Internal assessment resource number Chem/2/3 – E version 1
PAGE FOR TEACHER USE
2005
Internal Assessment Resource
Subject Reference: Chemistry 2.3
Internal assessment resource reference number:
Chem/2/3 – E version 1
“Quantitative Chemistry 2”
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 – E version 1 : Quantitative Chemistry 2
PAGE FOR STUDENT USE
2005
SUBJECT REFERENCE: CHEMISTRY 2.3
Internal assessment resource reference number:
Chem/2/3
“Quantitative Chemistry 2”
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
What is the mass of 0.126 moles of oxalic acid, H2C2O4? M(H2C2O4) = 90.0 g mol1
2
19.4 g of baking soda, NaHCO3, is dissolved in water to make exactly 1 litre of
solution.
What is the concentration in mol L1? M(NaHCO3) = 84.0 g mol1
3
4
How many moles of sucrose are present in 14.0 mL of 0.450 mol L 1 sucrose
olution?
Allicin is the compound responsible for the characteristic smell of garlic.
Its formula is C6H10S2O. What is the % of carbon in this compound?
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Internal assessment resource number Chem/2/3 – E version 1 : Quantitative Chemistry 2
PAGE FOR STUDENT USE
5(a) To carry out a colorimetric analysis a set of standard solutions are prepared from a
stock solution of 0.00500 mol L1 FeCl3.6H2O.
What mass of FeCl3.6H2O is required to make up 500 mL of this solution?
(b)
20.0 mL of a 0.00500 mol L1 solution from part (a) above was pipetted into a
volumetric flask and diluted to 100 mL. What is the concentration of the diluted
solution?
6
The porphryn molecule forms an important part of the haemoglobin structure. This
porphryn molecule is 78.5% carbon, 3.2% hydrogen and 18.3% nitrogen.
(i)
Calculate the empirical formula of porphryn.
(ii)
If the molar mass of porphryn is 306 g mol1 calculate the molecular formula.
7
A precipitate of 1.34 g of Fe2O3 was obtained by treating a vitamin supplement
dissolved in water with NaOH and then heating. What is the mass of iron in the
tablet?
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Internal assessment resource number Chem/2/3 – E version 1 : Quantitative Chemistry 2
PAGE FOR STUDENT USE
8
Hydrated magnesium chloride is heated in a crucible. The following data is
collected.
mass of crucible and lid = 26.47 g
mass of crucible, lid and hydrated magnesium chloride = 28.62 g
mass crucible, lid and magnesium chloride after first heating = 27.49 g
mass of crucible, lid and magnesium chloride after second heating = 27.48 g
Use these results to determine the formula of hydrated magnesium chloride.
M(MgCl2) = 95.0 g mol1 M(H2O) = 18.0 g mol1
9
Wine makers convert sugars such as glucose to ethanol and carbon dioxide:
C6H12O6
2 C2H5OH + 2 CO2
M / g mol1: C2H5OH = 46.0, C6H12O6 = 180
Starting with 540 g of glucose:
(a)
what is the mass of ethanol that can be obtained by this process?
(b)
what is the volume of ethanol that can be obtained by this process?
density ethanol = 0.789 g mL1
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Internal assessment resource number Chem/2/3 – E version 1 : Quantitative Chemistry 2
PAGE FOR STUDENT USE
10
10.0 g of iron is reacted with excess chlorine gas to produce iron chloride. The
resulting precipitate has a mass of 29.0 g.
Use this information to decide whether the product is iron(II) chloride FeCl2 or iron(III)
chloride FeCl3.
11
In an experiment sodium hydroxide was used to prepare a sample of Glauber's salt,
Na2SO4.10H2O, by neutralisation with sulfuric acid followed by recrystallisation from
aqueous solution.
H2SO4 + 2 NaOH
Na2SO4 + 2 H2O
Calculate the maximum mass of Glauber's salt which could be made from 5.00 g of
sodium hydroxide.
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Internal assessment resource number Chem/2/3 – E version 1 : Quantitative Chemistry 2
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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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 – E version 1
PAGE FOR TEACHER USE
AS90763 Version 1 Assessment Schedule: Quantitative Chemistry 2 Chem 2.3 E version 1
Q
Judgement for
Achievement
Evidence
Judgement for Merit
1
11.3 g
Correct
2
0.231 mol L1
Correct
3
6.30 x 103 mol
Correct
4
M(C6H10S2O) 162 g mol1
Correct molar mass of
C6H10S2O.
Correct %.
