Download Form 3 Chem. Term 1 Notes.FINAL.

CHEMISTRY
FORM THREE TERM ONE
CLASS NOTES UPDATED JANUARY 2014
BY ONYANGO NGOYE
[email protected]
For My Dear 2014 Form Three Students at Ruthimitu Mixed Secondary School
Called to Serve the Lord Jesus Christ eternally.
1
THE GAS LAWS
The factors that affect the behavior of gases are-:
(i)
(ii)
(iii)
Pressure
Temperature &
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
gas
piston
P1
V1
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.
Pressure
p
Volume (V)
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
P
mm/Hg
1/V (cm3)
3
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 cm3
40 atm
= 30 cm3
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 250cm3
= 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
Scale (metre rule)
rer
Stir
Capillary tube
Mercury pellet
Th
te r
me
o
erm
Water
Heat
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
volume
(cm3)
-2 7 3 o C
0oC
80oC
160oC
240oC
Temp
6
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 Kelvin
T1 = 15oC + 273 = 288K
T2 = 30oC + 273 = 303K
V1 = 100cm3
V2 = ?
100cm3 = V2
288K
303K
V2 = 100cm3 x 303K
288K
=105cm3
(b) V1 = 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
225cm3 x 373k =V2
303k
276.98cm3 = V2
8
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 constant
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
Volume
= 760 mmHg
=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 = 301K
T2 = 0 + 273K = 273K
9
P1V1 = P2V2
T1
T2
P1V1T2 = V2
T1P2
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 + = 309K
P1 = 85KPa
= 8500Pa
V2 = ?
T2 = 0oC + 273 = 273K
P2 = hdg = 0.76m x 13600 x 10 = 101325Pa
P1V1 = P2V2
T1
T2
P1V1T1 = V2
T1P2
85000pa x 75cm3 x 273K = V2
390k x 101325Pa
55.59cm3 = V2
GRAHAM’S LAW OF DIFFUSION
10
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.
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.
x
if
y
Nh 3
HCL
Co tto n wo o l
so a ke d in
Ammo n ia
wh ite rin g o f
Ammo n ia ch lo rid e
11
co tto n wo o l
so a ke d in HCL
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)
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
12
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
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
eA
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
eA
13
Since density is directly proportional to molecular mass, Grahams law can also be expressed as
Rate = constant
√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
MB – relative molecular mass of gas B
MA
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
RB
TA and TB are times of diffusion of gases A and B.
TA
But RA = √e B = √MB = √TB
RB
eA
MA
TA
TB = √e B
TA
eA
14
TB = √ MB
TA
MA
RA = √MB
RB
MA
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
2cm3/sec = √32
3cm/sec
4
9
=
Mx
32
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 gas
M2 Molecular mass of oxygen
15
t1 time taken by unknown gas = 6.3 sec
t2 time taken by oxygen = 5.6 sec
t1 unknown = √M1 unknown
t2 O2
M2 of oxygen
6.3 sec = √M1
5.6
32
(6.3sec) = M1
5.6 sec
32
39.69 = M1
31.36
32
1270.08 = 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 = 32
Molecular mass of CH4 = 12 + (4 X 1) = 16
Molecular mass of SO2 = (r x 16) = 64
t of CH4 = √MCH4
t of SO2
MS02
tCH4 = √16
80sec
64
2
t = 16
6400 64
64t2 = 16 x 640
16
t2 = 16 x 6400
64
= 1600sec
t = 40sec
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.
= √ my
mx
rx
ry
2
3
=
√my
mx
4 = √ my
9
mx
4my = 48 x 4
My = 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.
10 cm
x
HCL
Ammonia (Nh3)
white ring
HCL
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
17
HCl = 1 + 35.5 = 36.5
r2 = √m1
r1
m2
x =√ 17
10
36.5
X2 = 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 = e1
r2 e2
= √1.98
2.9
=0.83
(ii)
RA = √ex = √ Mx
RB
ey
My
= ex = √Mx
ey
My
1.98 = X
2.9
64
1.98 X 64 = X
2.9
43.69 = X
18
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 = 100cm3
11.1min
= 9.01cm/min
Rate of diffusion of hydrogen = 100cm3
2.42min
Rate of x = MH2
Rate of H2
Mx
= 41.32 cm3/min
9.01cm3/min = 2
41.32cm3/min
mx
Mx = 3415.07
81.1801
Mx = 42
\
THE MOLE
A 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.
19
Molar masses and Avagadro’s constant
Element
Hydrogen
Carbon
Magnesium
Sulphur
Iron
Calcium
Nitrogen
Symbol
H
C
Mg
S
Fe
Ca
N
R.A.M
1
12
24
32
56
40
14
Molar Mass
1
12
24
32
56
40
14
No. of Particles
6.023 x 1023
6.023 x 1023
6.023 x 1023
6.023 x 1023
6.023 x 1023
6.023 x 1023
6.023 x 1023
Calculating the relative molecular mass
Relative 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
20
(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
O
(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
mass
m
m
o
a
l a r
s s
N
m
o
o
.
o f
l e s
21
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
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 = 16
R.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)
22
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.
