Download 2 - MACscience

Internal assessment resource number Chem/2/_G2
PAGE FOR TEACHER USE
2010
Internal Assessment Resource
Subject Reference: Chemistry 2.3
Internal assessment resource reference number:
Chem/2/3_G2
Chemistry Calculations
Supports internal assessment for:
Achievement Standard 90763 v2
Solve simple quantitative chemical problems
Credits: 2
Date version published:
April 2008.
Modified March 2010
Ministry of Education
quality assurance status
For use in internal assessment
from 2008
© Crown 2008
1
Mt Aspiring College AS2.3 assessment task 2011
2010
Internal Assessment Resource
Subject Reference: Chemistry 2.3
Chemistry Calculations
Achievement Standard 90763 v2
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 6 to find the required molar masses of elements.
 Give answers to 3 significant figures where appropriate.
1
Calculate the amount (in moles) of methane, CH4, in 23.0 g of the gas?
M(CH4) = 16.0 g mol1
2
Calculate the mass of 6.25 mol of sodium hydroxide, NaOH.
M(NaOH) = 40.0 g mol1
3
25.4 mL of a solution contains 0.00257 mol of HCl. Calculate the concentration, in
mol L–1, of the HCl solution.
© Crown 2008
2
Mt Aspiring College AS2.3 assessment task 2011
4
Calculate the amount (in moles) of NaOH present in 15.6 mL of 0.125 mol L–1
solution.
5(a) Arginine is an aminoacid which has a mass composition of 41.5% carbon, 8.0%
hydrogen, 32.2% nitrogen and 18.3% oxygen.
Calculate the empirical formula of arginine.
(b)
If the molar mass of arginine is 174 g mol1, use your answer to part (a) above to
determine the molecular formula of arginine.
6.
Calculate the percentage of carbon and oxygen in glucose, C6H12O6.
© Crown 2008
3
Mt Aspiring College AS2.3 assessment task 2011
7
800 mL of 0.253 mol L1 NaHCO3 solution is mixed with 500 mL of 0.824 mol L1
NaHCO3 solution.
What is the concentration of the final solution?
8
A chemist found that 9.38 g of sulfur combined with fluorine to produce 31.62 g of
gas. Determine the empirical formula of the compound.
9
What mass of CO2 is produced in the complete combustion of 35.6 g of ethanol
according to the following equation?
C2H5OH +
© Crown 2008
3 O2
2 CO2
+
3 H2O
4
Mt Aspiring College AS2.3 assessment task 2011
10
What mass of iron can be produced if 25.0 g of carbon monoxide react with
iron(III) oxide according to the following equation?
Fe2O3
11
+
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 = 2.13 g
mass crucible, lid and magnesium sulfate after first heating = 27.55 g
mass of crucible, lid and magnesium sulfate after second heating = 27.53 g
Use these results to determine the formula of hydrated magnesium sulfate.
M(MgSO4) = 120.3 g mol1 M(H2O) = 18.0 g mol1
© Crown 2008
5
Mt Aspiring College AS2.3 assessment task 2011
PERIODIC TABLE OF THE ELEMENTS
18
Atomic Number
1
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
Molar Mass / g mol
–1
13
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
© Crown 2008
2
H
1.0
2
3
1
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
6
Internal assessment resource number Chem/2/3_G2
PAGE FOR TEACHER USE
AS90763 v2 Assessment Schedule: Chemistry Calculations 2 - Chem/2/3_G2
Q
1
23.0 g
= 1.44 mol
16.0 g mol 1
Correct
m = 6.25 mol x 40.0 g mol–1 = 250 g
Correct
n(CO2) =
2
3
c=
4
Judgement for
Achievement
Evidence
n = 0.125 mol L–1 x 15.6 x 10–3 L = 1.95 x 10–3 mol
(a) n(C) =
41.4 g
= 3.45 mol
12.0 g mol 1
32.2 g
n(N) =
= 2.3 mol
14.0 g mol 1
Judgement for
Excellence
Correct
0.00257 mol
= 0.101 mol L–1
25.4 x 10 3 L
5
Judgement for Merit
n(H) =
8g
= 8 mol
1.0 g mol 1
18.3 g
n(O) =
= 1.14 mol
16.0 g mol 1
Correct
Amount in moles of each
element correct.
