Chap 1: Practice Module Sem 1 KMKt 2022/23

1.1: Atoms and Molecules PRACTICE MODULE CHAPTER 1 SK015

CHAPTER 1 : MATTER

Mind map / I-Think Map

WORKSHEET 1.1

Isotopic Notation

1. Define isotope
Two or more atoms of the same element having same proton number but different nucleon
number @ number of neutrons

2. Here are three isotopes of an element:

162 163 164
1

PRACTICE MODULE CHAPTER 1 SK015

a. The element is Carbon
b. The number 6 refers to the proton number @ atomic number
c. The numbers 12, 13, and 14 refer to the atomic mass
d. How many protons and neutrons are in the first isotope? P=6 N=6
e. How many protons and neutrons are in the second isotope? P=6 N=7
f. How many protons and neutrons are in the third isotope? P=6 N=8

3. Give the isotope notation for the isotope of potassium with A=40
4109

4. Write down the isotope symbols of oxygen with proton number of 8 and nucleon number of
16.
186

5. Complete the table below using suitable answer

Atom Protons Neutrons Electrons Nucleon Charge Isotopic
Number Notation
Pb 82 125 80 +2
Se 34 45 34 207 0 28027Pb2+
Cr 24 28 21 79 +3
F 9 10 9 52 0
Nb 41 52 36 19 +5 +
P 15 16 18 93 -3
Rb 37 48 36 31 +1
Pd 46 60 46 85 0 4931Nb5+
Os 76 114 72 106 +4 1315P3−
K 19 20 19 190 0 +
Mo 42 54 36 39 +6
Sg 106 159 106 96 0 +
Fr 87 136 87 265 0
Hg 80 121 78 223 +2 +
Xe 54 77 54 201 0
131
+


2

PRACTICE MODULE CHAPTER 1 SK015

Interpret Mass Spectrum, Calculate Average Atomic Mass & Relative Atomic Mass

1. Based on the mass spectrum of zirconium above, calculate the average atomic mass of

zirconium. (91.3 amu)

= ∑( )

∑

= (51.5 90)+ (11.2 91)+ (17.1 92)+ (17.4 94)+ (2.8 96)

100

= 91.3 amu

2. Naturally occurring chlorine is 75.78% 35Cl, which has an atomic mass of 34.969 amu, and

24.22% 37Cl, which has an atomic mass of 36.966 amu. Calculate the average atomic mass of

chlorine. (35.45 amu)

= ∑( )

∑

= 75.78 34.969) + (24.22 36.966)

100

= 35.45 amu

3. Naturally occurring magnesium has the following isotopic abundances:

Isotope Abundance (%) Atomic Mass (amu)

24Mg 78.99 23.985
25Mg 10.00 24.985
26Mg 11.01 25.983

a) What is the average atomic mass of Mg? (24.30 amu)

= ∑( )

∑

= (78.99 23.985) + (10.00 24.985) + (11.01 25.983)

100

= 24.30 amu

b) Based on data from table above, sketch the mass spectrum of Mg.

3

PRACTICE MODULE CHAPTER 1 SK015

Abundance 100 78.99 mass spectrum of Mg
80
m/e
60

40
20 10 11.01

0
isotope mass

4. Boron has two naturally occurring isotopes with the natural abundances shown in the table
below:

Isotope Natural abundance (%)
10B 19.9
11B 80.1

Calculate the relative atomic mass of boron.

= ∑( )

∑

= (10 19.9) + (11 80.1)

100

= 10.80 amu

= ( )

1 − 12 ( )
12
10.80
=
1
12 12

= 10.80

5. Lithium has two naturally occurring isotopes: 6Li (7% abundance) and 7Li (93% abundance).

Calculate the relative atomic mass of lithium.

= ∑( )

∑

= (6 7) + (7 93)

100

= 6.93 amu

= ( )

1 − 12 ( )
12
6.93
=
1
12 12

= 6.93

4

PRACTICE MODULE CHAPTER 1 SK015

6. Rubidium has a relative atomic mass of 85.47 and consists of two naturally occurring

isotopes, 85Rb (Mass = 84.91 amu) and 87Rb (Mass = 86.91 amu). Calculate the percentage

composition of these isotopes in a naturally occurring sample of rubidium.