(Na = 23, C = 12, O = 16
R.M.M of Na2CO3
23X2 + 12 + 16X3 =106
1mole of Na2CO3 = 106
0.1 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
23
= 1.8069 x 1023 atoms
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.
Lid
Crucible
Magnesium ribbon
Heat
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
24
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
Magnesium
Mass (g)
Oxygen
0.84
0.56
Atomic mass
24
16
No. of moles
0.84
0.56
24
16
1
1
Mole ratio
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.
25
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.
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
26
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)
Element
Cu
O
% Composition by mass
80
20
Atomic mass
63.5
16
No. of moles
= 80
= 20
63.5
16
= 1.25
Mole ratio =
The empirical formula
2)
1
1.25
:
1
= CuO.
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
27
Mole ratio
1
1.5
To make whole number multiply the mole ratio by 2
2
:
3
Empirical formula of iron oxide is Fe2O3
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
12
1
87.8
12.2
12
1
7.312
12.2
R.A.M
No. of moles
Mole ratio
1
Empirical formula
CH2
(CH2) n
:
2
= 82
28
(12+2) n
= 82
14 n
= 82
N
= 82
=6
C6H12
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
% composition by mass 40
H
O
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
2
1
=
Mole ratio = 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
29
H2O
Mass given
3.21
Molecular mass
3.21
=
0.02
0.36
0.36
=
0.02
63.5 +32 + 64 18
Divided by smallest number 0.02
=1
0.02
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
0.08 =1
0.12 =1.5
Mole ratio
30
0.08
0.08
1
1.5
Multiply by 2 to make a whole number.
1 x2
2
1.5 x 2
:
3
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
31
The concentration is 4g/ 500cm3 of Solution.
NaOH
No. of moles = 4
40
0.1 Moles.
i)
ii)
The concentration is 0.1 moles per 500cm3 of solution.
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
32
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
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
33
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
1000cm3
= 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
=
Mass per litre
= molarity
molarity x molar mass
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 =
=
=
(ii)
volume in cm3 x Molarity
1000cm3
400cm3 x 0.25
1000cm3
0.1 moles
Mass per litre = Molarity x molar mass
34
=
=
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
= 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
35
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.
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
36
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)
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)
37
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-
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)
38
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.
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) + SO42- + H2O (l)
2H +CuO (s) → Cu2+ (aq) + H2O
39
IONIC EQUATIONS INVOLVING WATER AND GAS
Na2CO3 (aq) + 2HCl (aq) → 2 NaCl (aq) + CO2 (g) + H2O (l)
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
40
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.
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
41
Titration
Trial
1
2
3
Final burette
readings (cm3)
22.05
41.40
19.25
38.50
Initial burette
readings (cm3)
0.00
22.05
0.00
19.25
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)
1mole
1 mole
NaCl(aq) + H2O
1mole
1 mole
Calculations of the result
Moles of HCl used =
V X M
1000cm3
= 19.25cm3 x 0.1M
42
1000cm3
= 0.001925 moles
No. of moles of NaOH used = 0.001925moles
Molarity of NaOH
25cm3 of NaOH contain → 0.001925 moles
1cm3 of NaOH contain → 0.001925 moles
25cm3
1000cm3 of NaOH → 0.001925 X 1000cm3
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
V1 = 25 cm3
M1 = 0.02m
1 mole
1 mole
V2 = 20cm3
M2 = ?
V1M1 = V2M2
43
1 mole
V1M1 = M2
V2
25cm3 x 0.2M
= M2
20cm3
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)
1mole
→
2NaCl (aq) + CO2 + H2O
2 moles
V1 = 25cm3
V2 = 20cm3
M1 = 0.1M
M2 = ?
Moles of Na2CO3 = V1M1 = 1
Moles of HCl
V2M2
2
44
25cm3 x 0.1M =1
20cm3 x M2
2
2 X 25cm3 x 0.1M = M2
20cm3
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
25cm3 x 0.2M = 2
10cm3 x M2
1
45
25cm3 x 0.2M = M2
10cm3 x 2
0.25m = M2
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)
1mole
→
MNO3(aq) + H2O
1 mole
V1 = 25cm3
V2 = 20cm3
M1 = ?
M2 = 0.1m
Moles of MOH = V1M1 = 1
46
Moles of HNO3 = V2M2 = 1
25cm3 x M1 = 1
20cm3 x 0.01M 1
M = 20cm3 x 0.1M
25cm
= 0.08m
Concentration in g/dm3 = molarity x molar mass
3.2g/dm3
= 0.08m x molar mass
3.2g/dm3 = molar mass
0.08
40 = molar mass
The relative atomic mass of m = molar mass - (m + 16 + 1)
= 40 – 17
= 23
REVISION QUESTIONS
REVIEW QUESTIONS 1.1
Gas Laws
I.