Empirical and molecular
formulae correct.
Correct process, but
incorrect empirical formula
used.
Divide each by 1.14. Empirical formula C3H7N2O (M=87 g mol–1)
(b) Emp. formula x 2 = molecular formula C6H14N4O2
6
M(C6H12O6) = 180 g mol–1
% C=
%O=
© Crown 2008
6 x12 g
x 100 = 40.0 %
180 g mol 1
Process correct but
incorrect molar mass of
glucose used.
% of C and O correct
6 x16 g
x 100 = 53.3 %
180 g mol 1
7
Internal assessment resource number Chem/2/3_G2
PAGE FOR TEACHER USE
7
n(NaHCO3)
= (0.253 mol L1 x 0.800 L) + (0.824 mol L1 x 0.500 L)
OR
= 0.614 mol
Total volume = 1.30 L
concentration of final solution =
8
n(S) =
Correct total amount of
NaHCO3
0.6144 mol
= 0.473 mol L1
1 .3 L
9.38 g
= 0.292 mol
32.1 g mol 1
correct process for
calculating concentration
using a volume of 1.3 L.
Correct amount of S OR F.
Correct process with one
minor error.
Correct determination
empirical formula.
eg Incorrect mass of F.
m(F) = 31.62  9.38 = 22.2 g
n(F) =
Correct determination of
concentration of final
solution.
22.2 g
= 1.16 mol
19.0 g mol 1
Hence S : F = 1 : 4 and empirical formula is SF4.
9
M(C2H5OH) = 46.0 g mol1
(i)
n(C2H5OH) = 35.6 g / 46.0 g mol1 = 0.774 mol
(ii)
n(CO2) = 2 x n(C2H5OH) = 1.55 mol
Correct process used in
either step (i) or (iii).
Correct process with one
minor error.
Correct answer for mass
of CO2 produced.
eg An incorrect molar
mass.
(iii) m(CO2) = 1.55 mol x 44.0 g mol1 = 68.1g
10
25.0 g
= 0.893 mol
28.0 g mol 1
(i)
n(CO) =
(ii)
n(Fe) = 3 x n(CO) = 0.595 mol
2
Correct process used in
either step (i) or (iii).
Correct process with one
minor error.
Correct answer for mass
of Fe produced.
eg An incorrect molar
mass.
(iii) m(Fe) = 0.595 mol x 55.9 g mol1 = 33.3 g
© Crown 2008
8
Internal assessment resource number Chem/2/3_G2
PAGE FOR TEACHER USE
11
mass of anhydrous MgSO4 = 1.04 g
mass of H2O = 1.09 g
amount of MgSO4 =
amount H2O =
1.04 g
120 .3 g mol 1
1.09 g
18 .0 g mol 1
Correct determination of
masses of MgSO4 and
H2O.
Correct determination of
either amount of MgSO4 or
H2O.
Correct formula for
hydrated salt.
= 8.65 x 103 mol
= 6.06 x 102 mol
ratio n(H2O) / n(MgSO4) = 7
formula is MgSO4.7H2O
Sufficiency Statement
In the application of the assessment schedule, the learner's response to each question is only awarded one level of A, M or E.
The requirements for sufficiency are as follows:
Achievement
SIX questions answered correctly at achievement or higher.
6xA
Merit
SEVEN questions answered correctly with FOUR answers at merit level or higher PLUS THREE answers at achievement level or higher.
4xM + 3xA
Excellence
EIGHT questions answered correctly with THREE answers at excellence level PLUS TWO answers at merit level or higher PLUS THREE
other answers at achievement level or higher AND answers to all questions except 5, 8 and 11 require 3 significant figures and units.
3xE + 2xM + 3xA
© Crown 2008
9