Lets,

x = % of 85Rb

X – 100 = % of 87Rb

= ∑( )

∑

85.47 = ( 84.91) + ((100− ) 86.91)
100

X = 72%

So,
% abundance of 85Rb = 72%
% abundance of 87Rb = 100 – 72

= 28%

7. Iridium has a relative atomic mass of 192.22 and consists of Ir-191 and Ir-193 isotopes.

Calculate the percentage composition of a naturally occurring sample of iridium.

Lets,

x = % of 191Ir

X – 100 = % of 193Ir

= ∑( )

∑

192.22 = ( 191) + ((100− ) 193)
100

X = 39%

So,
% abundance of 191Ir = 39%
% abundance of 193Ir = 100 – 39

= 61%

5

PRACTICE MODULE CHAPTER 1 SK015

1.2 MOLE CONCEPT
I-think map (Definition)

Formula Formula

Simplest Empirical Atoms Exact Molecular atoms of
ratio formula of the numbers formula each
elements
elements

compound molecules

Concentration Measurement (I-Think Map)

Molarity, Percentag
M (mol L- e by

1) mass, w/w
(%)

Molality, Concentratio Percentag
m n e by

(mol Kg-1) Mole volume,
fraction of v/v (%)

A, XA.

*Note: You can write formula/definition in
the blank box above.

6

PRACTICE MODULE CHAPTER 1 SK015

WORKSHEET 1.2

Empirical Formula & Molecular Formula

1. Define empirical formula: An empirical formula represents the simplest whole-number ratio of
various atoms present in a compound.

2. Define molecular formula: The molecular formula shows the exact number of different types of
atoms present in a molecule of a compound.

3. What is the empirical formula of a compound that contains 0.783 g of C, 0.196 g of H and 0.521 g of O?
(RAM C = 12, H = 1, O = 16)

Substance Carbon Hydrogen Oxygen
Mass (g) 0.521
Number of moles (mol) 0.783 0.196 0.521 =0.0326
0.17283= 0.06525
0196 = 0.196 16
1
0.0326/0.0326 = 1
Smallest ratio 0.06525/0.0326 =2 0.196/0.0326 =6
Empirical formula C2H6O

4. An organic compound that consist of 58.8% carbon, 9.8% hydrogen and 31.4% oxygen has a molar
mass of 102 g mol-1. Determine the empirical and molecular formula of the compound. (C5H10O2)

Assume mass of compound = 100g

Element C H O

Mass (g) 58.8 9.8 31.4
Number of moles 58.8 9.8 31.4
(mol) 12 = 4.9 1 = 9.8 16 = 1.96
Smallest ratio 4.9 9.8
1.96 = 2.5 1.96 = 5 1.96
Ratio 2.5 x 2 = 5 5 x 2 = 10 1.96 = 1
1x2=2
Empirical formula C5H10O2 Molecular formula (C5H10O2)n = 102 g mol-1
n[5(12)+10(1)+2(16)] = 102
n = 102/102
n=1
Molecular formula = (C5H10O2)

5. Compound Y is made up of 39.52% of carbon, 13.14% hydrogen and nitrogen.

i. Determine the empirical formula of compound Y. (CH4N)

Assume mass of compound = 100g

Element Carbon Hydrogen Nitrogen

Mass (g) 39.52 13.14 100-39.52-13.14 = 47.34

Number of moles 39.52= 3.29 13.14 = 13.14 47.34 =3.38
(mol) 1 14
12

Smallest ratio 3.29 = 1 133.2.194= 3.99 ≈ 4 33,.3289= 1.03 ≈ 1
3.29

Empirical formula CH4N

7

PRACTICE MODULE CHAPTER 1 SK015

ii. the molar mass of compound Y is 30 g mol-1. Determine its Molecular formula.

