CHE 0315
TOPIC 1:
1
2
3
SEM 3, 2013/14
ATOMS, MOLECULES AND IONS
Complete the following table on sub-atomic particles.
[3]/ 1b
Subatomic particle
Mass (g)
Charge
Location
Proton
1.67 x 10-24
positive
nucleus
Electron
1.67 x 10-24
negative
Outside the nucleus
Neutron
9.11 x 10-24
No charge
nucleus
Complete the following table.
[5]/ 1d
Atoms/Ions
Symbol
Atomic No
Mass No
Sodium ion
Arsenic
Platinum
Cadmium
Strontium
Na+
As
Pt
Cd
Sr
11
33
78
48
38
23
75
195
112
87
No of
proton
11
33
78
48
38
No of
neutron
12
42
117
64
49
No of
electron
10
33
78
48
38
State the similarity and the difference between the following pair of isotopes.
(a)
235
Uranium and
238
Uranium
[1]/ 1f
They have same no of proton/atomic no which is 92 proton but
different number of neutron/mass no.
(b)
85
Rb and
87
Rb
[1]/1f
They have same no of proton/atomic no which is 37 proton but
different number of neutron/mass no.
4
Complete the following table by calculating the number of protons, neutrons and electrons
present in each isotope.
[3]/ 1f
S
25
S
55
Fe
56
Fe
14
N
15
N
23
Protons
16
16
26
26
7
7
Neutrons
7
9
29
30
7
8
Electrons
16
16
26
26
7
7
CHE 0315
5
SEM 3, 2013/14
An element X is present as 10X with an isotopic mass of 10.01 amu and as 11X with
isotopic mass of 11.01 amu. The natural abundance of 10X and 11X is 19.78% and
80.22%, respectively.
(a)
Calculate the average atomic mass of the element.
[2]/ 1i
Average atomic mass X = (10.01 x 0.1978) + (11.01 x 0.8022) =10.8122
amu
(b)
Identify the element X.
[2]/ 1i
Element X is Boron
6
The mass spectrum of a sample of chromium shows four peaks.
Mass
Percentage abundance (%)
(a)
50
4.3
52
83.8
53
9.5
54
2.4
Use the data above to calculate the relative atomic mass of chromium in
the sample.
[1]/ 1i
(50 x 0.043) + (52 x 0.838) + (53 x 0.095) + (54 x 0.024)
= 52.057
(b)
Draw a labeled mass spectrum that would be produced from the data
above.
[3]/1j
52Cr+
53Cr+
50Cr+
54Cr+
CHE 0315
7
SEM 3, 2013/14
A sample of silver is analyzed using a mass spectrometer. One isotopes makes up
51.35% of the sample and has an atomic mass of 106.905 amu. The other isotope
has a mass of 108.905 amu. Calculate the average atomic mass of the sample.
[2]/ 1i
average atomic mass = (0.5135 x 106.905) + (0.4865 x 108.905)
= 107.9amu
8
The two naturally occurring isotopes of chlorine are 35Cl with a mass of 34.9689 amu and
37
Cl with a mass of 36.9659 amu. The atomic mass of elemental chlorine is found to be
35.46 amu. Calculate the percent abundance of each of the two chlorine isotopes.
x = 35Cl
1-x = 37Cl
34.9689 (x) + 36.9659 (1-x) = 35.46 amu
34.9689x + 36.9659 – 36.9659x =35.46amu
-1.997x = -1.5059
x= 0.75
1-x = 0.25
Cl = 75% ,
35
9
[3]/ 1i
Cl=25%
37
Bromine consist of two isotopes, 79Br and 81Br. Apart from the peak at 79 and 81 due
to Br+ ions from these two isotopes, the mass spectrum of bromine molecule also
show peaks at 158, 160 and 162.
(a)
List the ions that give rise to these three peaks in the spectrum of bromine.
Br2+,
79
(b)
Br81Br +,
79
81
Br2+
[1]
Sketch a labeled mass spectrum for bromine molecule obtained considering
fragmentation does occur.
[3]/1j
81Br+
79Br+
79B +
r2
79Br81Br+
81B +
r2
CHE 0315
10
Given isotopes of hydrogen are 1H and 2H and isotopes of fluorine are 18F and 20F.
When HF is vaporized and analyzed in mass spectrometer, the following spectrum is
found. Identify the species for each peak.
[6]1j
P=1H+
11
SEM 3, 2013/14
Q= 2H+
R=
F
18 +
S= 1H18F+ U=1H20F+
V=2H20F+
Determine the number of moles in each of the following compound to 3 significant figures:
1n
(a)
82.5 g N2O4
[1]
82.5 g N2O4 x 1 mol N2O4 = 0.897 mol N2O4
92.02 g N2O4
(b)
2.60 x 1020 molecules NO2
2.6 x 1020 molecules NO2 x 1 mol NO2
6.022 X 1023 molecule NO2
[1]
= 4.32 x 10-4 mole NO2
CHE 0315
12
SEM 3, 2013/14
Determine the mass of the following compounds:
1f
(a)
[1]
2.60 mol Cr2(SO4)3
2.60 mol Cr2(SO4)3 x 392.21 g
= 1.02 x 108 g Cr2(SO4)3
1 mol Cr2(SO4)3
(b)
8.55x1022 molecules S2F4
[1]
8.55x1022 molecules S2F4 x 1 mol S2F4
x
6.022 x 10 23 molecules S2F4
13
140.14 g = 19.9 g S2F4
1 mol S2F4
Aspartame is an artificial sweetener that is 160 times sweeter than sucrose when
dissolved in water. The molecular formula of aspartame is C14H18N2O5.
