Download Quantities in Chemistry

Unit 1 Chemistry
succeeding in the vce, 2017
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QUANTITIES IN CHEMISTRY
SIGNIFICANT FIGURES
When performing calculations, you are required to state answers to the appropriate number
of significant figures.
A significant figure indicates the precision to which a value or measurement is given.
A significant figure is:


A non-zero digit.
A zero that follows a digit.
Therefore:




All non-zero digits are significant. For example: 1, 2, 3, 23, 341
Zeros which follow a non-zero digit are significant. For example: 200, 0.0001500
Zeros which come before a non-zero digit are insignificant. For example: 0.0001500
Any zero between two digits is significant. For example: 20.00046
How many significant figures?
Count the number of figures from the first non-zero digit.
Examples:
38.4
has 3 significant figures.
20.03
has 4 significant figures.
500
has 3 significant figures.
0.007
has 1 significant figure.
0.00700 has 3 significant figures.
© The School For Excellence 2017
Succeeding in the VCE – Unit 1 Chemistry
Page 1
MATHEMATICAL OPERATIONS
Addition and Subtraction
When adding or subtracting numbers, count the number of decimal places to determine how
many significant figures should appear in the answer. The answer cannot contain more
decimal places than the smallest number of decimal places in the numbers being added or
subtracted.
For example:
The sum of the numbers 0.25, 1.3324, 23.112233 is:
23.112233
1.3324
1.25
6 decimal places
4 decimal places
2 decimal places
8 significant figures
5 significant figures
3 significant figures
The sum of these numbers (25.694633) must be written to the same number of decimal
places as the value with the smallest number of decimal places i.e. 1.25
Therefore, the answer is 25.69, which has 4 significant figures.
Multiplication and Division
When multiplying or dividing numbers, the answer must be written to the same number of
significant figures as the piece of data with the smallest number of significant figures.
For example:
39.100
1.0
12.00
48.0
3 decimal places
1 decimal place
2 decimal places
1 decimal place
5 significant figures
2 significant figures
4 significant figures
3 significant figures
The sum of these numbers must be written correct to 1 decimal place i.e. 100.1 (which has 4
significant figures!)
The product of these numbers (8445.6) must be written to 2 significant figures i.e. 8.4 10 .
3
Note:

Do NOT consider counting numbers in significant figures.

Significant figures are only considered in the FINAL answer.

Do not round off values until the final step (i.e. round when stating the final answer).
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Page 2
STANDARD FORM
A value written in standard form lies between 1 and 10, and is multiplied by a power of 10.
For example:
0.102 is written as 1.02  10 1
102 is written as 1.02  10 2
0.000102 is written as 1.02  104
400 is written as 4.00  10 2
Standard form is used to express a large or very small number value to a small number of
significant figures.
For example: 371.74 becomes 3.7  10 2 , correct to 2 significant figures.
Note:
If you are able to eliminate zeros by writing a number in scientific notation, then these zeros
4
are not significant. For example: 0.00040 can be written as 4.0 10 , and hence has 2
significant figures.
QUESTION 1
Consider the following numbers: 107.9, 16.42, 7.02, 8.30, 9.1
(a)
The least number of significant figures in this data is __________________.
(b)
If these figures were added the answer would be: __________________.
(c)
The product of these numbers is: __________________.
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Page 3
RELATIONSHIPS USED IN STOICHIOMETRY
Stoichiometry involves the calculation of the amounts of chemical substances.
THE MOLE
The “mole” is defined as the amount of substance contains the same number of formula
units as there are atoms in 12 grams of the carbon-12 isotope i.e. 6.0226  10 23 formula
units. This quantity is known as Avogadro's Constant ( N A ).
1 mole of a substance contains 6.0226  10 23 formula
units of that substance
The symbol for the amount of substance (mole) is n and the unit of measurement is mol .
RATIOS IN CHEMICAL FORMULAE
In order to solve quantity based questions in chemistry, we may need to use ratios in
chemical formulae.
As an example, a molecule of water, H 2O , is made up of 2 atoms of hydrogen and 1 atom
of oxygen. We say that these atoms have combined in a 2 to 1 ratio (2 : 1).
We can use these ratios to determine the amount of any chemical species in any given
scenario.
For example:

100 molecules of water are made up of 200 atoms of hydrogen and 100 atoms of
oxygen.