(CH4N)n = 30 g mol-1
n[12 + 4(1) + 14] = 30

n = 1 ∴ CH4N

6. What is empirical formula if compound consist of 21.2% N, 6.1% H, 24.2% S and 48.5% O? (RAM: N
=14, H= 1, S =32, O= 16). (N2H8SO4)

Assume mass of compound = 100g

Element Nitrogen Hydrogen Sulphur Oxygen

Mass (g) 21.2 6.1 24.2 48.5
6.1 24.2 48.6
No. of moles (mol) 21.2 1 = 6.1 32 = 0.756 16 = 3.03
Simplest ratio 14 = 1.514 6.1
1.514 0.756 = 8 24.2 3.03
0.756 = 2 0.756 = 1 0.756 = 4

Empirical formula N2H8SO4

7. Determine the molecular formula of hydrocarbon which contains 85.6% carbon and has a molar mass of
84 g mol-1. (C6H12)

Assume mass of hydrocarbon = 100 g

Element C H

Mass (g) 85.6 100-85.6=14.4
Number of moles (mol)
85.6 14.4
Smallest ratio 12 = 7.13 1 = 14.4
7.13
Empirical formula 7.13 = 1 14.4
Molecular formula CH2 7.13 = 2.01 ≈ 2

(CH2)n = 84 g mol-1

n[12+2(1)] = 84

n = 84/14

n=6

Molecular formula = C6H12

8. 1.44 g sample of an oxide of copper contains 1.28 g of copper. What is the empirical formula of this
oxide of copper? (Ar Cu= 64, O=16). (Cu2O)

Element Cu O
Mass (g)
Number of moles (mol) 1.28 0.16
1.28 0.16
Smallest ratio 64 = 0.02 16 = 0.01

Empirical formula 0.02 0.01
0.01 = 2 0.01 = 1

Cu2O

8

PRACTICE MODULE CHAPTER 1 SK015

9. A sample of hydrated salt was found to contain 12.00g Cu, 6.00 g of S, 12.00g of O and 17.00g of
water. Determine the empirical formula. [Ar : Cu = 63.5, O= 16, S=32, H=1] ()

Element Cu S O H2O
Mass (g)
No. of moles (mol) 12.00 6.00 12.00 17.00
12.00 6.00 12.00 17.00
Simplest ratio 63.5 = 0.19 32 = 0.19
16 = 0.75 18 = 0.94
Empirical formula 0.19 0.19 0.75 0.94
0.19 = 1 0.19 = 1 0.19 = 3.9 ≈ 4 0.19 = 4.9 ≈ 5
CuSO4.5H2O

Empirical and molecular formula involving combustion.

1. After combustion with excess oxygen, a 12.501 g of a petroleum compound produced 38.196 g of
carbon dioxide and 18.752 of water. A previous analysis determined that the compound does not contain
oxygen. Establish the empirical formula of the compound. (C5H12)

Mass of carbon = 38.196 x12 /44 =10.4171 g
Mass of hydrogen = 18.752 x2x1/18 = 2.0836 g

Element Carbon Hydrogen
Mass (g) 10.42 2.08
Mol (mol) 10.42/12 = 0.87 2.08/1 =2.08
Simplest ratio 0.87/0.87 = 1 2.08/0.87 = 2.40
Simplest mol 1x5=5 2.4 x 5 = 12
Empirical formula C5H12

2. Combustion of 0.202 g of an organic compound produce 0.361 g CO2 and 0.147 g H2O. Calculate the
empirical formula for this compound. [Ar: C=12.01, H=1, O=16.00]. (C3H6O2).

Mass of carbon = 0.361 x 12/44 =0.0985 g
Mass of hydrogen = 0.147 x2x1/18 = 0.0163 g

Element C H O
Mass (g)
0.0985 0.0163 0.202 – 0.0985 – 0.0163 =
Number of moles 0.0872
(mol) 0.0985 0.0163
12.01 1 = 0.0163 0.0872 = 5.45 × 10−3
Smallest ratio = 8.20 × 10−3
8.20 × 10−3 0.0163 16.00
Ratio 5.45 × 10−3 = 1.5 5.45 × 10−3 = 2.99
Empirical formula 2.99 ≈ 5.45 × 10−3
1.5 x 2 = 3 2.99 x 2 = 5.98≈ 5.45 × 10−3 = 1
C3H6O2
1x2=2

