SHE 1315
MATTER AND MEASUREMENT
TOPIC 1: MATTER AND MEASUREMENT
1.
Fill in the boxes with appropriate answers.
Matter
Mixture
Substance
Homogeneous
Heterogeneous
Ex: salt water
Ex: oil and water
Atom
Ex: C
Element
Molecule
Ex: H2
Compound
Molecule
Ex: H2O
Formula Unit
Ex: NaCl
2.
Classify the following as physical or chemical change.
a)
Burning of candle – chemical change
b)
Tree leaves are green in the spring and summer, but brown during autumn chemical and physical change
c)
Condensation of steam -physical change
d)
Souring of milk -chemical change
e)
Electrolysis of water -chemical change
f)
Melting of gold -physical change
g)
Gasoline evaporates from a container-physical change
h)
Rusting of steel -chemical change
i)
Boiling an egg -physical change
j)
Boiling a liquid -physical change
k)
Grinding sugar into fine powder-physical change
CHEMISTRY DEPARTMENT
1
SHE 1315
3.
4.
5.
6.
MATTER AND MEASUREMENT
Classify the following as an element, a compound, homogeneous or heterogenous mixture.
a)
A gold chain - homogeneous
b)
Liquid eye drops - homogeneous
c)
Chunky peanut butter- heterogeneous
d)
Cooking oil - homogeneous
e)
Italian salad dressing - heterogeneous
f)
Glass window- homogeneous
g)
Distilled water -compound
h)
Steel (an alloy of several metals) - homogeneous
i)
Helium – element
j)
Mud - heterogeneous
k)
Carbon dioxide -compound
Identify the following properties as intensive or extensive.
a)
Mass – extensive
e)
Melting point– intensive
b)
Density – intensive
f)
Length – extensive
c)
Color – intensive
g)
Temperature– extensive
d)
Volume– extensive
Classify each of the following statements as CORRECT or FALSE.
a)
Every substance is a compound
FALSE
b)
All mixtures are homogeneous
FALSE
c)
Every mixture contains two or more substances
CORRECT
d)
Every compound contains two or more elements
CORRECT
e)
Every mixture contains two or more compounds
FALSE
The world’s tallest waterfall is Angel Falls. It is 3212 ft high. What is the height of Angel
Falls:
[1 mi = 5280 ft = 1.609 km]
a)
in miles?
3212 ft x 1 miles = 0. 6083 miles
5280 ft
b)
in km?
3212 ft x 1.609 km = 0.9788 km
5280 ft
c)
in m?
3212 ft x 1.609 km x 1000 m = 978.8 m
5280 ft
1 km
CHEMISTRY DEPARTMENT
2
SHE 1315
7.
MATTER AND MEASUREMENT
A slow jogger runs a mile in 13 minutes. Calculate the speed in:
[1 mi = 1609 m ; 1 in = 2.54 cm]
a)
in miles?
1 mile x 1609 m x 100 cm x 1in
13 min 1 mile
1m
2.54 cm
b)
x 1min
60 s
=
81. 2134 in/s
in km?
1 mile x 1609 m = 123. 7692 m/min
13 min 1 mile
c)
in m?
1 mile x 1609 m x 1 km x
13 min 1 mile 1000 m
8.
=
7.4262 km/hr
Liquid nitrogen is obtained from liquefied air. It is used in preparation of frozen goods and in
low-temperature research. The density of the liquid at its boiling point (-196oC or 77 K) is
0.808 g/cm3. Convert the density to units of kg/m3.
0.808 g X 1 kg X 100 cm3
cm3 1000 g 1 m3
9.
60 min
1 hr
= 0.0808 kg/m3
A person’s average daily intake of glucose is 0.0833 lb. Calculate the mass in mg?
[1 lb = 453.6 g]
0.0833 lb X 453.6 g X 453.6 g X 1000 mg = 3.778 X 104 mg
1 lb
1 lb
1g
CHEMISTRY DEPARTMENT
3
SHE 1315
10.
MATTER AND MEASUREMENT
Convert the average speed of H2 molecules at 25oC (1.70 x 103 m/s) to miles/hr, keeping
track of significant figures.
[1 mi = 1.6093 km]
1.70 X 103 m X 1 km X 1 mi
X 60 s X 60 m = 3.778 X103 mi/hr
s
1000 m 1.6093 km
1m
1 h
11.
Carry out the following operations and give your answers to the correct number of
significant figures.
a)
(12.37-11.07)3 X (16.58) ÷ ((3.5575 + 3.520)
5.15
b)
(892 ÷ 986.7) + 5.44
6.34
c)
(78.7 X 105 ÷ 88.529) + 5.44
8.89 X 104
d)
(78.4 – 44.889) ÷ 0.0087
3.9 X 103
e)
(9.2 x 10‐4) (2.1 x 104)
1.9 x 101
f)
(12.06 – 11.84)
0.271
0.81
CHEMISTRY DEPARTMENT
4
SHE 1315
ATOMS, MOLECULES AND IONS
TOPIC 2: ATOMS, MOLECULES AND IONS
1.
Argon has three naturally occurring isotopes,
a)
Ar,
Ar, and
38
Ar.
40
What is the mass number of each isotope?
Ar
36
Ar
38
36
38
Ar
40
40
b)
36
How many protons, neutrons and electrons are present in each?
protons
neutrons
electrons
Ar
18
18
18
Ar
18
20
18
Ar
18
22
18
36
38
40
A
2.
Write the
a)
Z
X notation for each atomic depiction :
18p+ ,
20n0
18 e-
38
18Ar
b)
25p+ ,
30n0
55
25e-
Mn
25
47p+ , 62n0
c)
47e-
109
Ag
47
CHEMISTRY DEPARTMENT
5
SHE 1315
3.
4.
ATOMS, MOLECULES AND IONS
Define:
a)
Mole
b)
Relative atomic mass unit
c)
Atomic mass
d)
Formula mass
e)
Molecular mass
A mole is defined as the amount of substance which contains equal
number of particles (atoms / molecules / ions) as there re atoms in
exactly 12.000g of carbon-12.
A mass exactly equal to 1/12 the mass of a carbon-12 atom.
The average of the masses of the naturally occurring isotopes
of an element weighted according to their abundances.
The sum (in amu) of the atomic masses of a formula unit of a
compound.
The sum (in amu) of the atomic masses of the elements in a
molecule of a compound.
Chlorine has two naturally occurring isotopes, 35Cl (isotopic mass 34.968 amu, abundance
75.53%) and 37Cl (isotopic mass 36.95 amu , abundance 24.47%). Calculate the atomic
mass of chlorine.
Atomic mass
of Chlorine = (34.968 amu)(75.53%) + (36.956 amu)(24.47%)
(100%)
(100%)
= 35.45 amu
5.
Li has two naturally occurring isotopes, 6Li (isotopic mass 6.0151 amu) and 7Li (isotopic
mass 7.0160 amu). Calculate the percent abundance of each isotope if the atomic mass of
lithium is 6.941 amu.
Average atomic
Mass of Li
: 6.941 amu = x(6.0151 amu) + (1-x)(7.0160 amu)
6.941 = -1.0009x + 7.0160
1.0009x = 0.075
X= 0.075
So
Li = 7.5%
Li = 92.5%
6
7
CHEMISTRY DEPARTMENT
6
SHE 1315
6.
ATOMS, MOLECULES AND IONS
An oxide of nitrogen contains 30.45% nitrogen according to mass. When the substance is
vaporized and analyzed in mass spectrometer, the following spectrum is found:
R
Q
Percent
abundance
S
P
14 16
30
46
Identify the species of peaks P, Q, R and S.
P = N+ ; Q = O+ ; R = NO+ ;
S =
7.
m/e
NO2+
The table below gives the m/e values and the peak size of a sample atom X.
Use periodic table to answer the following questions:
m/e
82
83
84
85
86
Peak size (%)
12
12
58
0
18
a)
Calculate the atomic mass of X, in amu.
X= (12x82) + (12x83) + (58x84) + (18x86)
100
= 84
8.
b)
Based on your answer to the question above, suggest the name for atom X.
Krypton
a)
How many grams of NaHCO3 in 0.0626 mol?
Mass of NaHCO3 = 22.99 + 1.008 + 12.01 + (3 x 16) = 84 g
Molar mass = 84 g/mol
0.0626 mol NaHCO3 x 84 g NaHCO3 = 5.26 g NaHCO3
1 mole NaHCO3
b)
Calculate the number of molecules in 28.0 g of NaHCO3?
28 g NaHCO3 x 1 mole NaHCO3 x 6.022 x 1023 molecules
84 g NaHCO3
1 mole NaHCO3
=
2.007 x 1023 molecules
CHEMISTRY DEPARTMENT
7
SHE 1315
9.
a)
ATOMS, MOLECULES AND IONS
Calculate the molecular mass and molar mass of sucrose, C12H22O11.
Molecular mass : (12 x 12.01) + (22 x 1.008) + (11 x 16.00)
= 342.296 g
Molar mass : 342.296 g/mol
b)
By using Factor Label Method/Dimensional Analysis, calculate the following number
of atoms or molecules in 171 g of C12H22O11.