2 mole of water is made up of 4 mole of hydrogen atoms and 2 mole of oxygen atoms.
Note:
Ratios cannot be used directly to determine the mass of each element in a compound.
For example: 1 gram of H 2 SO4 does not contain 2 grams of H  , 1 gram of S atoms and
4 grams of O atoms.
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QUESTION 2
In one mol of aluminium hydroxide, Al(OH)3.3H2O, there are:
A
3 mol of aluminium atoms.
B
5 mol of hydrogen atoms.
C
16 mol of atoms.
D
4 mol of oxygen atoms.
CALCULATIONS INVOLVING AVOGADRO’S CONSTANT
Number of Particles  n( particles )  N A
Step 1: Calculate the amount, in mol, of the given substance.
Step 2: Use mole ratios to determine the amount, in mol, of the particle being counted.
For example:
1 mole of H 2 SO4 contains 6.023  10 23 H 2 SO4 molecules.
1 mole of H 2 SO4 contains 2  6.023  10 23 atoms of hydrogen.
1 mole of H 2 SO4 contains 3  6.023  10 23 ions (when in an aqueous solution).
Step 3: Multiply the amount, in mol, by Avogadro's Constant.
÷ NA
Amount of
Substance (n)
Number of
Particles (N)
 NA
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QUESTION 3
How many particles of CuSO4 are present in 2.00 mole CuSO4 ?
Solution
Number of Particles  n( particles )  N A
N A  2.00  6.023  10 23  1  1.21  10 24 particles
QUESTION 4
Given 3.0 mol of ethanol, C2 H 5OH , calculate:
(a)
The number of mol of carbon atoms.
(b)
The number of mol of hydrogen atoms.
(c)
The number of mol of oxygen atoms.
(d)
The number of carbon atoms.
(e)
The number of hydrogen atoms.
(f)
The number of oxygen atoms.
(g) The number of atoms.
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QUESTION 5
Calculate the number of chloride ions in 0.50 mol of calcium chloride ( CaCl2 ).
Solution
Number of Particles  n( particles )  N A
QUESTION 6
Calculate the amount of substance (mol) corresponding to 3.7  10 27 molecules of ethanol.
Solution
Number of Particles  n( particles )  N A
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RELATIVE MASSES
Chemists have used experimental data to determine the relative masses of different isotopes.
The standard of relative mass is the common isotope of carbon, 12C, which has a relative
mass of 12 units.
The relative isotopic mass (RIM) of an isotope is the mass of an atom of the isotope
relative to the mass of an atom of 12C = 12 units.
A naturally occurring sample of an element contains the same isotopes present in the same
proportions, regardless of its source. The average relative mass (relative to the 12C atom) of
these naturally occurring mixtures of isotopes is called the relative atomic mass (RAM) of
an element, and its symbol is Ar . Since masses are relative, there are no units for RAM.
The relative molecular mass (RMM, M r ) of a compound is the mass of a molecule of that
substance relative to the mass of 12C, i.e. 12. The M r is calculated by adding the relative
atomic masses of the elements in the molecular formula.
For Example: CH 4 = 12 + (4 x 1) = 16.
For non-molecular compounds, relative formula mass (RFM, M r ) is used. This mass is
calculated by adding the relative atomic masses of the elements in the formula.
For Example: NaCl = 23 + 35.5 = 58.5.
As relative atomic masses describe the relative masses between particles, we can say that:
12C
= 12
1 mol of 12C = 12 g
Therefore:

The mass of 1 mol of any atom = Ar in g.

The mass of 1 mol of any molecule = M r in g.
i.e. The mole is a measure of the amount of substance that has the same number of particles
as there are atoms in 12 g of 12C.
The mass of one mol of an element or compound is its molar mass. It is the relative mass of
the element or compound expressed in g. The symbol for molar mass is M and the unit of
measurement is g mol-1.
Example: If M  CO2  = 44 g mol-1 we may assume that:

The molar mass of CO2 = 44 g mol-1.

One mol of CO2 has a mass of 44 g.