State:
(a)
(b)
(i)
Boyle’s Law
(ii)
Charles’ Law
Write mathematical expressions for each of the laws stated in (a) above and combine them to
give the combined gas equation.
47
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.
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 dm 3 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?
48
12. Study the set-up below and answer the questions that follow.
glass tube
cotton wool soaked
in concentrated HCL
cotton wool soaked
in concentrated NH3
(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?
49
18.
A certain mass of a gas occupies 250cm3 at 25°C and 750mmHg pressure. Calculate its volume at 10°C and
760mmHg.
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.
2.
Define each of the following:
(a)
A mole
(b)
Relative atomic mass
What is the mass of 1 mole of :
50
(a) Sodium
3.
(b) Copper
(c) Carbon
(d) Zinc?
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.
5.
6.
7.
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
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
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?
Which one weighs more?
51
8.
9.
10.
11.
(a)
20 atoms of sodium or 10 atoms of zinc?
(b)
100 atoms of hydrogen or 25 atoms of helium?
(c)
10 molecules of oxygen or 10 atoms of sulphur?
(d)
207 atoms of hydrogen or 1 atom of lead?
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.
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.
Calculate the number of atoms present in:
(a)
112g of iron
(b)
128g of sulphur
(c)
40g of helium
(d)
22.3g of francium
Calculate the number of moles of molecules present in:
(a)
128 g of sulphur (IV) oxide
52
12.
13.
14.
15.
(b)
98 g of oxygen gas
(c)
170 g of ammonia
(d)
30 g of ethane gas.
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
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.
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?
How many moles of ions (i.e. total of all the ions) are there in:
(a)
0.2 mole FeSO4 .7H2O
(b)
20gNaOH
53
(c)
16.
17.
l0g CaCO3.
Calculate the mass of the following:
(a)
3 moles of H2
(b)
0.1 mole of HC1
(c)
0.04 mole of CO2
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.
2.
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.
Calculate the formula mass or molecular mass of the following:
(a)
CuSO4.5H2O
(b)
NH3
(c)
H2SO4
(d)
NiSO4
(e)
PbO2
(f)
Fe2O3
54
(g)
3.
Phd4
(h)
A12O6
(i)
FeCl3
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.
55
7.
10 g of calcium reacted completely with 9.5 g of fluorine atoms.
Work out the empirical formula of the compound.
8.
9.
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.
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.nH 2O. 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 Cu 2O, 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
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(a) Cl2O
(b) CH2Cl2
(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?
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REVIEW QUESTIONS 2.3
1.
2.
3
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.
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.
What is the molarity of a solution containing;
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4.
(a)
0.05 moles in 150 cm3
(b)
0.75 moles in 250 cm3
(c)
0.025 moles in 3 dm3
(d)
0.05 moles iii 500 cm3
(e)
0.04 moles in 200 cm3
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:
7.
(a)
3.0 M KI solution.
(b)
1.50 M Na2SO4 solution
(c)
1.1 M K3PO4 solution
(d)
1.25 M BCl2 Solution?
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?
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8.
What mass of the substance would you weigh out to make the molarity
of the substances stated below:
9.
(a)
0.05 M KMnO4
(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?
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.
2.
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
Calculate the maximum mass of sodium chloride that could be obtained when 11.5g of sodium are burnt
in excess chlorine.
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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)
(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
8.
9.
→ 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)?
When dry chlorine is passed over heated aluminium, aluminium chloride, A12C16 is obtained.
(a)
Write the equation for this reaction.
(b)
Calculate;
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)
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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:
2NaOH(s) + CO2 (g) + 9H2O (g) → Na2CO3.10H2O(s)
12.
(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).
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.
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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;
(a)
18.
19.
Calcium Oxide
(b)
Magnesium Oxide obtained.
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.
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
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(c) 5.8 g of zinc oxide
(d) 20cm3 of 1M lead (II) nitrate solution
(e) 15 g of potassium hydrogen carbonate
(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.
40 cm3 of 2M sodium chloride solution was required to precipitate all the lead ions in 200cm 3 of a solution
of lead (II) nitrate. Calculate the concentration of the lead (II) nitrate solution in g / I.
4.
(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
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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
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 A 2 and B2 used and the volume
ofAB2 formed.
65
2
(iii)
Write a balanced formula equation for the reaction between A2 and B2 to form AB3.
(a)
State Avogadro’s law.
(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
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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
(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.
67
(c) Find the volume at s.t.p. of the gas evolved.
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 cm 3
(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
68
(Na = 23, -= 12, 0 = 16, H = 1).
10.
Determine the molecular formula of hydrocarbon X, given that it consists of 85.7% carbon and that 4 dm 3
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 150 0C
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).
69
(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.
14,
A mixture of zinc granules and zinc oxide were reacted with excess Sulphuric acid. 400 cm 3 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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