(i) C atoms
171 g C12H22O11 x
1 mole C12H22O11 x
342.296 g C12H22O11
12 mole atoms C
1 mole C12H22O11
x 6.022 x 1023 atoms C = 3.6132 x 1024 atoms C
1 mole atom C
(ii) H atoms
171 g C12H22O11 x
1 mole C12H22O11 x
342.296 g C12H22O11
22 mole atoms H
1 mole C12H22O11
x 6.022 x 1023 atoms H = 6.6242 x 1024 atoms H
1 mole atom H
(iii) C12H22O11 molecules
171 g C12H22O11 x 1 mole C12H22O11 x 6.022 x 1023 molecules
342.296 g C12H22O11
1 mole C12H22O11
= 3.011 x 1023 molecules C12H22O11
10.
Calculate the mass, in g for 6.022 X 1023 molecules of N2.
6.022 X 1023 N2 molecules x
1 mole N2
x
28 g = 28 g
6.022 X 1023 N2 molecules
1 mole N2
CHEMISTRY DEPARTMENT
8
SHE 1315
11.
Calculate the number of water molecules in a drop of water containing 0.030 mol water.
0.03
12.
ATOMS, MOLECULES AND IONS
moles H2O x 6.022 X 1023 H2O molecules = 1.8066 x 1022 H2O
1 mol H2O
molecules
Calculate the mass of nitrogen, N2 which contains the same number of molecules as in 4.4 g
of carbon dioxide, CO2.
4.4g CO2 x 1 mole CO2 x
44 g
13.
a)
28 g N2 = 2.8 g N2
1 mole N2
Given 27.5 g of MgCl2, calculate the number of:
(i)
cations
27.5 g MgCl2 x 1 mole MgCl2 x 1 mole Mg2+ x 6.022 X 1023 Mg2+ ions
95.2 g
1 mole MgCl2
1 mole Mg2+
= 1.7395 x 1023 Mg2+
(ii)
anions
27.5 g MgCl2 x 1 mole MgCl2 x 2 moles Cl+ x 6.022 X 1023 Cl+ ions
95.2 g
1 mole MgCl2
1 mole Cl+
= 3.4790 x 1023 Cl+
(iii) ions
27.5 g MgCl2 x 1 mole MgCl2 x 3 moles ions x 6.022 X 1023 ions
95.2 g
1 mole MgCl2
1 mole ions
= 5.2185 x 1023 ions
CHEMISTRY DEPARTMENT
9
SHE 1315
14.
ATOMS, MOLECULES AND IONS
Calculate the percent by mass of each element in sodium chloride, NaCl.
Mass = 22.99 + 35.45 = 58.44 g
%Na = (22.99 ÷ 58.44) x 100 = 39.34%
%Cl = (35.45 ÷ 58.44) x 100 = 60.66%
15.
Calculate the percent by mass of each element present in ammonium phosphate, (NH4)3PO4.
Mass ammonium phosphate [(NH4)3PO4] = 149.096 g
%N = (42.03 ÷ 149.096) x 100 = 28.19%
%H = (12.096 ÷ 149.096) x 100 = 8.11%
%P = (30.97 ÷ 149.096) x 100 = 20.77%
%O = (64.00 ÷ 149.096) x 100 = 42.93%
16.
A compound with a mass of 48.72 g is found to contain 32.69 g of zinc, Zn and 16.03 g of
sulphur, S. Calculate the percentage composition of each element in the compound.
% Zn =
=
=
%S
=
=
=
17.
mass of Zn
Total mass
32.69 g
48.72 g
67.1 %
X
100 %
X
100 %
mass of S
mass total
16.03 g
48.72 g
32.90 %
X
100 %
X
100 %
Sulphuric acid, H2SO4 is an important acid in laboratories and industries. Determine the
percentage composition of each element in sulphuric acid.
Mass of H2SO4 = 98.06 g
% H = (2 x 1.008) g x 100 % = 2%
98.06 g
% S = 32.06 g x 100 % = 33%
98.06 g
% O = (4 x 16.0) g x 100% = 65 %
98.06 g
CHEMISTRY DEPARTMENT
10
SHE 1315
18.
19.
20.
21.
ATOMS, MOLECULES AND IONS
a)
Calculate the percent by mass of aluminum in aluminum oxide, Al2O3.
Mass of Al in Al2O3 = 26.98g x 2 = 53.960g
% Al in Al2O3 = 53.960g/ 101.96 g x 100% = 52.9 %
b)
Calculate the mass of aluminum in 28.0 g of aluminum oxide, Al2O3.
52.9 % x 28 g = 14. 818 g
Define:
a)
Empirical formula
Empirical formula is a chemical formula that shows the simplest ratio of all
elements in a molecule.
b)
Molecular formula
Molecular formula is a formula that shows the actual number of
of each element in a molecule
atoms
State the empirical formulae for the following molecular formulae.
a)
C8H18 :C4H9
c)
Na2C2O4 :NaCO2
b)
H2O2
d)
C7H12 :C7H12
:HO
Determine the empirical formula for a sample containing 25.9% N and 74.1% O.
Element
N
O
25.9
74.1
25.9 / 14.01 =1.85
74.1/16.00
=46.3
1.85/1.85
=1 x 2 =2
46.3/1.85
=2.5 x 2 =5
Mass (g)
Moles (mol)
Mole ratio
N2O5
Empirical Formula
CHEMISTRY DEPARTMENT
11
SHE 1315
22.
ATOMS, MOLECULES AND IONS
A
compound
is
found
to contain
47.25%
Calculate the empirical formula for this compound.
element
copper
and
Cu
Cl
47.25
52.75
47.25/63.6
52.75/35.5
= 0.7355
= 1.4880
0.7355/0.7355
1.4880/0.7355
=1
=2.001 ≈2
52.75%
chlorine.
Mass (g)
Moles (mol)
Mole ratio
CuCl2
Empirical Formula
23.
A compound with a molecular mass of 34.0 g/mol is known to contain 5.88% hydrogen and
94.12% oxygen. Determine the molecular formula for the compound.
element
H
O
5.88
94.12
5.88/1.008
94.12/16.00
= 5.8333
= 5.8825
5.8333/5.8333
5.8825/5.8333
=1
=1.008 ≈1
Mass (g)
Moles (mol)
Mole ratio
HO
Empirical Formula
Empirical formula mass: 1.0 + 16.0 = 17.0 g/mol
Molecular Mass = n x empirical formula mass
34.0 = n x 17.0
n = 34.0 ÷ 17.0 = 2
Molecular Formula: 2 x (HO) = H 2O2
CHEMISTRY DEPARTMENT
12
SHE 1315
24.
ATOMS, MOLECULES AND IONS
A substance is decomposed and found to consist of 53.2% C, 11.2% H, and 35.6% O by
mass. Calculate the molecular formula of the unknown if its molar mass is 90.0 g/mol.
Assume 100 g total.
Thus: 53.2 g C, 11.2 g H, and 35.6 g O
Element
Mass (g)
Moles (mol)
Mole ratio
Empirical Formula
C
53.2
53.2/12.01
=4.430
4.430/2.225
=1.99  2
C2H5O
H
11.2
11.2/1.01
=11.09
11.09/2.225
=4.98 5
O
35.6
35.6/16.00
=2.225
2.225/2.225
=1
Calculate the empirical formula mass: 45.1 g/mol
Molecular Mass = n x empirical formula mass
90.0 = n x 45.1
n=2
Molecular Formula: 2 x (C2H5O) = C4H10O2
25.
In an experiment, 17.471 g of trioxane is burned. 10.477 g H2O and 25.612 g CO2 are
formed from the combustion. The molecular mass of trioxane is found to be 90.079. What is
molecular formula of trioxane?
? g O = 17.471 g trioxane - 6.9899 g C - 1.1724 g H = 9.3093 g O
Element
Mass (g)
Moles (mol)
Mole ratio
Empirical Formula
C
6.9893
6.9893/12.01
=0.5820
0.5820/0.5818
=1.0003 1
CH2O
H
1.1724
1.1724/1.008
=1.1631
1.1631/0.5818
=1.999 2
O
9.3093
9.3093/16.00
=0.5818
0.5818/0.5818
=1
Empirical formula mass = 1(12.011) + 2(1.00794) + 1(15.9994)
= 30.026
molecular formula= C3H6O3
CHEMISTRY DEPARTMENT
13
SHE 1315
26.
ATOMS, MOLECULES AND IONS
Combustion of 14.765 g of Dianabol produced 43.257 g CO2 and 12.395 g H2O. Determine
the molecular formula for Dianabol, if the molar mass is 300.44 g/mol.
? g O = 14.765 g total - 11.805 g C - 1.3870 g H = 1.573 g O
element
C
H
O
11.805
1.3870
1.573
s (g)
Moles (mol)
Mole ratio
11.805/12.01
1.3870/1.008
1.573/16.00
=0.9829
= 1.3760
=2.225
0.9829 /0.09832
1.3760/0.09832
0.09832/0.09832
=9.9969 ≈10
=13.9951≈14
=1
C10H14O
Empirical Formula
Empirical formula mass = 10(12.011) + 14(1.00794) + 1(15.9994)
= 150.22
molecular formula = C20H28O2
CHEMISTRY DEPARTMENT
14
SHE 1315
REACTIONS AND STOICHIOMETRY
TOPIC 3: REACTIONS AND STOICHIOMETRY
1.
