Internal assessment resource number Chem/2/3 – D version 1
PAGE FOR TEACHER USE
2005
Internal Assessment Resource
Subject Reference: Chemistry 2.3
Internal assessment resource reference number:
Chem/2/3 – D version 1
“Quantitative Chemistry 1”
Supports internal assessment for:
Achievement Standard 90763 Version 1
Solve simple quantitative chemical problems
Credits: 2
Date version published:
November 2004
Ministry of Education
quality assurance status
For use in internal assessment
from 2005.
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Internal assessment resource number Chem/2.3 – D version 1 : Quantitative Chemistry 1
PAGE FOR STUDENT USE
2005
SUBJECT REFERENCE: CHEMISTRY 2.3
Internal assessment resource reference number:
Chem/2/3 D version 1
“Quantitative Chemistry 1”
Achievement Standard 90763 Version 1
Solve simple quantitative chemical problems
Credits: 2
Time allowed: 1 hour
This is a closed book activity.
Student Instructions Sheet
 For some of the following calculations you will need to refer to the Periodic Table on
page 5 to find the required molar masses of elements.
 Give answers to 3 significant figures where appropriate.
1(a) Calculate the molar mass of aluminium oxide, Al2O3. Include units in your answer.
(b) Calculate the percentage mass of oxygen in aluminium oxide.
2
Calculate the amount (in moles) of methane, CH4, in 12.5 g of the gas?
M(CH4) = 16.0 g mol1
3
Calculate the mass of 5.20 mol of sodium hydroxide, NaOH. M(NaOH) = 40.0 g mol1
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Internal assessment resource number Chem/2.3 – D version 1 : Quantitative Chemistry 1
PAGE FOR STUDENT USE
4(a) Caffeine is a stimulant which has a mass composition of 49.5% carbon, 5.20%
hydrogen, 28.9% nitrogen and 16.5% oxygen.
Calculate the empirical formula of caffeine.
(b) If the molar mass of caffeine is 194 g mol1, use your answer to part (a) above to
determine
the molecular formula of caffeine.
5
A colorimetric analysis of a solution containing Cu2+(aq) is to be carried out.
Calculate the mass of CuSO45H2O that would be needed to prepare 100 mL of a
stock solution of concentration 1.50 x 102 mol L1.
M(CuSO45H2O) = 249.5 g mol1
6
What mass of CO2 is produced in the complete combustion of 34.5 g of ethanol
according to
the following equation?
C2H5OH +
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3 O2
2 CO2
+
3 H2O
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Internal assessment resource number Chem/2.3 – D version 1 : Quantitative Chemistry 1
PAGE FOR STUDENT USE
7
What mass of iron can be produced from the reduction of 1.50 kg of iron(III) oxide
according to the following equation?
Fe2O3
8
+
3 CO
2 Fe
+
3 CO2
Hydrated magnesium sulfate is heated in a crucible. The following data is collected.
mass of crucible and lid = 26.49 g
mass of hydrated magnesium sulfate = 2.12 g
mass crucible, lid and magnesium sulfate after first heating = 27.55 g
mass of crucible, lid and magnesium sulfate after second heating = 27.52 g
Use these results to determine the formula of hydrated magnesium sulfate.
M(MgSO4) = 120 g mol1 M(H2O) = 18.0 g mol1
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Internal assessment resource number Chem/2.3 – D version 1 : Quantitative Chemistry 1
PAGE FOR STUDENT USE
9
20.0 mL of an aqueous solution of 0.105 mol L 1 sodium carbonate is titrated with
0.250 mol L1 hydrochloric acid solution.
What volume of hydrochloric acid is required to reach the equivalence point of the
titration?
2 HCl(aq) +
Na2CO3(aq)
2 NaCl(aq) + CO2(g) + H2O(l)
10 500 mL of 0.253 mol L1 NaHCO3 solution is mixed with 800 mL of 0.824 mol L1
NaHCO3 solution.
What is the concentration of the final solution?
11 A chemist found that 4.69 g of sulfur combined with fluorine to produce 15.81 g of gas.
Determine the empirical formula of the compound.
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Internal assessment resource number Chem/2.3 – D version 1 : Quantitative Chemistry 1
PAGE FOR STUDENT USE
PERIODIC TABLE OF THE ELEMENTS
18
Atomic Number 1
1
2
3
4
Li
6.9
11
Na
23.0
19
K
39.1
37
Rb
85.5
55
Cs
133
87
Fr
223
Be
9.0
12
Mg
24.3
20
Ca
40.1
38
Sr
87.6
56
Ba
137
88
Ra
226
13
Atomic Mass
5
B
10.8
13
Al
3
4
5
6
7
8
9
10
11
12
27.0
21
22
23
24
25
26
27
28
29
30
31
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
45.0
47.9
50.9
52.0
54.9
55.9
58.9
58.7
63.5
65.4
69.7
39
40
41
42
43
44
45
46
47
48
49
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
88.9
91.2
92.9
95.9
98.9
101
103
106
108
112
115
71
72
73
74
75
76
77
78
79
80
81
Lu
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
175
179
181
184
186
190
192
195
197
201
204
103
104
105
106
107
108
109
Lr
Rf
Db
Sg
Bh
Hs
Mt
262
57
Lanthanide Series
Actinide Series
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2
H
1.0
58
59
La
Ce
Pr
139
140
141
89
90
91
Ac
Th
Pa
227
232
231
60
61
62
63
Nd
Pm
Sm
Eu
144
147
150
152
92
93
94
95
U
Np
Pu
Am
238
237
239
241
14
15
16
17
6
7
8
9
C
12.0
14
Si
28.1
32
Ge
72.6
50
Sn
119
82
Pb
207
N
14.0
15
P
31.0
33
As
74.9
51
Sb
122
83
Bi
209
O
16.0
16
S
32.1
34
Se
79.0
52
Te
128
84
Po
210
F
19.0
17
Cl
35.5
35
Br
79.9
53
I
127
85
At
210
64
65
66
67
Gd
Tb
Dy
Ho
157
159
163
165
96
97
98
99
Cm
Bk
Cf
Es
247
249
251
254
He
4.0
10
Ne
20.2
18
Ar
40.0
36
Kr
83.8
54
Xe
131
86
Rn
222
68
69
70
Er
Tm
Yb
167
169
173
100
101
102
Fm
Md
No
257
258
255
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Internal assessment resource number Chem/2/3 – D version 1
PAGE FOR TEACHER USE
AS90763 Version 1 Assessment Schedule: Quantitative Chemistry 1 Chem 2.3
Q
1
Judgement for
Achievement
Evidence
(a)
102 g mol1
(b)
% O = 47.1%
Correct molar mass.
