Uploaded by Rania Samir Shaaban Mohamed

Formative-Assessment-MYP-Chemistry

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Criterion A: Knowing and Understanding
Unit Title: Stoichiometry
Name:
The questions in this test will be used to give you a criterion A level, it is divided into three sections.
Section A: two questions that will assess how you explain your scientific knowledge.
Section B: two questions that will assess your application of scientific knowledge and how you use it
to solve problems.
Section C: one questions that will assess the way you analyses, evaluate and judge scientific
information.
Each question is designed to be hierarchical, this means that as you answer each question you will
find it will get progressively more difficult.
Question
Level
1-2
Section A
Section B
Section C
3-4
5-6
7-8
1
2
3
4
5
6
Section A: Strand One
Question One
This question is about Mole and Chemical Formula
(a) State the missing words from this paragraph, using only the words from the list. You should
only use a word once and you will need to use all of the words. [1-2]
Mole
Stoichiometry
Same
Particle
Electron
Proton
6,022 x 1023
Neutron
Stoichiometry is a method for chemist do calculation. The heart of stoichiometry is mole. It
is actually a unit of amount of particles . One mole contains of 6,0022 x1023 number of
particles. The number of atoms or other particles in a mole is the same for all substances. In
comparison, one mole of oxygen consists, by definition, of the same number of atoms as
carbon-12, but it has a mass of 15.999 grams. This mass difference due to the different
number of Proton and Neutron In both atoms. Electron is not affecting mass of atom
because it is too small.
(b) Outline the different between empirical and molecular formula [3-4]
Empirical describes the simplest ratio among atoms in the compound, while molecular
describe the real ratio, and it is used to calculate molar mass of the compound.
(c) Using your scientific knowledge, describe and explain the determination of magnesium
oxide. Your level will be determined, be the detail in your response: explain to give a detailed
account including reasons or causes. [5-6] and [7-8]
A known mass of magnesium ribbon is heated and let
it be exposed with oxygen until it’s burned and turned
into white ash. The mass before and after heating then
used to calculate mole ratio of Mg and O.
The fire needs to contact the crucible so the
temperature is high enough to burn Mg, the lid must be
open a little to supply oxygen
Section A: Strand One
Question Two
a. State two component of atom that determine the mass of atom [1-2]
Proton and neutron
b. Outline two types of chemical formula [3-4]
Empirical formula is the simplest ratio
Molecular formula is the actual ratio
c. Describe the composition of carbon atom [5-6]
Atom carbon has 6 neutron and proton in the nucleus/center and it has 12 electron moving
around in two layers
d. Explain why the relative mass does not has unit [7-8]
Relative mass is mass of atom relative to 1/12 mass of carbon, since the unit of gram will be
canceled out, there is no unit at the end.
1
π΄π‘Ÿ = π‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘Žπ‘› π‘Žπ‘‘π‘œπ‘š (π‘”π‘Ÿπ‘Žπ‘š) π‘₯
12 π‘₯ π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘π‘œπ‘› (π‘”π‘Ÿπ‘Žπ‘š)
Section B: Strand two
Question Three
In this question you must solve problems related "Mole calculation”
(a) Calculate the number of mole of 300 g of iron [1-2]
1 π‘šπ‘œπ‘™ 𝑓𝑒
300 π‘”π‘Ÿπ‘Žπ‘š 𝐹𝑒 π‘₯
= 5,357 π‘šπ‘œπ‘™
56 π‘”π‘Ÿπ‘Žπ‘š 𝑓𝑒
This Information below is for question B to D
An organic compound A contains 62.0% by mass of carbon, 24.1% by mass of nitrogen,
the remainder being hydrogen.