Define:
a)
Acid-base, according to Arrhenius and Bronsted-Lowry theory
According to Arrhenius:
Acid is a substance that has H in its formula and dissociates in water to
yield H3O+.
Base is a substance that has OH in its formula and produces OH - in water.
According to Bronsted-Lowry Theory:
Acid is a proton donor, any species that donates an H + ion.
Base is a proton acceptor, any species that accepts an H+ ion.
b)
Redox reaction, in terms of electron transfer and oxidation number
Oxidation is a loss of electrons by a substance.
Reduction is a gain of electrons by a substance.
Oxidation occurs when there is an increase in oxidation number.
Reduction occurs when there is a decrease in oxidation number.
c)
Precipitation reaction
a reaction in which two soluble ionic compounds form an insoluble product,
a precipitate
2.
a)
List three examples for each strong and weak acids.
Strong acids: HCl, HClO4 and HNO3
Weak acids: HF, H3PO4 and CH3COOH
b)
List three examples of strong bases from group 2A and one example of weak base.
Strong bases: Ca(OH)2, Sr(OH)2 and Ba(OH)2
Weak base: NH3
3.
Balance the following equations:
a)
N2O5 (g) +
H2O (l) 
HNO3 (aq)
N2O5 (g) + H2O (l)  2HNO3 (aq)
b)
NH4NO3 (s)  N2 (g) +
O2 (g) + H2O (g)
2NH4NO3 (s)  2N2 (g) + O2 (g) + 4H2O (g)
c)
CH3NH2
(g)
4CH3NH2
+
(g)
O2
(g)
+ 9O2

(g)
CO2
 4CO2
(g)
+
H2O
(g)
+ 10H2O
(g)
(g)
+
N2
+ 2N2
(g)
(g)
CHEMISTRY DEPARTMENT
15
SHE 1315
4.
REACTIONS AND STOICHIOMETRY
Write the net ionic equations and determine the spectator ions for the following reactions :
a)
Potassium phosphate solution is mixed with calcium nitrate solution.
Molecular equation
2 K3PO4 (aq) + 3 Ca(NO3)2 (aq)  6KNO3 (aq) + Ca3(PO4)2
Ionic equation
6K+ (aq) + 2PO43-(aq)+ 3Ca2+(aq)+ 6NO3-(aq)  6K+
Net ionic equation
2PO43-(aq) + 3Ca2+(aq)  Ca3(PO4)2
Spectator ions: K+ and NO3-.
b)
(aq)
(s)
+ 6NO3-(aq) + Ca3(PO4)2
(s)
(s)
Mixing Al(NO3)3 solution with NaOH solution.
Molecular equation
Al(NO3)3 (aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaNO3(aq)
Ionic equation
Al3+(aq) + 3NO3-(aq) + 3Na+(aq) + 3OH-(aq)  Al(OH)3(s) + 3Na+(aq) + 3NO3-(aq)
Net ionic equation
Al3+(aq) + 3OH-(aq)  Al(OH)3(s)
Spectator ions: NO3- and Na+
5.
For the following complete redox reactions:
a)
b)
write the half equations
identify the oxidizing and reducing agents
(i)
2 Sr + O2  2 SrO
(a)
(ii)
(b) Oxidizing agent: O2
Reducing agent: Sr
2 Li + H2  2 LiH
(a)
(iii)
Oxidation: 2 Sr  2 Sr2+ + 4 eReduction: O2 + 4 e-  2 O2-
Oxidation: 2 Li  2 Li+ + 2 eReduction: H2 + 2 e-  2 H-
(b) Oxidizing agent: H2
Reducing agent: Li
3 Mg + N2  Mg3N2
(a)
Oxidation: 3 Mg  3 Mg2+ + 6 eReduction: N2 + 6 e-  2 N3-
(b) Oxidizing agent: N2
Reducing agent: Mg
CHEMISTRY DEPARTMENT
16
SHE 1315
6.
REACTIONS AND STOICHIOMETRY
Write a balanced chemical equation for each reaction and identify whether precipitation
occur.
a)
Ni(NO3)2 and NaOH
Ni(NO3)2(aq) + 2 NaOH(aq)  Ni(OH)2 (s) + 2 NaNO3 (aq)
b)
NaOH and K2SO4
No precipitation
c)
Na2S and Cu(C2H3O2)2
Na2S(aq) + Cu(C2H3O2)2(aq)  2 Na(C2H3O2)2(aq) + CuS(s)
7.
8.
9.
Identify the type of reaction for the following reactions.
a)
Cu(OH)2 (s) + 2 HNO3
b)
Fe2O3
c)
Sr(NO3)2
d)
4 Zn(s) + 10 H+(aq) + 2 NO3-(aq)  4 Zn2+(aq) + N2O(g) + 5H2O
(s)
+ 3 CO
(aq)
(g)
(aq)
 Cu(NO3)2
 2 Fe
+ H2SO4
(aq)
(s)
(aq)
+ 3 CO2
 SrSO4
(s)
+ 2 H2O
Acid-base
(l)
Redox
(g)
+ 2 HNO3
Precipitation
(aq)
Redox
(l)
Using solubility guidelines, predict whether each of the following compounds is soluble or
insoluble in water.
(a)
NiCl2
soluble
(c)
Cs3PO4
soluble
(b)
Ag2S
insoluble
(d)
SrCO3
insoluble
An antacid tablet containing 0.70 g NaHCO3 is dissolved in water and the total volume of
solution is 250 mL. Calculate the molarity of this NaHCO3 solution.
Mol NaHCO3 = 0.70 g
= 0.00833 mol NaHCO3
84.01 g mol-1 NaHCO3
M=
mol
Volume
= 0.00833 mol = 0.0333 M
0.250 L
CHEMISTRY DEPARTMENT
17
SHE 1315
10.
REACTIONS AND STOICHIOMETRY
In a laboratory, 6.58 g of barium nitrate Ba(NO3)2 (molar mass = 261.36) is dissolved in
enough water to form 0.850 L of solution. A 0.100 L sample is transferred from this stock
and titrated with 0.051 M solution of K2SO4. What volume of K2SO4 solution is required to
form BaSO4 precipitate?
Ba(NO3)2 + K2SO4
BaSO4 + 2 KNO3
Mol Ba(NO3)2 =
6.58 g
261.36 g mol-1
= 0.0252 mol
M Ba(NO3)2 = 0.0252 mol = 0.0296 M
0.850 L
MaVa = a
MbVb
b
= 0.100 L x 0.0296 M = 0.051 M x Vb
Vb = 0.0580L
11.
54.7 g of propane, C3H8 react with 89.6 g of oxygen, O2 as shown in the equation below:
C3H8 + 5 O2
a)
3 CO2 + 4 H2O
Define limiting reactant.
Reactant that limits the amount of the other reactant that can react, and
thus the amount of product that can form/ Reactant that yields the lower
amount of product.
b)
Determine the limiting reactant.
54.7 g C3H8
x 1 mol C3H8 x 3 mol CO2 x 44.01 g CO2
44.094 g C3H8 1 mol C3H8
1 mol CO2
= 163.787gCO2
89.6 g O2 x 1 mol O2 x 3 mol CO2 x 44.01 g CO2
32.00 g O2 5 mol O2
1 mol CO2
= 73.937 g CO2
O2 is the limiting reactant. O2 form fewer products CO2.
c)
Calculate the mass of CO2 produced, in grams.
Theoretical yield O2 = 73.937 g CO2
CHEMISTRY DEPARTMENT
18
SHE 1315
12.
REACTIONS AND STOICHIOMETRY
A reaction has a theoretical yield of 124.3 g SF 6, but only 113.7 g SF6 are obtained in the
lab. What is the percent yield of SF6 for this reaction?
% yield of SF6 = actual yield/theoretical yield x 100
= 113.7 g/ 124.3 g x 100
= 91.47%
13.
The combustion of methane produces carbon dioxide and water. Assume that 2.00 mol of
CH4 burned in the presence of excess air. Calculate the percentage yield if the reaction
produces 87.0 g of CO2.
CH4 + 2 O2
CO2 + 2 H2O
Theoretical yield = 2 mol CH4 x 1 mol CO2 x 44 g CO2 = 88g CO2
1mol CH4
% yield = 87.0g/88.0g x 100 = 99%
14.
70.0 g of manganese dioxide, MnO2 is mixed with 3.50 mol of hydrochloric acid. How many
grams of Cl2 will be produced from this reaction if the % yield for the process is 42.0%?
MnO2 + 4 HCl  MnCl2 + 2 H2O + Cl2
70 g MnO2 x 1 mol MnO2 x 1 mol Cl2 x 70.9 g Cl2 = 57.085 g
86.94 g mol-1 1 mol MnO
1 mol Cl2
3.5 mol HCl x 1 mol Cl2 x 70.9 g Cl2 = 62.038 g
4 mol HCl 1 mol Cl2
MnO2 is the limiting reactant because it forms less Cl 2.
% yield = actual yield / theoretical yield x 100
Actual yield = 57.085 = 23.98 g Cl2
42%
CHEMISTRY DEPARTMENT
19
SHE 1315
15.