2
0.781 mol
Correct
3
208 g
Correct
4(a)
n(C) =
49 .5 g
12 .0 g mol
n(N) =
1
49 .5 g
12 .0 g mol
1
= 4.13 mol
n(H) =
49 .5 g
12 .0 g mol
= 2.06 mol
n(O) =
1
49 .5 g
12 .0 g mol 1
= 5.20 mol
Judgement for Merit
Judgement for
Excellence
Correct %.
Amount in moles of
each element correct.
Empirical formula
correct.
Correct process, but
incorrect empirical
formula used.
Molecular formula
correct.
One step of calculation
correct.
Correct determination of
mass of CuSO4.5H2O
Correct process used in
either step (i) or (iii).
Correct process with
one error.
= 1.03 mol
empirical formula C4H5N2O
4(b)
5
molecular formula C8H10N4O2
(i)
n(CuSO45H2O) = 0.0150 mol L1 x 0.100 L
= 1.50 x 102 mol
(ii) m(CuSO45H2O) = 1.5 x 10-2 mol x 249.5 g mol1
= 0.374 g
6
M(C2H5OH) = 46.0 g mol1
(i)
n(C2H5OH) = 34.5 g / 46.0 g mol1 = 0.750 mol
(ii) n(CO2) = 2 x n(C2H5OH) = 1.50 mol
(iii) m(CO2) = 1.50 mol x 44.0 g mol1 = 66.0 g
© Crown 2004
Correct answer for mass
of CO2 produced.
eg An incorrect molar
mass or incorrect mole
ratio used.
6
Internal assessment resource number Chem/2/3 – D version 1
PAGE FOR TEACHER USE
7
M(Fe2O3) = 160 g mol1
(i)
n(Fe2O3) =1500 g / 160 g mol1 = 9.38 mol
Correct process used in
either step (i) or (iii).
Correct process with
one error.
Correct answer for mass
of Fe produced.
eg An incorrect molar
mass or incorrect mole
ratio used or incorrect
conversion from kg to g.
Answers for individual
steps may have more
than 3 sf but final
answer given to 3sf.
Correct determination of
masses of MgSO4 and
H2O.
Correct determination of
either amount of MgSO4
or H2O.
Correct formula for
hydrated salt.
Correct determination of
n(Na2CO3).
Correct determination of
n(HCl).
Correct determination of
volume of hydrochloric
acid.
One step of calculation
correct.
Correct process with
one error.
Correct determination of
concentration of final
solution.
(ii) n(Fe) = 2 x n(Fe2O3) = 18.8 mol
(iii) m(Fe) = 18.8 mol x 55.9 g mol1 = 1050 g (1.05 kg)
8
mass of anhydrous MgSO4 = 1.03 g
mass of H2O = 1.09 g
amount of MgSO4 = 1.03 g / 120 g mol1 = 8.58 x 103 mol
amount H2O = 1.09 g / 18.0 g mol1 = 6.06 x 102 mol
ratio n(H2O) / n(MgSO4) = 7
formula is MgSO4.7H2O
9
n(Na2CO3) = 0.105 mol L1 x 0.0200 L = 2.10 x 103 mol
n(HCl) = 2 x 2.10 x 103 mol = 4.20 x 103 mol
V(HCl) = 4.20 x 103 mol / 0.250 mol L1 = 1.68 x 102 L
= 16.8 mL
10
(i)
n(NaHCO3)
= (0.253 mol L1 x 0.500 L) + (0.824 mol L1 x 0.800 L)
= 0.786 mol
(ii) total volume = 1.30 L
(iii) concentration of final solution = 0.786 mol / 1.3 L
= 0.604 mol L1
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Internal assessment resource number Chem/2/3 – D version 1
PAGE FOR TEACHER USE
11
(i)
n(S) =
4.69 g
32 .0 g mol
1
= 0.1465 mol
One step of calculation
correct.
(ii) m(F) = 15.81  4.69 = 11.12 g
(iii) n(F) =
11 .12 g
19 .0 g mol 1
Correct process with
one error.
Correct determination
empirical formula.
eg Incorrect mass of F.
= 0.5853 mol
(iv) Hence S : F = 1 : 4 and empirical formula is SF4.
Correct use of significant figures and units is required for answers at excellence level.
Sufficiency Statement
In the application of the assessment schedule, the student's response to each question is only awarded one level of A, M or E.
The requirements for sufficiency are as follows:
Achievement
7 answers at achievement level.
Merit
6 answers at merit level PLUS 3 other answers at achievement level.
Excellence
4 answers at excellence level PLUS 4 other answers at merit level AND 3 other answers at achievement level.
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