Molecular mass of the acid.

5. What mass of silver chloride will be precipitate when excess silver nitrate solution is added to a solution containing 3.4 g of zinc chloride?’

6. A concentrated solution of sulphuric acid has a density of 1.98 and two per cent of the acid is water. 10 cm3 of the acid is measured and then diluted with water to make a total solution of 500 cm3

(a) Calculate the concentration of the concentrated sulphuric acid in

(i) grammes per litre

(ii) Moles per litre.s

(b) Calculate the concentration of the sulphuric acid solution

7. 0.4 g of magnesium was reacted with excess sulphuric acid.

(a) Determine the volume of hydrogen gas evolved measured at s.t.p

(b) If the solution was evaporated to dryness, what mass of anhydrous powder would be left?

(Mg= 24, S = 32,O= 16. Molar gas volume = 22.4 dm3 at s.t.p)

8. 25 cm3 of 0.25M sodium hydroxide reacted exactly with 10 cm3 of a solution of nitric acid. Determine the molarity of the acid.

9. 7.15 g of sodium carbonate crystals produced 2.65 g of anhydrous solid, Na2CO3.xH2O on heating. Determine the value of x in the compound

(Na = 23, -= 12, 0 = 16, H = 1).

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10. Determine the molecular formula of hydrocarbon X, given that it consists of 85.7% carbon and that 4 dm3 of the gas at r.t.p had a mass of 7g (C=12,H=1)

11. A mixture (10.0 g) of magnesium oxide and magnesium chloride were reacted with 2M dilute hydrochloric acid. 100km3 of the acid was required for the reaction.

(a) Calculate the moles of acid that reacted.

(b) Calculate the moles of magnesium oxide present in the mixture

(c) Calculate the mass of magnesium oxide present in the original mixture.

(Mg = 24, O= 16)

(d) Calculate the percentage of magnesium chloride present in the mixture.

(e) The resulting solution was carefully evaporated to form magnesium chloride crystals, MgCI2.7H2O.

Determine the mass of crystals formed. (H= g, O = 16).

12. A sample of a gas consists of 40% methane (CH4), 20% butane (C4H10) and the rest hydrogen gas by volume. Calculate

(a) the volume of oxygen required to completely combust 50 dm3 of the gas.

(b) the volume of carbon dioxide produced measured at 760 mm Hg and 1500C

13. 2.5 g of zinc carbonate were reacted with 300 cm3 of 0.2 hydrochloric acid

(a)Determine the moles of acid which reacted.

(Zn = 65, C= 12, O=16).

(b).How many moles of the acid remained unreacted?

(c)Determine the volume of 0.5 M sodium hydroxide required to neutralize the excess acid.

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14, A mixture of zinc granules and zinc oxide were reacted with excess Sulphuric acid. 400 cm3 of hydrogen gas were produced (measured at r.t.p.). If the mixture had a mass of 2 g, determine the percentage of zinc oxide in it. (Zn = 65,O = 16).

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