REACTIONS AND STOICHIOMETRY
Determine the mass of sodium carbonate, Na2CO3 present in 50.00 mL of 0.750 M solution.
Mol = 0.750 M x 0.050 L = 0.0375 mol
Mass = 106.0 g mol-1 x 0.0375 mol = 3.975 g
16.
Calculate the volume of 1.50 mol/L HCl solution in 10.0 g of hydrogen chloride.
M = mol__
Volume
mol = 10.0 g
= 0.27397 mol
36.5 g mol-1
Volume = 0.27397 mol = 0.183 L
1.50 mol/L
17.
Complete the following table by giving the definition for each concentration terms.
Concentration term
Definition
Amount (Mol) of solute
Molarity (M)
Volume (L) of solution
Amount (mol) of solute
Molality (m)
Mass (kg) of solvent
Mass of solute
Percentage by mass
(%w/w)
Mass of solution
Volume of solute
Percentage by volume
(%v/v)
Volume of solution
CHEMISTRY DEPARTMENT
20
SHE 1315
GASES AND KINETIC MOLECULAR THEORY
TOPIC 4: GASES AND KINETIC MOLECULAR THEORY
1.
a)
State the Avogadro’s Law.
At fixed temperature and pressure, equal volumes of
any ideal gas contain equal no. of particles.
b)
Consider a 1.0 L flask containing neon gas and a 1.5 L flask containing xenon gas.
Both gases are at the same pressure and temperature. According to Avogadro’s Law,
what can be said about the ratio of the number of atoms in the two flasks?
V α n (T & P constant)
Since volume of Ne : Xe = 1 : 1.5
So, no. of atoms Ne : Xe = 1: 1.5
2.
Calculate the total mass (grams) of oxygen in a room measuring, 4.0 m x 5.0 m x 2.5 m.
Assume that the gas is at STP and the air contains 20.95 % oxygen by volume.
[1m3 = 1000 dm3]
PV = nRT
; P= 1 atm
T= 273 K
V room= 4.0 m x 5.0 m x 2.5 m = 50 m 3
V O2 = 20.95%x 50 m3 = 10.475 m3
n= PV/RT
=
1 atm (10475 L)
0.0821 Latm mol-1 K-1 (273 K)
= 467.36 mol
Mass O2 = 467.36 mol x 32.00 g = 1.5 x 104 g
1 mol
3.
One mole of any gas at STP has a volume of 22.4 L. Calculate the densities of the following
gases, in grams per liter at STP.
a)
CH4
d= 16.05 gmol-1/22.4 Lmol-1 = 0.717 g/L
b)
CO2
d= 44.01 gmol-1/22.4 Lmol-1 = 1.96 g/L
c)
UF6
d= 352.03 gmol-1/22.4 Lmol-1 = 15.7 g/L
CHEMISTRY DEPARTMENT
21
SHE 1315
4.
GASES AND KINETIC MOLECULAR THEORY
A 0.98478 g of unknown gas is placed in a 1.500 L bulb at a pressure of 356 mmHg and a
temperature of 22.5 0C. Calculate the molar mass of the gas.
V= 1.500 L
P= 356 mmHg x 1 atm = 0.468 atm
760 mmHg
T= 295.5 K
m= 0.9847 g
M= mRT/PV
= 0.9847 g (0.0821 L atm mol-1 K-1)(295.5 K)
0.468 atm (1.500 L)
= 34.03 g/mol
5.
Hydrogen gas can be prepared by a reaction of zinc metal with aqueous HCl.
Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
a)
Calculate how many liters of H2 would be formed at 742 mmHg and 15.0 0C, if 25.5 g
of zinc was allowed to react.
P= 742 mmHg x 1 atm = 0.976 atm
760 mmHg
T= 15 0C + 273 = 288 K
n Zn= 25.5 g x 1 mol = 0.390 mol Zn
65.39 g
From the balanced equation, 1 mol Zn reacts to produce 1 mol H 2.
Therefore, 0.390 mol Zn reacts to produce 0.390 mol H 2.
V H2= nRT/P
= 0.390 mol (0.0821 L atm mol-1 K-1) (288 K)
0.976 atm
b)
= 9.45 L
Calculate how many grams of zinc would you need to produce 5.00 L of H2 at 350
mmHg and 30.0 0C.
V H2=5.00 L
P = 350 mmHg x
1 atm = 0.461 atm
760 mmHg
T= 30.0 0C + 273 = 303.0 K
n H2 = PV/RT
= 0.461 atm (5.00 L)
(0.0821 L atm mol-1 K-1)(303K)
= 0.0926 mol H2
From the balanced equation, 1 mol H2 is produced from the reaction of 1
molZn.
Therefore, 0.0926 mol H2 is produced from the reaction of 0.0926 mol Zn.
Mass Zn = 0.0926 mol x 65.39 g = 6.05 g Zn
1 mol
CHEMISTRY DEPARTMENT
22
SHE 1315
6.
a)
GASES AND KINETIC MOLECULAR THEORY
State Dalton’s Law of partial pressure.
In a mixture of gases, the total pressure of the mixture is the sum of the
partial pressure of the constituent gases.
PT = PA + PB + PC +…
Where PA, PB , PC = partial pressuere of A, B and C.
PT =total pressure of the mixture.
b)
What do you understand by the term partial pressure?
Is the pressure that each gas would exert if it alone occupy the container
under the same temperature
7.
8.
Complete the following table for an ideal gas:
P
V
n
T
2.00 atm
1.00 L
0.500 mol
48.7 K
0.300 atm
0.250 L
3.05 x 10-3 mol
27 0C
650 torr
11.2 dm3
0.333 mol
350 K
1.04 x 106 Pa
585 mL
0.250 mol
295 K
Indicate which of the following statements regarding the kinetic-molecular theory of gases
are correct.
a)
The gas molecules are assumed to exert no forces on each other.
Correct
b)
All the molecules of a gas at a given temperature have the same kinetic energy.
Incorrect. The molecules have the same average kinetic energy at constant
temperature
c)
The volume of the gas molecules is negligible in comparison to the total volume in
which the gas is contained.
Correct
CHEMISTRY DEPARTMENT
23
SHE 1315
GASES AND KINETIC MOLECULAR THEORY
9.
The stopcock between the two containers is opened and the gases are allowed to mix.
[Assume no reaction between N2 and O2]
a)
Calculate the partial pressures of N2 and O2 after mixing.
P N2= 1.0 atm (2.0L)= 0.4 atm
5.0 L
P O2= 2.0 atm (3.0L)= 1.2 atm
5.0 L
b)
Calculate the total pressure in the container after the gases mix.
PT = PN2 + PO2
= 0.4 atm + 1.2 atm = 1.6 atm
c)
Determine the mole fraction of each gas.
X N2 = 0.4/1.6
= 0.25
X O2 = 1.2/1.6
= 0.75
CHEMISTRY DEPARTMENT
24
SHE 1315
10.
GASES AND KINETIC MOLECULAR THEORY
A closed bulb contains 0.0100 mol of inert helium gas and a solid white ammonium chloride,
NH4Cl. Assume that the volume of the solid NH4Cl is negligible compared to the volume of
the bulb. The pressure of the helium gas at 27.0 °C is found to be 114 mmHg. The bulb is
then heated at 327 °C. All the NH4Cl decomposes according to the equation:
NH4Cl (s)  NH3 (g) + HCl (g)
The final total pressure in the bulb after complete decomposition of the solid is 908 mmHg.
a)
Calculate the pressure of helium gas at 327 °C.
P1 = 114 mmHg
T1 = 27 + 273 K
= 300 K
P2 = P1 x
T2
T1
= 114 mmHg
b)
P2 = ?
T2 = 327 + 273 K
= 600 K
x
600 K =
300 K
228 mmHg
What is the partial pressure of HCl (g)?
From the eq:
Mole of NH3 = Mole of HCl
Therefore:
P (NH3) = P (HCl) = x
Then:
Ptotal = P (He) + P (NH3) + P (HCl)
908 = 228
+ x
+ x
0r
2x = 908 – 228 mmHg
= 680 mmHg
Hence:
x = 680 mmHg = 340 mmHg
2
c)
Calculate the number moles of HCl (g) produced in the reaction.
Since V and T are the same for all gases (He & HCl)
Hence:
V = n (HCl) = n (He)
RT
P (HCl)
P (He)
Therefore:
n (HCl) = n (He) x P (HCl)
P (He)
= 0.01 mol x 340 mmHg = 0.0149 mol
228 mmHg
d)
How many grams of NH4Cl (s) were in the bulb at 27.0 °C?
Since all of NH4Cl was converted to HCl & NH3,
therefore:
n (NH4Cl) = n (HCl)
= 0.0149 mol
Hence:
Mass of
= 0.0149 mol x 53.5 g/mol = 0.798 g
NH4Cl
CHEMISTRY DEPARTMENT
25
SHE 1315
11.
GASES AND KINETIC MOLECULAR THEORY
A 0.705 g mixture of KClO3 and a catalyst was placed in a tube and heated vigorously to
drive off all the oxygen as O2. The O2 was collected over water at 27.0 0C. The volume of
gas collected was 150.0 mL and a pressure of 610 torr.