(b) Determine the percentage by mass of hydrogen and the empirical formula of A. [3-4]
%H = 100-(62+24.1)
= 13.9 %
C. :
N. : H
62 24.1 13,9
∢
∢
12 14
1
3 :
1 : 8
C3NH8
(c) The relative molecular mass of A is 116. Determine the molecular formula of A [5-6]
(C3NH8)n= 116
((12 x3)+14+(8x1))n=116
58n=116
n=2
C6N2H16
(d) How many number of hydrogen’s atom are present in 90 gram of this molecules. [7-8]
= 90 π‘”π‘Ÿπ‘Žπ‘š π‘šπ‘œπ‘™π‘’π‘π‘’π‘™π‘’π‘  π‘₯
23
75.33 x 10 Hydrogen
CD.E FGHI JKLGMFNO
CPP FGHI IMQNRSQNT
π‘₯
C IMQ JULGMFNO
C IMQN JULGMFNO
π‘₯
V.PWW X CPYZ JULGMFNO
C IMQN [ULGMFNO
Section B: Strand two
Question Four
A solution containing NaCl 0.1M
a. How many moles of NaCl present in 2 L of this solution [1-2]
0.1 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™
= 2 π‘™π‘–π‘‘π‘’π‘Ÿ π‘π‘ŽπΆπ‘™ π‘₯
= 0.2 π‘šπ‘œπ‘™
1 π‘™π‘–π‘‘π‘’π‘Ÿ π‘π‘ŽπΆπ‘™
b. How many gram of NaCl present in 2 L of this solution [3-4]
= 2 π‘™π‘–π‘‘π‘’π‘Ÿ π‘π‘ŽπΆπ‘™ π‘₯
0.1 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™ 58.5 π‘”π‘Ÿπ‘Žπ‘š π‘π‘ŽπΆπ‘™
π‘₯
= 11.7 π‘”π‘Ÿπ‘Žπ‘š
1 π‘™π‘–π‘‘π‘’π‘Ÿ π‘π‘ŽπΆπ‘™
1 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™
=
c. Explain how to make 0,1 M NaCl [5-6]
For one litter of solution, weigh 5,85 gram of NaCl. This is because in 0.1 M NaCl, it consist of
0,1 mol of NaCl or 5.85. Then dissolve it in distilled water. Note that for dissolving the NaCl
the final volume must be 1 litter.
d. If 100 ml of NaCl 0,1M is heated until the volume of solution become 50 ml because of
evaporation of water, calculate the molarity after heating [7-8]
Amount of NaCl is not changed after heating = 0.01 mol
𝑛
𝑉
0.01 π‘šπ‘œπ‘™π‘’
𝑀=
= 𝟎. 𝟐 𝑴
0.05 π‘™π‘–π‘‘π‘‘π‘’π‘Ÿ
𝑀=
Section C: Strand Three
Question Five
The table below list some data synthesis of a compound X from sulphur and oxygen,
No
Mass of Sulphur
Mass of Oxygen
Mass of compound
Mass of reactant
Formed
remaining
1
6
6
10
2
2
6
7
10.7
1.3
3
6
8
13.3
0.7
4
6
9
15
0
5
6
10
15
1
a. How does the mass of compound form change as the mass of oxygen increases? [1-2]
Increase then constant
b. If you were to express the data above using visual format, what form would it take? You
can sketch this below if you wish. [3-4]
c. Einstein is asked to make conclusion from the data above. [5-6] [7-8]
“I think some data are not valid. The data from 9 gram of oxygen and above showing a
constant result of compound. These two data should be ignored because it’s not aligned
with the other three data. So over all the conclusion from this experiment is the more
oxygen mass added the more compound will be formed”.
Do you agree with Einstein’s conclusion? You should use the idea you have learnt in
class, and these results, to answer this question
No, I disagree. Atoms in a compound have certain ratio. The ratio in this experiment is
2:3 . if the mass reacted does not follow the ratio, than there will be an excess reagent
not reacted. In the experiment the mass compound increases as the mas oxygen
increase but constant after 9 and 10 gram.
The data of 9 and 10 gram of oxygen is still valid. The compound mass are constant
because it has reached the maximum possible ratio.
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