6.02 x 1023 molecules of CO2 have a mass of 44 g.
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QUESTION 7
The relative formula mass of calcium hydrogen carbonate, Ca(HCO3)2 is closest to:
A
B
C
D
149.1
161.1
162.1
202.2
QUESTION 8
(a)
What is the molar mass of zinc?
(b)
How many zinc atoms are contained in one mole of zinc?
(c)
Calculate the mass in grams of one zinc atom.
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MOLAR MASS (M) AND MOLE
To link a given mass of substance ( m ), its molar mass ( M ) and the amount of substance
( n ), we apply the following rule:
Amount of substance (mol) = mass of substance (g) / molar mass (g mol-1)
n
Where:
m
M
n
m
M
= The mass of substance in grams
= The molar mass (g/mol)
= The amount of substance in mol
M
Amount of Substance
(mol)
Mass (g)
M
Note: The molar mass used must correspond to the substance for which a mass has been
given or is the subject of the question.
QUESTION 9
Calculate the amount, in mol, of substance in 0.20 g of hydrogen fluoride ( HF ).
Solution
n
m
0.20

 0.010 mol
M 19.9984
QUESTION 10
Calculate the amount, in mol, of CO2 molecules present in 8.8 g of CO2 .
Solution
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QUESTION 11
Calculate the mass of 0.10 mol of CO2 molecules.
Solution
QUESTION 12
What amount, in mol, of oxygen atoms are there in 36 g of glucose (C 6 H 12 O6 ) ?
M (C 6 H 12 O6 )  180 g / mol .
Solution
n  C6 H12O6  
m 36

 0.20 mol
M 180
1 mole C6 H12O6 contains 6 mol O atoms
 nO   6  nC 6 H 12 O6   6  0.2  1.2 mol
QUESTION 13
The amount, in mol, of chloride ions in 3.85 g of FeCl3 is closest to:
A
B
C
D
0.0079
0.024
0.036
0.072
M ( FeCl3 )  162.206 g / mol
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QUESTION 14
Find the amount of H 2 O in 20.0 g of CuSO 4 .5 H 2 O .
M (CuSO4 .5 H 2 O )  249.5 g / mol
M (CuSO 4 )  159.5 g / mol
Solution
QUESTION 15
A sample of Fe2O3 contains 0.60 mole of oxide ions. The total mass of the sample Fe2 O3
is closest to:
A
B
C
D
29 g
32 g
64 g
144 g
M ( Fe2 O3 )  159.694 g / mol
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QUESTION 16
In 1.7 g of NH3, find:
(a)
The total number of atoms.
(b)
The number of H atoms.
(c)
The number of N atoms.
(d)
The mass of H.
(e)
The number of mol of N.
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QUESTION 17
A cup of tea contains 5.6 g of sucrose C12H22O11. Calculate the:
(a)
Number of sucrose molecules present.
(b)
Number of H atoms present.
(c)
Mass of C present.
(d)
Number of mol of O atoms present.
(e)
Total number of atoms present.
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QUESTION 18
In exactly 2.0 g of hydrogen gas, H 2 , there are approximately:
A
6.023 x 1023 hydrogen molecules.
B
1.2 x 1024 hydrogen molecules.
C
6.023 x 1023 hydrogen atoms.
D
6.023 x 1023 protons.
QUESTION 19
What is the mass of 6.02 x 1023 molecules of CO2 ?
Solution
Number of Particles  n( particles )  N A
n(CO2 ) 
6.02 1023
 1.00 mol
NA
m(CO2 )  n  M  1 44  44.0 g
QUESTION 20
Calculate the number of ions in 5.85 g of NaCl.
Solution
Number of Particles  n( particles )  N A
n
m 5.85

 0.100 mol
M 58.5
n(ions)  2  n(NaCl)  2  0.100  0.200 mol
Number of ions  0.200  N A  1.20 1023
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QUESTION 21
Calculate the number of chloride ions in 4.75 g of AlCl3.
Solution
QUESTION 22
Calculate the number of O atoms in 5.7 x 1024 H 2 SO 4 molecules.
Solution
QUESTION 23
Calculate the mass of O in 1.73 x 1024 molecules of H2SO4.
Solution
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QUESTION 24
Calculate the mass of one P4 O10 molecule.
Solution
QUESTION 25
An atom weighs 4.49 x 10-23 g. The element is most likely:
A
B
C
D
Hydrogen
Carbon
Aluminium
Oxygen
QUESTION 26
How many molecules of water are present in 20.00 g CuSO 4 .5 H 2O ?
Solution
no H 2O  nCuSO4. .5H 2O  N A  C.S.
nCuSO4. .5H 2 O  