[P H2O at 27.0 0C = 22.4 torr]
a)
How many moles of O2 were produced?
Pt = PO2 + PH2O
PO2 = 610 – 22.4 = 587.6 torr = 0.773 atm
nO2 = PV/RT =
0.773 atm x 0.150 L
0.0821 atmLmol-1K-1 x 300 K
= 4.71 x 10-3 mol
b)
How many moles of KClO3 were in the original mixture?
2KClO3 
2KCl + 3O2

From eqn.,
3 mol O2  2 mol KClO3
thus 4.71 x 10-3 mol O2  2/3 x 4.71 x 10-3
= 3.14 x 10-3 mol KClO3
c)
What was the mass percent of KClO3 in the original mixture?
Mass of KClO3 = 122.5 x 3.14 x 10-3
= 0.385 g KClO3
Mass % = 0.385 / 0.705 x 100 %
= 54.6 %
12.
State the conditions needed for a gas to behave most ideally.
High temperature and low pressure
13.
Under the same conditions of temperature and pressure, which of the following gases would
behave most ideally? Explain.
Ne
N2
CH4
CH4. All molecules are non-polar. Since CH4 has the lowest molecular mass, so the
attractive forces between CH4 molecules (London dispersion forces)are very weak
compared to the others.
CHEMISTRY DEPARTMENT
26
SHE 1315
ATOMIC STRUCTURE
TOPIC 5: ATOMIC STRUCTURE
1.
Draw the following orbital with the correct orientation:
a)
1s
c)
2py
y
x
x
y
z
b)
2s
d)
2px
y
z
x
x
y
2.
Draw an orbital diagram showing valence electrons and write the condensed ground-state
electron configuration for:
a)
Ba
,
[xe] 6s2
↑↓
6s
b)
Co
↑↓
↑↓
↓
4s
c)
3.
Ag
↑↓
↑
↑
↑↓
↑↓
↑
,
[Ar] 4s2 3d7
3d
↑
↑↓
5s
↑
↑↓
↑↓
, [Kr] 5s1 4d10
4d
How many orbitals in an atom can have the following designations?
a)
5f
Seven
c)
5d
Five
b)
4p
Three
d)
n=2
Four
CHEMISTRY DEPARTMENT
27
SHE 1315
4.
5.
ATOMIC STRUCTURE
Give all possible ml values for orbital that have each of the following :
a)
l=3
ml= -3,-2,-1,0,+1,+2,+3
b)
n=2
c)
n=1; l=1
l=0 , ml=0
l=1 , ml= -1,0,+1
ml=-1,0,+1
Write a full set of quantum numbers (n, l, ml, ms) for:
a)
the outermost electron in a Li atom
n=2 ; l=0 ; ml=0 ; ms=+1/2
b)
the electron gained when a Br atom becomes a Br- ion
n=4 ; l=1 ; ml=+1
c)
;
ms=+1/2
the highest energy electron in the ground- state B atom
n=2
6.
ms=-1/2
the electron lost when a Cs atom ionizes
n=6 ; l=0 ; ml=0
d)
;
;
l=1
;
ml=-1
;
ms=+1/2
Write the full electronic configuration for:
a)
Br
1s22s22p63s23p64s23d104p5
b)
Mg
1s22s22p63s2
c)
Se
1s22s23s23p64s23d104p4
d)
Cu
1s22s23s23p64s13d10
CHEMISTRY DEPARTMENT
28
SHE 1315
7.
8.
9.
10.
ATOMIC STRUCTURE
How many inner, outer and valence electrons are present in each of the following elements?
No.
Elements
Inner electrons
Outer electrons
Valence electrons
a)
Br
28
7
7
b)
Cs
54
1
1
c)
Cr
18
1
6
d)
Sr
36
2
2
e)
F
2
7
7
Identify each element below.
a)
[Ar ] 4s2 3d104p4
Se
b)
[Xe ] 6s2 4f145d2
Hf
c)
[Ar ] 4s2 3d5
Mn
For each of the following, state the sublevel designation, the ml values, and the number of
orbital:
sublevel
ml value
no. of orbital
a)
n=2 , l=0
s(l=0)
0
1
b)
n=3 , l=2
d(l=2)
-2,-1,0,+1,+2
5
c)
n=5 , l=1
p(l=1)
-1,0,+1
3
Each of the following represents the electron configurations of an atom after one electron
has been removed. Identify the element, and write its condensed ground-state electronic
configuration.
a)
b)
1s22s22p63s1
1s22s22p63s23p5
Mg
;
Ar
;
[Ne]3s2
[Ne]3s23p6
CHEMISTRY DEPARTMENT
29
SHE 1315
ATOMIC STRUCTURE
11.
Identify the types of orbital given below:
12.
Define:
a)
Atomic radii
The size of an atom as measured by the distance from the
nucleus to the boundary of the surrounding cloud of electrons.
13.
b)
Ionic radii
c)
Effective nuclear charge (Zeff)
The size of an ion as measured by the distance between the
nuclei of adjacent ions in a crystalline ionic compound.
The nuclear charge an electron actually experiences as a result
of shielding effects due to the presence of other electrons.
Briefly explain the general trend of atomic radii across the period, from left to right.
Across the period, Zeff is increasing, nucleus attraction is stronger, thus the
outermost electron in group 8A (18) is more tightly held than that in group 1A(1).
Group 1A(1) atoms are much larger than those of group 8A(18) atoms.
14.
Compare the sizes of cation and anion to their parent atoms.
Cation: has lost an electron, therefore electrostatic repulsion between electrons is
decreased and the size of the ion shrinks. The cation of an element is smaller than
the neutral atom of the same element.
Anion: has gained an electron, therefore electrostatic repulsion between electrons
is increased and the size of the ion increases. The anion of an element is larger
than the neutral atom of the same element.
CHEMISTRY DEPARTMENT
30
SHE 1315
15.
a)
ATOMIC STRUCTURE
Define isoelectronic species.
Atoms or ions with the same electronic configuration.
b)
Given Na+, Mg2+, Al3+, P3- , S2- and Cl- ;
(i)
List all isoelectronic ions.
Na+, Mg2+, Al3+ , Si4+ & P3- , S2- , Cl-
(ii)
Arrange the isoelectronic ions in the order of increasing size.
P3- > S2- > Cl- > Na+ > Mg2+ > Al3+> Si4+
(iii)
Explain your answer in b(ii).
All the cations have ionic radius smaller than the anions. This is
because the anions have higher principle quantum number for
the valence electron (n=3) compared to cations (n=2).
16.
a)
Define first ionization energy, IE1.
Ionization energy is the minimum energy needed to remove an
electron from a gaseous atom, usually expressed on a per mole
basis.
b)
Write a balanced chemical equation to represent the first ionization energy
process for element X.
X (g)  X+(g) + e¯
17.
Given sodium and potassium, which would have a larger first ionization energy? Explain.
Na has the larger 1st ionization energy because the valence electron is located in
the period 3 compared to K. Valence electron in K is located in period 4. Valence
electron in Na feels stronger nucleus attraction than in the K atom. Thus harder
to remove an electron from Na atom than in K atom.
18.
Explain the difference in first ionization energy between Beryllium, Be and Boron, B.
Across the period, the proton number is increasing. With the increase in
proton,
comes an increase in nuclear charge. The
nucleus of the B atom has a greater
positive charge, thus hold on more tightly to the electron located in the 2p
sublevel. Therefore, B requires more ionization energy to remove the first
electron in the 2p sublevel compared to Be.
CHEMISTRY DEPARTMENT
31
SHE 1315
ATOMIC STRUCTURE
19.
Chlorine, selenium and bromine are located near each other in the periodic table. Which of
these elements is the smallest atom and which has the highest ionization energy?
Chlorine has the smallest atomic radius and the highest ionization
energy
20.
The first four successive ionization energies of Period 3 element, A is shown below:
a)
IE
1
2
3
4
(kJmol-1)
576
1810
2740
11600
In which group does A appear in the periodic table? Explain.
3A (13). Because this is where the highest jump/significant jump in IE
occurred. Attempting to remove inner core electron requires larger amounts
of energy.
b)
Write the valence electronic configuration for element A.
3s23p1
21.
Five successive ionization energies (kJ mol-1) for atom M is shown below:
IE
1
2
3
4
5
(kJmol-1)
800
1580
3230
4360
16000
Determine:
a)
Valence electron configuration for M.
IE 2
1580

 1.98
IE1
800
IE 3
3230

 2.04
IE 2
1580
IE 4
4360

 1.34
IE 3
3230
IE 5
16000

 3.67
IE 4
4360
The first, second, third and fourth electron are removed from
the
same energy subshell. The fifth electron is removed from an inner
subshell, hence it requires a higher IE5 (3.67 times)
than IE4.
Since IE5 / IE4 have the highest ratio, 4 valence
electrons are
2
2
present. Valence electron configuration: ns np
b)
Group number of M in the periodic table.
Group 14
CHEMISTRY DEPARTMENT
32
SHE 1315
22.
a)
ATOMIC STRUCTURE
Define first electron affinity, EA1.
Electron affinity is is the amount of energy release or absorbed
when an electron added to a gaseous atom or ion.
b)
Write a balanced chemical equation to represent the first and second electron
affinity process for element Y.