(a) Calculate the molar mass of aspartame.
[1]/1m
(12.01 x 14) + (1.01 x 18) + (14.01 x 2) + (16.00 x 5) = 294.02
g/mol
(b) Calculate the number of moles in 10.0 g of aspartame.
[1] /1n
10.0 g x 1 mole C14H18N2O5 = 3.40 x 10-2 moles
294.02 g mol-1
(c) Calculate the mass of 1.56 moles of aspartame.
[2]/1n
1.56 moles C14H18N2O5 x 294.02 g = 458.67 grams
1 mole C14H18N2O5
(d) Calculate the number of molecules in 5.00 mg of aspartame?
5.00 mg x 1g
1000 mg
[2]/1n
x 1 moles x 6.02 x 1023 molecules = 1.02 x 10
294.02g
19
1 mole
(e) Calculate the number of nitrogen atoms in 1.20 grams of aspartame?
1.20 g C14H18N2O5 x 1 mole x 6.02 X 1023 molecules x 2 atom N
294.02 g
1 mole
1 molecule
21
= 4.91 x 10
atoms N
[2]/1n
CHE 0315
14
SEM 3, 2013/14
Determine the percent composition of each atom in the following compounds:
1p
(a)
[3]
H3PO4
Element
Mass
Percent
(b)
P
3.03/98 X 100
=3.06%
30.97/98 X 100
=31.6%
64/98 X 100
=65.3%
NaF
[2]
19.00/41.99 X
100
=45.25%
22.99/41.99 X
100
=54.75%
Mass
Percent
(c)
F
Na
Element
15
O
H
MgCO3
[3]
O
Element
Mg
C
Mass
Percent
24.31/84.32 X100
=28.88%
12.01/84.32 X
100
=14.24%
Complete the table below.
Empirical
[4]/1q
Mr
formula
Molecular
formula
CH2
42
C3H6
CH2
70
C3H8
44
C5H10
formula
C3H8
HO
34
H2O2
sdddddformula
48.00/84.32 X
100
=56.93%
CHE 0315
16
SEM 3, 2013/14
When 2.500 g of an oxide of mercury, is decomposed into the elements by heating,
2.405 g of mercury are produced. Calculate the empirical formula for the oxide. [3]
Mass of O = 2.500 – 2.405 = 0.095 g
Element
Hg
O
Mass
2.405
0.095
Mole
2.405 = 0.01199
200.59
0.01199 = 2
0.005938
Hg2O
0.095 =0.005938
16.00
0.005938 = 1
0.005938
Mole Ratio
Empirical formula
17
1r
An analysis of a compound consisting of carbon, hydrogen and oxygen showed
that it contains 0.273 g C, 0.046 g H, and 0.182 g O.
[3]/ 1r
(a)
Calculate its empirical formula.
C
Element
Mass
Mole
Mole
ratio
Empirical
Formula
(b)
H
0.273
0.046
0.273 = 0.0227
12.01
0.046 = 0.0455
1.01
0.0227 =2
0.0114
0.0455 = 4
0.0114
O
0.182
0.182 =0.0114
16.00
0.0114 = 1
0.0114
C2H4O
Given that the relative molecular mass of the compound is 88, calculate its
molecular formula.
[1]/1r
n= 88 = 2
44
C4H8O2
CHE 0315
18
SEM 3, 2013/14
A compound having a molar mass of 168 g/mole has the following percentage
composition: 42.87% carbon, 3.598% hydrogen, 28.55% oxygen and 25.00%
nitrogen. Determine the molecular formula of the compound.
[3]/1p
Assume 100g of sample
C
H
O
N
Element
Mass
Mole
Mole
ratio
Empirical
Formula
19
42.87
3.598
28.55
42.87 = 3.570
12.01
3.598 = 3.562
1.01
28.55 =1.784
16.00
3.570=2
1.784
3.562=2
1.784
1.784 = 1
1.784
25.00
25.00=1.786
14.00
1.786=1
1.784
C2H2ON
1.50 g sample of a hydrocarbon (molar mass = 90.0 g/mol) undergoes complete
combustion to produce 4.40 g of CO2 and 2.70 g of H2O. Calculate the empirical and
molecular formula of this compound.
[5]/1r
Mass of C = 4.40 g CO2 x 1 mol CO2 x 1mol C x 12.01g C =1.20 g C
44.01g CO2 1 mol CO2 1 mol C
Mass of H = 2.70 g H2O x 1 mol H2O x 2 mol H x 1.01g H = 0.303g H
18.01 g H2O 1 mol H2O 1 mol H
Element
Mass
Mole
Mole Ratio
Empirical formula
C
1.20
1.20/12.01=0.1
0.1/0.1 = 1
CH3
n= 90.0/15.0 =6
Molecular formula = C6H18
H
0.303
0.303/1.01=0.3
0.3/0.1= 3