m
M
20.00
 0.080096 mol
249.7
 no H 2 O  0.080096 6.02310  5
23
 2 .4121  10 23
 2 .412  10 23 molecules
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QUESTION 27
Calculate the mass in grams of:
(a)
(b)
1.204  10 24 barium atoms, correct to 3 significant figures.
One barium atom, correct to 3 significant figures.
Solution
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DETERMINING FORMULAE OF COMPOUNDS
PERCENTAGE COMPOSITION
The percentage composition of an element in a compound is simply the mass of that
element in 100 grams of a sample.
The percentage by mass of an element in a compound is constant for all pure samples irrespective of their source. We can therefore use Relative Atomic Masses to calculate the
percentage composition of an element in a compound.
% of an element = mass of that element in one mole of the given compound x 100
by mass
mass of one mole of the given compound
or more simply:
% composition =
(M of element) x The number of atoms of that element in the given compound x 100
M of the given compound
QUESTION 28
What is the percentage (by mass) of nitrogen in ammonium sulphate?
Solution
Mr (( NH 4 ) 2 SO4 )  132
Mr ( N )  14
 %N 
14  2
 100%  21.2%
132
QUESTION 29
The percentage by mass of carbon in butane, C4 H10 , is approximately:
A
B
C
D
40%
41%
80%
83%
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QUESTION 30
Calculate the percentage composition by mass of compounds with the formulae:
(a)
H 2O
(b)
Al2 (CO3 )3
QUESTION 31
(a)
The relative atomic mass of carbon is 12.0. Explain the meaning of relative atomic
mass and show clearly how it is different from relative isotopic mass.
(b)
Calculate the percentage composition of carbon in carbonic acid (H2CO3).
Solution
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QUESTION 32
A sample of fertiliser was analysed and found to contain 80% by mass of ammonium nitrate
( NH 4 NO3 ) and 20% by mass of potassium chloride ( KCl ). The mass of nitrogen in a
1.00 kg packet of the fertiliser is:
A
B
C
D
140 g
175 g
280 g
350 g
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QUESTION 33
An industrial chemist working for a mining company analysed a nickel ore sample by
dissolving it in acid, then precipitating the nickel as the compound Ni(C4H7N2O2)2.
A 6.372 g sample of ore yielded 2.288 g of precipitate after drying. What is the nickel content
of this ore sample expressed as a percentage by mass?
Solution
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Page 22
EMPIRICAL FORMULAE
The empirical formula of a compound is the smallest whole number ratio of the atoms of the
different elements that make up the compound.
Empirical formulae are determined experimentally, usually by determining the mass of each
element present in a given mass of compound. Therefore, to determine the empirical formula
of a compound, an experimentally determined ratio by mass must be converted to a ratio by
numbers of atoms. This is done by calculating the amount, in mol, of each element.
METHOD:
Step 1: Set up the elements present as a ratio.
Step 2: Calculate the mass or percentage composition of each element.
Set up a ratio by mass.
Step 3: Set up a ratio by mol (convert the answers to amounts in mole).
Step 4: Simplify the ratio by dividing through by the smallest number.
Step 5: If required, multiply answers by an appropriate number to obtain the simplest
whole number ratio.
QUESTION 34
Which of the following is not an empirical formula?
A
CH2O
B
CHO
C
C2H5O4
D
(COOH)2
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QUESTION 35
Find the empirical formula of a compound containing:
8.80 g of Cu
2.20 g of O
Solution
Cu
:
O
8.80g
:
2.20g
Find n
8.80
63.5
:
2.20
16
Simplify
0.139
:
0.138
Divide by
smallest ratio
0.139
0.138
:
0.138
0.138
Mass (g)
Simplify
1
1
The empirical formula of the compound is CuO .
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QUESTION 36
Find the empirical formula of a compound containing:
15 g of P
19.35 g of O
Solution
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QUESTION 37
Zinc iodate contains 61.20% iodine and 23.13% oxygen. Calculate the empirical formula of
zinc iodate.
Solution
Zn
:
I
:
O
15 .67 %
15.67 g
:
61.20%
61.20 g
:
23.13%
23.13 g
Find n
15.67
65.37
:
61.20
126.90
:
23.13
15.99
Simplify
0.2397
:
0.4823
:
1.4465
Divide by
smallest ratio
0.2397
0.2397
:
0.4823
0.2397
:
1.4465
0.2397
1
1
:
:
6.03
6
Mass in 100 g
Simplify
2.01
2
The empirical formula of the compound is ZnI 2 O6 .
QUESTION 38
A compound containing only C and O contains 27.3% C. What is the compound’s empirical
formula?
Solution
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QUESTION 39
What is the empirical formula of a compound containing:
32.4 % Na
22.6 % S
45.0 % O
Solution
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QUESTION 40
0.8 g of magnesium is heated and reacts completely with chlorine. 2.26 g of a white powder
forms. What is the empirical formula of the compound?
Solution
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MOLECULAR FORMULAE
The molecular formula describes the actual number of atoms of each element present in a
molecule of a compound.
The molecular formula is always a whole number multiple of the empirical formula, and may
be obtained from the empirical formula if the molar mass of a compound is known.
METHOD:
Step 1: Find the molar mass of the empirical formula unit.
Step 2: Find the molar mass of the compound.
Step 3: Divide the molar mass of the compound by the molar mass of the empirical
formula unit to give the number of empirical formula units.
Step 4: Multiply the empirical formula by the number of empirical formula units.
Molecular Formula  Empirical Formula n
Where n 
RMM
Empirical Formula ( mass )
QUESTION 41
A compound has an empirical formula of CH. If the compound’s molar mass is 78 g mol-1,
what is its molecular formula?
Solution
QUESTION 42
Hydrazine is used as a rocket fuel. Its empirical formula is NH2 and it has a relative
molecular mass of 32. Find its molecular formula.
Solution
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QUESTION 43
A hydrocarbon contains 7.2 g of C and 1.5 g of H . The molar mass of the compound was
found to be 58 gmol 1 . What is the molecular formula of the compound?
Solution
C
:
H
7.2 g
:
1.5 g
Find n
7.2
12
:
1.5
1
Simplify
0.6
:
1.5
Divide by
smallest ratio
0.6
0.6
:
1.5
0.6
Mass (g)
Simplify
1
2.5
2
5
The empirical formula of the compound is C2 H10 .
Molecular Formula   Empirical Formula n
M  C2 H 5   29 gmol 1
M  compound   58 gmol 1
Number of empirical formula units  n 
58
RMM