Y (g) + e¯  Y-(g)
c)
Second electron affinity always has a positive value (an endothermic
process). Explain.
When an electron is added to a negative ion (anion), a strong
repulsion is felt, and the energy of the system increases. Thus,
heat is absorbed to overcome the electrostatic repulsion
between the electron and the negative charge on the
anion. This phenomenon leads to a positive value for 2nd
electron affinity.
CHEMISTRY DEPARTMENT
33
SHE 1315
CHEMICAL BONDING
TOPIC 6: CHEMICAL BONDING
1.
Use Lewis electron dot symbols to depict ions formed from the following atoms and predict
the chemical formula of their compound.
a) Na and F
2.
c) Sr and Br
d) K and O
c) O
d) O2-
c) BI3
d) SeF4
Write the Lewis dot symbols for:
a) Ca
3.
b) Ca and O
b) Ca2+
Write the Lewis structures for:
a) AsCl3
b) O3
CHEMISTRY DEPARTMENT
34
SHE 1315
4.
a)
CHEMICAL BONDING
What is a coordinate covalent bond?
Coordinate covalent bond : A covalent bond formed when a pair of electron
is contributed by only one of the bonded atoms
b)
State two requirements for the formation of a coordinate covalent bond.
Two requirements:
a) The donor atom must have a lone pair of electrons
b) The acceptor atom must have an empty orbital to accommodate the pair
of electrons.
5.
6.
Boron trifluoride, BF3 accepts an electron pair from ammonia, NH3 to form BF3NH3.
a)
Draw Lewis structures to illustrate the formation of BF 3NH3 from boron trifluoride and
ammonia.
b)
Show which of the bond is the coordinate covalent bond.
The following species do not obey the octet rule:
a)
ICl2-
b)
NO2
c)
BBr3
Draw a Lewis structure for each species and state the type of octet rule exception.
CHEMISTRY DEPARTMENT
35
SHE 1315
7.
CHEMICAL BONDING
Rank the bonds in each set in order of increasing
a) C-I, C-F, C-Br, C-Cl
b) C-S, C=O, C-O
c) N-H, N-S, N-O
bond length
a) C–F< C–Cl < C–Br<C-I
b) C=O <C–O < C–S
c) N–H < N–O < N–S
bond length and bond energy.
bond energy
a) C-I<C-Br<C-Cl<C-F
b) C–S < C–O < C=O
c) N–S < N–O < N–H
8.
Match the following bonds with its bond length by using arrows to show your answers.
9.
Calculate the formal charges of each atom in the following molecules and determine
whether it represents a neutral molecule or ion.
Structure A
Structure B
H : 1-1 = 0
H-O : 6-6 = 0
N=O : 6-6 = 0
N : 5-4 = +1
O : 6-7 = -1
Thus, it is a neutral molecule
H : 1-1 = 0
C : 4-4 = 0
O= : 6-6 = 0
O- : 6-7 = -1
Thus, it is -1 ion.
10.
Define resonance structures.
Structures with the same placements of atoms but different locations of bonding
pair and lone pair electrons.
11.
Draw the resonance structures of:
a)
SO3
b)
N3-
CHEMISTRY DEPARTMENT
36
SHE 1315
12.
CHEMICAL BONDING
a)
Draw a Lewis structure of methyl cyanide, CH3CN
b)
Calculate the formal charge for each atom in methyl cyanide.
Formal charges on H : 0
F. C on C-C : 0
F. C on CΞ C : 0
F. C on N : 0
13.
A Lewis structure of CS2 usually is written as S=C=S rather than S-CS. Explain.
Formal charges:
Formal charges:
Sulphur : 6-6 = 0
Carbon : 4-4 = 0
Sulphur (singly bonded) : 6-7 = -1
Sulphur (triply bonded) : 6-5 = +1
Carbon : 4-4 = 0
The structure S=C=S is preferred because each atom has a formal charge
14.
Use the VSEPR model to predict the geometry of the following molecules and ions.
a) AsH3
d) I3-
g) IF2-
b) OF2
e) C2H4
h) SF6
c) AlCl4-
f) SnCl3-
i) TeF5-
Trigonal pyramidal
linear
octahedral
square pyramidal
CHEMISTRY DEPARTMENT
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SHE 1315
15.
CHEMICAL BONDING
Nitrogen and phosphorus are the first two elements in Group 15 of the periodic table.
Nitrogen can form nitrogen trichloride, NCl3 only while phosphorus can forms both
phosphorus trichloride, PCl3 and phosphorus pentachloride, PCl5.
a)
Write the electronic configuration of the nitrogen and phosphorus atom.
N : 1s22s22p3 , P : 1s22s22p63s23p3
b)
Draw the Lewis structures for NCl3 and PCl5. Predict their shapes.
Trigonal pyramidal
c)
Trigonal Bipyramidal
Explain why phosphorus can form the PCl 5 molecule, but not nitrogen.
N atom cannot expand its octet, P atom can expand its octet.
16.
Define electronegativity.
Electronegativity is the relative ability of a bonded atom to attract shared
electron/ Electronegativity is the tendency of an atom to attract electrons
17.
State the factors that affect the polarity of a molecule.
Shape of molecule & bond polarity / the difference in electronegativities between
atoms in a bond
18.
Use figure 9.21 (pg 349), to indicate the polarity of each bond.
a)
N—B
b)
N—O
c)
C—S
d)
S—O
CHEMISTRY DEPARTMENT
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SHE 1315
19.
CHEMICAL BONDING
BF3 is a non-polar molecule, whereas PF3 is a polar molecule. Explain
The molecular shape of BF3 is trigonal planar which is symmetrical. Although each
bond is polar, the bond dipole moment cancels each other. Therefore, BF3 is non
polar molecule. The molecular of PF 3 is trigonal pyramidal which is not
symmetrical. The bond dipole moments do not cancel each other and PF 3 is a polar
molecule.
20.
Arrange the following molecules in order of increasing dipole moment:
BH3 , NH3 , NF3
BH3<NH3<NF3
21.
Determine whether these molecules is polar or non polar.
g) PCl5 non polar
a) Br2 non-polar
d) HBr
b) XeF4 non-polar
e) SO2 polar
h) SF6
non polar
c) PH3
f) SO3 non polar
i) ClF3
polar
polar
polar
22.
a)
What
is
the
hybridization of each C and O atom
in the molecule?
b)
How many C atoms have a tetrahedral shape?
3
c)
How many  and π bonds are in the molecule? = 14, π= 3
CHEMISTRY DEPARTMENT
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SHE 1315
23.
24.
CHEMICAL BONDING
a)
Describe the hybridization process of the central atom in AlI3.
b)
Draw the orbital overlapping diagram of AlI3.
State the hybrid orbital of the underline atoms and the bond angle for the following
substance:
a)
SO32-
c)
sp3, 109.5
b)
CH3—CH=CH-Cl
sp3, 109.5 ; sp2, 120 ; sp2, 120
CH3—CΞN
sp3, 109.5 ; sp, 180
d)
Br—CH2—CH2—COOH
sp3, 109.5 ; sp3, 109.5 ; sp2, 120
CHEMISTRY DEPARTMENT
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SHE 1315
25.
Describe the hybridization process of the atoms in N2.
26.
Define, with the aid of appropriate model, the term metallic bond.
CHEMICAL BONDING
A metallic bond is the electrostatic attraction between the positively charged
metal and the sea of delocalized electron.
27.
The melting point of aluminum is higher than sodium. Explain.
Both aluminium and sodium are metallic. The strength of metallic bond depends
on the number of valence electrons. Since aluminium has three valence electrons
and sodium has one valence electron, the metallic bonds in aluminium are
stronger than sodium. Therefore, aluminium has a higher melting point than that
of sodium
CHEMISTRY DEPARTMENT
41
SHE 1315
CHEMICAL BONDING
CHEMISTRY DEPARTMENT
42
SHE 1315
INTERMOLECULAR FORCES, LIQUIDS AND SOLIDS
TOPIC 7: INTERMOLECULAR FORCES, LIQUIDS AND SOLIDS
1.
Water molecules can form hydrogen bonding.
a)
Define hydrogen bonding.
Hydrogen bonding is the attraction between hydrogen atom, which is
covalently bonded to a small and highly electronegative atom (like F, O and
N), and the lone pair of electrons of another very electronegative atom.
b)
Explain how hydrogen bonding affects the boiling point of water.
More energy is required to break the strong hydrogen bonding between
water molecules. This makes the boiling point of water higher than
expected from its small molecular size.
2.
3.
Name the intermolecular forces that exist between the particles of the following liquids:
a)
trichloromethane, CHCl3
Dipole dipole forces and London dispersion forces
b)
ethanol, CH3CH2OH
hydrogen bonding, dipole dipole forces and London dispersion
c)
aluminium trichloride
London dispersion forces
Explain the following observations in terms of intermolecular forces:
a)
The melting point of water, H2O is 0 °C while hydrogen sulphide, H2S is -83 °C.
H2O : hydrogen bonding
H2S : London Dispersion forces and Dipole dipole forces
Hydrogen bond is stronger than van der Waals forces
b)
The molecular mass of butane and propanone are both 58. However, the boiling
point of butane is 273 K while that of propanone is 330 K.