Amount of FeCl3.6H2O
correct.
Correct mass
determined.
Amount of FeCl3.6H2O
correct.
Correct determination of
concentration of diluted
solution.
Amount in moles of
each element correct.
Empirical formula
correct.
Correct process, but
incorrect empirical
formula used.
Molecular formula
correct.
% C = 44.4%
5(a)
n(FeCl3.6H2O) = 5.00 x 103 mol L1 x 0.500 L
= 2.50 x 103 mol
Judgement for
Excellence
M(FeCl3.6H2O) = 270.4.g mol1
m(FeCl3.6H2O) = 270.4 g mol1 x 2.50 x 103 mol = 0.676 g
(b)
n(FeCl3.6H2O) = 5.00 x 103 mol L1 x 0.0200 L
= 1.00 x 104 mol
c(FeCl3.6H2O) = 1.00 x 103 mol L1
6(a)
n(C) =
78 .5 g
12 .0 g mol
n(N) =
1
18.3 g
14.0 g mol 1
= 6.54 mol
n(H) =
3.20 g
1.0 g mol
1
= 3.20 mol
= 1.31 mol
empirical formula C10H5N2
(b)
M(empirical formula) = 153 g mol1
molecular formula C20H10N4
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Internal assessment resource number Chem/2/3 – E version 1
PAGE FOR TEACHER USE
7
method A
(i)
n(Fe2O3) = 1.34 g /159.8 g mol1
Correct process used in
either step (i) or (iii).
Correct answer for mass
of Fe produced.
eg An incorrect molar
mass or wrong mole
ratio used.
= 8.39 x 103 mol
n(Fe) = 2 x 8.39 x 103 mol
(ii)
Correct process with
one error.
= 1.68 x 102 mol
(iii) m(Fe) = 1.68 x 102 mol x 55.9 g mol1
= 0.938 g
OR
method B
(i)
% Fe in Fe2O3 = 55.6 x 2 / 159.8 = 70.0 %
(ii) 70.0% of 1.34 g = 0.938 g
8
mass of anhydrous MgCl2 = 1.01 g
mass of H2O = 1.14 g
amount of MgCl2 = 1.01 g / 95 g mol1 = 1.06 x 102 mol
Correct determination of
masses of MgCl2 and
H2O.
Correct determination of
either amount of MgCl2
or H2O.
Correct formula for
hydrated salt.
Correct process used in
either step (i) or (iii).
Correct determination
for mass of ethanol.
Correct determination of
answer for volume of
ethanol produced.
amount H2O = 1.14 g / 18.0 g mol1 = 6.33 x 102 mol
ratio n(H2O) / n(MgCl2) = 6
formula is MgCl2.6H2O
9(a)
(i)
n(glucose) = 540 g / 180 g mol1 = 3.00 mol
(ii) n(ethanol) = 2 x n(glucose) = 6.00 mol
(iii) m(ethanol) = 6.00 mol x 46.0 g mol1 = 276 g
(b)
volume ethanol = 276 g / .789 g mL1 = 350 mL
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Internal assessment resource number Chem/2/3 – E version 1
PAGE FOR TEACHER USE
10
(i)
n(Fe) = 10.0 g / 55.9 g mol1 = 0.179 mol
(ii) M(FeCl2) = 126.9 g mol-1 and M(FeCl2) = 162.3 g mol1
Correct amount of Fe
determined.
Correct process with
one error.
eg Incorrect molar
mass.
(iii) If product is FeCl2
Correct identification of
product with appropriate
calculations.
m(FeCl2) = 0.179 mol x126.9 g mol1 = 22.7g
but if product is FeCl3
m(FeCl3) = 0.179 mol x 162.3 g mol1 = 29.1 g
therefore product is FeCl3
This calculation can also be done by determining the ratio of
n(Fe) to n(Cl) in the 29.0 g of product.
11
(i)
n(NaOH) = 5.00 g / 40.0 g mol1 = 0.125 mol
(ii) n(Na2SO4) = 1/2 x n(NaOH) = 0.0625 mol
Amount of NaOH
correctly determined.
Correct process with
one error.
Correct determination of
mass of salt.
(iii) n(Na2SO4.10H2O) = 0.0625 mol
M(Na2SO4.10H2O) = 322 g mol1
(iv) m(Na2SO4.10H2O) = 0.0625 mol x 322 g mol1 = 20.1 g
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
3 answers at excellence level PLUS 4 other answers at merit level AND 3 other answers at achievement level..
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