2
Empirical Formula ( mass ) 29
Therefore, Molecular Formula   C2 H5  2  C4 H10
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QUESTION 44
Ethylene glycol is a compound used as antifreeze in car motors in cold weather. Its relative
molecular mass is 62 g/mol and its percentage composition is 38.7% carbon, 9.7% hydrogen
and 51.6% oxygen. Find its molecular formula.
Solution
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QUESTION 45
The odour of mothballs is due to naphthalene, a hydrocarbon containing 93.8 % carbon by
mass. 0.200 mol of the compound has a mass of 25.6 g. Find the molecular formula of
naphthalene.
Solution
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QUESTION 46
The chemical analysis of a sugar that is found in blood shows that it consists of 53.34%
oxygen, 40.00% carbon and 6.66% hydrogen by mass.
(a)
Use this information to determine the empirical formula of this sugar.
(b)
Given that 0.034 mol of this sugar has a mass of 6.12 g, determine the molecular
formula of the sugar.
Solution
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QUESTION 47
A compound is known to contain only sodium, sulphur and oxygen. A 100 g sample of the
compound contains 17.04 g of sodium and 47.1 g of sulphur. Calculate the empirical formula
of the compound. If the compound has a relative molecular mass of 270, find its molecular
formula.
Solution
1.
Determine the mass of each element
Mass of sodium = 17.04 g in 100 g of the compound
Mass of sulphur = 47.41 g in 100 g of the compound
Mass of oxygen = 100 g – 17.04 g – 47.41 g = 35.55 g
2.
Calculate the amount, in mole, of each element:
Na
3.
:
mass
molar mass
17.04
:
23.0
moles
0.741 :
S
47.41
32.1
1.48
:
O
:
35.55
16.0
:
2.22
Divide through by the smaller number to find the simplest whole number ratio.
Na
:
0.741
:
0.741
moles
1
ratio
S
1.48
0.741
:
2
:
:
:
O
2.22
0.741
3
The empirical formula of the compound is NaS2O4.
4.
Note the empirical formula, and add the relative atomic masses.
 RAM NaS O   23.0  2  32.1  3  16.0  135.2 amu
2
5.
3
Divide the relative molecular mass by the relative mass of the empirical formula.
Groups of NaS2O3 in the compound 