Butane : Non-polar molecule. Intermolecular force is the weak London
Dispersion forces
Propanone : polar molecule. Intermolecular force is Dipole-dipole forces.
As dipole-dipole forces is stronger than London dispersion forces, higher
boiling point.
c)
CI4 decomposes at a lower temperature than CCl 4
The bond energy of the C-I bond is lower because of the larger size of the
iodine atom
CHEMISTRY DEPARTMENT
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SHE 1315
4.
INTERMOLECULAR FORCES, LIQUIDS AND SOLIDS
Ethanol, CH3CH2OH is completely miscible with water while phenol, C 6H5OH is partially
miscible with water. Explain.
Both ethanol and phenol can form hydrogen bonding with water molecules. But
the presence of a large non-polar hydrocarbon group (C6H5-) makes phenol less
soluble compared to the small C2H5- group in ethanol.
5.
Determine the type of crystalline solids based on their properties.
a)
This material forms very hard crystals, very high melting point and usually poor
thermal and electrical conductors.
Network covalent
b)
This material is hard, brittle, high melting point and dissolves well in water.
Ionic solid
c)
This material is soft, very low melting point and poor thermal and electrical
conductors.
Atomic solid
d)
This material is soft, malleable, ductile, has low to very high melting point and
excellent thermal and electrical conductors.
Metallic
6.
Which of the two liquids below has greater surface tension? Explain your answer.
CH3CH2OH or CH3OCH3
Ethanol (CH3CH2OH) has greater surface tension than dimethyl ether (CH 3OCH3).
Ethanol has stronger intermolecular forces (Hydrogen bond). Surface tensions
tend to increase as the strength of intermolecular forces increase. Both have
identical molar mass so attractions resulting from London dispersion forces will be
equal.
CHEMISTRY DEPARTMENT
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SHE 1315
7.
INTERMOLECULAR FORCES, LIQUIDS AND SOLIDS
Arrange the substances in order of increasing boiling point. State your reasons.
a)
CH3OCH3, CH3CH2OH, and CH3CH2CH3
CH3CH2CH3 < CH3OCH3< CH3CH2OH
All have similar molar masses: 46.07g/mol, 46.07g/mol and 44.09g/mol
respectively. Dimethyl ether cannot form hydrogen bonds (no O-H bond),
but is polar and has dipole-dipole forces. Ethanol can form hydrogen bonds.
propane is nonpolar, so it has only London dispersion forces. The boiling
point increases as the strength of the intermolecular forces increase:
London dispersion < dipole-dipole forces < hydrogen bond
b)
Br2, Cl2 and I2
Cl2 < Br2 < I2
All are non polar molecules so only London dispersion forces are present.
London dispersion forces get stronger as molar mass increases
c)
CaCO3, CH4, CH3OH and CH3OCH3
By using intermolecular forces, we can tell that these compounds will rank:
methane (Van der Waals forces), dimethyl ether (dipole-dipole forces),
methanol (hydrogen bonding), calcium carbonate (ionic electrostatic forces
that are much stronger than intermolecular forces).
8.
For each pair of substance, identify the substance that is likely to have a higher vapor
pressure. Explain your answers.
a)
CO2 or SO2
CO2 will have the higher vapor pressure. Vapor pressure tends to decrease
as the strength of the intermolecular forces increase. Carbon dioxide is nonpolar (dispersion forces only). Sulfur dioxide is polar (dipole-dipole forces
are present).
b)
CH3OH or CH3-O-CH3
CH3OCH3 will have the higher vapor pressure. Vapor pressure tends to
decrease as the strength of the intermolecular forces increase. CH 3OH can
form hydrogen bond. CH3OCH3 is polar (bent shape around the oxygen), so
dipole-dipole forces are the strongest forces in this compound.
CHEMISTRY DEPARTMENT
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SHE 1315
9.
INTERMOLECULAR FORCES, LIQUIDS AND SOLIDS
Define “boiling point” and "normal boiling point".
Boiling point: The temperature at which the liquid vapor pressure equals to the
external pressure.
Normal boiling point: The temperature at which the liquid vapor pressure equals to
1 atm
10.
State the difference between crystalline solid and amorphous solid.
Crystalline solid : the particles are arranged in a well defined shape.
Amorphous solid: the particles are not arranged in a well defined shape
11.
Define the term “lattice” and “unit cell”.
Lattice - the points form a regular pattern throughout the crystal.
Unit cell – the smallest portion of the crystal that, if repeated in all three
directions gives the crystal
12.
A
B
a)
Identify element A and B.
A -diamond
B -graphite
b)
Complete the table below for structure A and B.
Physical properties
Hardness
Melting point
Color
Density
Electrical conductivity
Structure A
Very hard
High
Colorless
2.27
None
Structure B
Very soft
High
Shiny black
3.51
High
CHEMISTRY DEPARTMENT
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SHE 1315
13.
INTERMOLECULAR FORCES, LIQUIDS AND SOLIDS
Graphite and diamond are two allotropes of carbon. Graphite is a conductor while diamond
is a non-conductor. Explain their difference in terms of structure and bonding.
In graphite, each carbon atom uses only three of its valence electron to form
covalent bonds. The other one electron is delocalized and can move freely. Thus,
graphite is a conductor.
In diamond, all the four valence electrons are used to form covalent bonds with
other carbon atoms. There is no delocalized electron in diamond. Hence, it is not a
conductor.
14.
Draw the crystal structure of simple cubic, body centered cubic and face centered cubic.
Calculate the number of atom/unit cell for each structure.
Simple cubic
No atom/unit cell: 1/8(8) = 1
15.
Body centered
1/8 (8) + 1 = 2
Face centered
1/8(8) + ½ (6) = 4
C and Si both react with oxygen to form CO2 and SiO2. Carbon dioxide is a gas at room
conditions, but silicon dioxide is a hard solid. Explain the difference between CO2 and SiO2
in terms of bonding and structure.
CO2: non polar covalent molecule with weak van der Waals forces between
molecules.
SiO2: Network covalent structure with strong covalent bonds holding the atoms
together.
CHEMISTRY DEPARTMENT
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SHE 1315
REACTION KINETICS
TOPIC 8: REACTION KINETIC
1.
Define the rate of reaction.
The reaction rate is the change in the concentration of reactants or products per
unit time
2.
Explain the difference between average rate, instantaneous rate and initial rate.
Average rate is the total change in concentration over a given period of time.
Instantaneous rate is the rate at particular instant during the reaction.
Initial rate is the instantaneous rate at the moment the reactants are mixed.
3.
Consider the following reaction,
2 N2O5 (g)
a)
4NO2 (g) + O2(g)
Write the rate expression in terms of changes in concentration.
Rate = - ½ ∆[ N2O5]/∆t = 1/4 ∆[ NO2]/∆t = ∆[ O2]/∆t
b)
If the rate of disappearance of N2O5 gas is 1.0 x 102 mol dm-3s-1 , what is the rate of
formation of NO2 under the same condition?
- ½ ∆[ N2O5]/∆t = 1/4 ∆[ NO2]/∆t
∆[ NO2]/∆t =- 4/2∆[ N2O5]/∆t
= 2.0 x 102 mol dm-3 s-1
4.
Write the chemical equation for the rate expression below.
Rate = - ∆[ CH4]/∆t
CH4 + 2 O2
= -½ ∆[ O2]/∆t
= ½ ∆ [ H2O]/∆t = ∆[CO2]/∆t
2 H2O + CO2
CHEMISTRY DEPARTMENT
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SHE 1315
5.
REACTION KINETICS
The rate of formation of ammonia shown below was measured as 10.0 mmol dm−3 s−1.
N2 (g) + 3 H2 (g) → 2 NH3 (g)
Calculate the rate of disappearance of hydrogen.
Rate = - ∆[ N2]/∆t = -1/3 ∆[ H2]/∆t = 1/2∆[ NH3]/∆t
∆[ H2]/∆t= 3/2∆[ NH3]/∆t
=3/2 x 10 mmol
=15 mmol dm−3 s−1
6.
2 NO (g) + 2 H2 (g)
2 H2O (g) + N2 (g)
The reaction above is second order with respect to NO and first order with respect to H2.
a)
Write the rate law and determine the overall order.
Rate = k[NO]2[H2]
Overall order = 3
b)
How does the reaction rate change if:
i)
the NO concentration doubles?
Rate increases by a factor of 4
ii)
the NO concentration triples?
Rate increases by a factor of 9
iii)
the H2 concentration doubles?
Rate increases by a factor of 2
iv)
both reactant concentrations are doubles?
Rate increases by a factor of 8
CHEMISTRY DEPARTMENT
49
SHE 1315
7.
REACTION KINETICS
The initial rate of a reaction A + B + C  D was measured for several starting concentrations of
A, B, and C. The results are as follows:
Experiment
[A] (M)
[B] (M)
[C] (M)
Initial Rate
(M/s)
0.0500
0.0500
0.0100
6.25 x 10-3
0.1000
0.0500
0.0100
1.25 x 10-2
3
0.1000
0.1000
0.0100
5.00 x 10-2
4
0.0500
0.0500
0.0200
6.25 x 10-3
1
2
a)
What is the order with respect to each reactant?
The rate law is rate = k [A]m [B]n [C]p.