RMM  compound 
RMM  NaS 2O3

270
 2 groups of NaS2O3
135.2
There are 2 groups of NaS2O3 in each molecule, so the molecular formula of the compound
is Na2S4O6.
© The School For Excellence 2017
Succeeding in the VCE – Unit 1 Chemistry
Page 34
QUESTION 48
1.00 g of vitamin C was burnt forming 1.50 g of CO2 and 0.408 g of H2O. Vitamin C contains
C, O and H.
(a)
Calculate the mass of carbon in the sample.
(b)
Calculate the mass of hydrogen in the sample.
(c)
Use your answers from (a) and (b) to determine the mass of oxygen in vitamin C.
(d)
Determine the empirical formula of vitamin C.
© The School For Excellence 2017
Succeeding in the VCE – Unit 1 Chemistry
Page 35
SUCCEEDING IN THE VCE 2017
UNIT 1 CHEMISTRY
STUDENT SOLUTIONS
FOR ERRORS AND UPDATES, PLEASE VISIT
WWW.TSFX.COM.AU/VCE-UPDATES
QUESTION 1
(a)
(b)
(c)
2
1.48.7
9.4  105
QUESTION 2
Answer is C
QUESTION 4
(a)
6.0 mol
(b)
18 mol
(c)
3.0 mol
(d)
3.6  1024 atoms
(e)
1.1  1025 atoms
(f)
1.8  1024 atoms
(g)
1.6  1025 atoms
QUESTION 5
6.02 x 1023
QUESTION 6
6.2  103 mol
QUESTION 7
Answer is C
QUESTION 8
(a)
65.37 g mol-1
(b)
6.02 1023
(c)
1.086  10-22g
QUESTION 10
0.20 mol
QUESTION 11
4.4 g
QUESTION 13
Answer is C
QUESTION 14
0.401 mol
© The School For Excellence 2017
Succeeding in the VCE – Unit 1 Chemistry
Page 1
QUESTION 15
Answer is B
QUESTION 16
(a)
2.4  1023
(b)
1.8  1023
(c)
6.0  1022
(d)
0.30 g
(e)
0.10 mol
QUESTION 17
(a)
9.9  1021
(b)
2.2  1023
(c)
2.4 g
(d)
0.18 mol
(e)
4.4  1023
QUESTION 18
Answer is A
QUESTION 19
44.0 g
QUESTION 20
1.20  1023
QUESTION 21
6.43  1022
QUESTION 22
2.3  1025
QUESTION 23
184 g
QUESTION 24
4.72  10-22 g
QUESTION 25
Answer is C
QUESTION 27
(a)
(b)
275 g
2.28  10-22 g
QUESTION 29
Answer is D
QUESTION 30
(a)
%H = 11.1%
%O = 88.9%
(b)
%Al = 23.1%
%C = 15.4%
%O = 61.5%
© The School For Excellence 2017
Succeeding in the VCE – Unit 1 Chemistry
Page 2
QUESTION 31
(a)
The RIM is the mass of an isotope relative to 12C which is 12 exactly (1), this is the
reference point (1). Naturally occurring C is a mixture of isotopes with a different RIM
and abundances (1). The RAM of C is a weighted average of these RIM (1).
(b)
% C = 19.4%
QUESTION 32
Answer is C
QUESTION 33
% Mass(Ni) = 7.302%
QUESTION 34
Answer is D
QUESTION 36
P2 O5
QUESTION 38
CO2
QUESTION 39
Na2 SO4
QUESTION 40
Mg4Cl5
QUESTION 41
C6H6
QUESTION 42
N2H4
QUESTION 44
C2 H6 O2
QUESTION 45
C10 H8
QUESTION 46
(a)
CH2O
(b)
C6H12O6
QUESTION 48
(a)
184 g
(b)
184 g
(c)
184 g
(d)
C3 H4 O4
© The School For Excellence 2017
Succeeding in the VCE – Unit 1 Chemistry
Page 3