CHEMISTRY DEPARTMENT
50
SHE 1315
b)
REACTION KINETICS
Write the rate law for the reaction above.
Rate = k [A][B]2
8.
c)
Calculate the value of k.
d)
How does the rate of reaction change if the concentration of C is increased to
0.500 M?
The rate of reaction does not change.
The rate of reaction between CO and NO2 was studied at 540 K starting with various
concentrations of CO and NO2. The data was collected as shown below.
Experiment
[CO] (M)
[NO2] (M)
Initial Rate (M/s)
1
5.10 x 10-4
0.350 x 10-4
3.4 x 10-8
2
5.10 x 10-4
0.700 x 10-4
6.8 x 10-8
3
5.10 x 10-4
0.175 x 10-4
1.7 x 10-8
4
1.02 x 10-3
0.350 x 10-4
6.8 x 10-8
5
1.53 x 10-4
0.350 x 10-4
10.2 x 10-8
a)
Predict (without calculation) the reaction order with respect to CO and NO2.
[CO] constant [NO2] doubled, rate doubled so rate order of NO 2 is 1st order
[CO] doubled [NO2] constant, rate doubled so rate order of CO is 1st order
b)
Calculate the value of k.
k=
Rate
[CO]1[NO2]1
=
3.4 x 10-8
(5.10 x 10-4)(0.350 x 10-4)
= 1.9 M-1s-1
c)
Calculate the rate of reaction if the concentration of CO is 2.1 x 10 -2 M and the
concentration of NO2 is 6.25 x 10-3M.
Rate = k[CO]1 [NO2]1
=(1.9 M-1s-1)(2.1 x 10-2)(6.25 x 10-3 )
=2.49 x 10-4 Ms-1
d)
How does the reaction rate change if the temperature is changed to 480 K, while the
concentration remains constant?
The rate of reaction decreases.
CHEMISTRY DEPARTMENT
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SHE 1315
9.
REACTION KINETICS
Fill in the table below.
Conc vs time graph
Reaction Order
First order
Integrated
Rate Law
Unit for Rate
constant (k)
ln [A]0 = kt
[A]t
s-1
ln[A]
time
Zero order
[A]t - [A]0= -kt
mol L-1s-1
[A]
time
1 – 1 = kt
[A]t [A]0
Lmol-1s-1
1/[A]
Second order
time
10.
In a first order reaction, a sample of 1.0 g of radium-224 was found to weigh 0.25 g after
7 days. Calculate the half life of radium.
1.0 g
t1/2
0.5 g
t1/2
0.25 g
2 t1/2 = 7 days
t1/2 = 3.5 days
11.
A decomposition reaction has a rate constant of 0.0012 yr-1. Calculate the half life of the
reaction.
(since the unit is yr-1, so, this is the first order reaction)
t1/2 = ln 2 / k = ln 2 / 0.0012 yr -1 = 577.62 yr = 5.8 x 102 yr
CHEMISTRY DEPARTMENT
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SHE 1315
12.
REACTION KINETICS
The rate constant for the decay of
a)
What is the half-life for
Rn is 0.1810 day-1.
222
Rn expressed in days?
222
(since the unit is day-1, so, this is the first order reaction)
t1/2 = ln 2/k
= 3.83 days
b)
Calculate the fraction of a sample of 222Rn decays in a period of exactly one week.
[Assume initial concentration is 100]
ln[A]t = ln[A]0 – kt
ln[A] t = ln[100] – (0.1810 day-1)(7days)
[A] t = 28.1675
Fraction : 28.17/100
13.
In a first order decomposition reaction, 50% of a compound decomposes in 10.5 min.
a)
What is the rate constant of the reaction?
t1/2 = 10.5 min
k = ln 2
t1/2
= 0.693
10.5
= 0.066 min-1
b)
How long does it take for 75% of the compound to decompose?
ln [A]0 – ln [A]t = kt
t = ln 100 – ln 25
0.066
= 21 min
CHEMISTRY DEPARTMENT
53
SHE 1315
14.
REACTION KINETICS
Concentration versus time data for the thermal decomposition of ethyl bromide at 700K is
given below.
C2H5Br(g)
Time (min)
0
1
2
3
4
C2H4 (g) + HBr (g)
[C2H5Br]
1.00 M
0.82 M
0.67 M
0.55 M
0.45 M
ln[C2H5Br]
0
-0.20
-0.40
-0.60
-0.80
1/[C2H5Br]
1
1.22
1.49
1.82
2.22
a)
Complete the table by filling in the ln [C2H5Br] and 1/[C2H5Br] values.
b)
Determine the order of the reaction by plotting graph.
time
[C2H5Br]
ln
[C2H5Br]
1
[C2H5Br]
time
time
time
time
The linear graph is observed from the second graph prove that the reaction is 1 st order
reaction.
c)
Calculate the rate constant.
Slope = -k
-k = -0.4 – (-0.6)
2-3
= -0.2
k = 0.2 min-1
d)
Calculate the half life of ethyl bromide in this reaction.
t1/2 = ln 2
k
k = ln 2
0.2
= 3.47 min
CHEMISTRY DEPARTMENT
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SHE 1315
15.
REACTION KINETICS
The rate constant for the reaction below is 30.0 Lmol-1min-1.
2 HI (g)
H2 (g) +2 I2 (g)
a)
What is the overall order of the reaction?
Second order
b)
Calculate the time taken for the concentration of HI to drop from 0.010 M to 0.005 M.
1– 1
= kt
[A]t [A]0
1
- 1
= (30.0) t
0.005
0.010
t
16.
= 3.3 min
The table below represent a reaction with the rate law of Rate = k.
Time (min)
[A]
a)
(M)
0
1
2
3
1.0
0.79
0.58
0.38
What is the overall order of the reaction?
Zero order reaction.
b)
Calculate the rate constant.
[A]t - [A]0= -kt
k = 1.0 – 0.79
1
k = 0.21 M min-1
17.
Define reaction mechanism and molecularity.
Reaction mechanism is a sequence of single reaction steps that sum to overall
chemical reaction.
Molecularity is the number of reactant particles in the step
CHEMISTRY DEPARTMENT
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SHE 1315
18.
REACTION KINETICS
Consider the energy profile diagram below.
a)
Write the mechanism and overall equation for the reaction.
Step 1H2+I2
H2 + I + I
Step 2H2 + I + I
H2I + I
Step 3H2I + I
HI +HI
Overall
H2+I2
2HI
b)
State the molecularity for each elementary step.
Step 1
Step 2
Step 3
c)
Unimolecular
Bimolecular
Bimolecular
Define intermediate.
Substances that is formed and used up during the overall reaction.
d)
Identify intermediate(s).
I and H2I
e)
Write the rate law for the reaction.
Rate = k[H2][I2]
CHEMISTRY DEPARTMENT
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SHE 1315
19.
REACTION KINETICS
The rate law for the endothermic reaction below is Rate = k[H2][N],
2 H2(g) + 2 NO(g)
2 H2O(g) + N2 (g)
The following equations represent a proposed mechanism for the reaction.
Step 1:
Step 2:
Step 3:
a)
H2(g) + NO(g)
N(g) + NO(g)
O(g) + H2(g)
H2O(g) + N(g)
N2(g) + O(g)
H2O(g)
Ea = 1690 kJ
Ea = 625 kJ
Ea = 436 kJ
Define elementary step.
The individual steps that make up a reaction mechanism.
b)
Which elementary steps is the rate determining step?
Step 1
c)
Sketch the potential energy diagram (energy profile) for the reaction.
Potential
energy
H2O
Reaction progress
H2 + NO
d)
Is the above mechanism valid? State the reason.
Not valid because the given rate law is not tally with the rate law from RDS
in the proposed mechanism above.
CHEMISTRY DEPARTMENT
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SHE 1315
20.
REACTION KINETICS
State four factor that influence reaction rate.
Concentration, temperature, catalyst and physical state.
21.
Based on Arrhenius equation, state how the rate of reaction can be increased by:
a)
increasing the temperature
The higher the temperature, the smaller the activation energy, the larger
the value of k and leads to the higher reaction rate.
b)
adding a catalyst
Catalyst causes lower activation energy, the larger the value of k and leads
to the higher reaction rate.
22.
Define catalyst and list all types of catalyst.
Catalyst is substances that speed up the reaction without being consumed. Two
type of catalyst are homogenous and heterogeneous catalyst
CHEMISTRY DEPARTMENT
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SHE 1315
23.
REACTION KINETICS
Explain how the rate of reaction can be affected by:
a)
adding a catalyst. (Add a suitable diagram)
Catalyst effects by providing an alternative pathway with lower Ea leads to
increasing of reaction rates.
I
Potentialb.
energy
without catalyst
Reaction progress
b)
increasing the temperature. (Add a suitable diagram)
As the temperature increases, the kinetic energy of the reactant molecules
increases. The fraction of reactant molecules with energy same or greater
than Ea increases leads to increasing of reaction rates.
Maxwell-Boltzmann Distribution Curve
T1 < T 2
c)
increasing the reactant concentration.
As the concentration increase the number of reactant particles increases.
Frequency of effective collision increase leads to increasing of reaction
rates.
CHEMISTRY DEPARTMENT
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