9

PRACTICE MODULE CHAPTER 1 SK015

3. A 0.1 g sample of ethyl alcohol known to contain only carbon, hydrogen and oxygen, was burnt
completely in excess oxygen to form the products 0.1910g CO2 and 0.1172 g H2O. What is the empirical
and molecular formula of the compound if the molecular mass is 138 g mol-1. State the number of
hydrogen atoms per molecule present in the above sample. (Empirical formula = C2H6O, Molecular
Formula= C6H18O3, H=18 atoms)

Mass of carbon = 0.191g x 12/44 = 0.0521 g

Mass of hydrogen = 0.1172 x 2x1/18 = 0.0130 g
Mass of oxygen = 0.1 g – 0.0521 g – 0.0130 g = 0.035 g

Element CH O
Mass (g)
Number of moles 0.0521 0.0130 0.0349
(mol)
Smallest ratio 0.0521 0.013 0.035 = 2.181 × 10−3
Empirical formula 12 1 = 0.0130 16
Molecular formula
= 4.342 × 10−3
Number of H atoms
4.342 × 10−3 0.0130 2.188 × 10−3
2.188 × 10−3 = 2 2.188 × 10−3 = 6 2.188 × 10−3 = 1

C2H6O

(C2H6O)n = 138 g mol-1

n[2(12)+6(1)+16] = 138

n = 138/46

n=3

C6H18O3
18 atoms

4. Sometimes, athletes illegally use anabolic steroids to increase muscle strength. A forensic chemist
analyzes some tablets suspected of being a popular steroid. He determined that the substance in the tablets
contains only carbon, hydrogen, and oxygen in and has a molar mass of 300.14 g/mole. When a 1.200 g
sample of this study by combustion analysis, 3.516 g of CO2 and 1.007 g of H2O are collected. What is the
molecular formula for the substance? (C22H28O2)

Mass of carbon = 3.516 x12 /44 =0.9589 g
Mass of hydrogen = 1.007 x2x1/18 = 0.1118 g
Mass of oxygen = 1.2-0.9589-0.1118 = 0.1293 g

Substance Carbon Hydrogen oxygen
Mass (g) 0.1293
Mol (mol) 0.9589 0.1118 0.1293/16=0.008
Simplest mol 1
Empirical formula 0.9589/12 = 0.0799 0.1118/1 =0.1118

Molecular formula 10 14

C10H14O

(C10H14O) n = 300.14g
150n =300.14

n=2

(C10H14O)2 = C20H28O2
∴ Molecular formula = C20H28O2

10

PRACTICE MODULE CHAPTER 1 SK015

Concentration Measurement

Molarity

1. Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulphate Na2SO4 in
enough water to form 125 mL of solution? (1.32 M).

= ( ) , = ( )
( )
( )

= 23.4 = 0.1648
2(23)+32+4( 1 6 )
0.125

= 0.1648 mol = 1.184 M

2. Calculate the molarity of each of the following solutions:
a. 16.4 g CaCl2 in 0.614 L solution. (0.24 M)

= ( ) , = ( )
( )
( )
16.4
= 40.1+2(35.5) = 0.1476

0.614

= 0.1476 mol = 0.24 M

b. 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L. (1.15 M)

DILUTION, M1V1 = M2V2

= = ( . )( . ) =

.

3. Calculate the mass of sodium carbonate, Na2CO3 required to prepare 250 mL of 0.5 M solution.
(13.25)
Mass of solute(Na2CO3) = ? , Vsolution = 250 mL, Molarity = 0.5 M

, = ( )

( )

Number of moles of solute = M × V
= (0.5 mol L-1)(250 mL ÷ 1000)
= 0.125 mol

Mass of solute(Na2CO3) = number of moles × molar mass
= 0.125 mol × [2(23)+12+3(16)]g mol-1
= 13.25 g

4. What volume (in mL) of sulphuric acid which concentration is 2M and contain 12.6 g of the acid?
(64.2)

11

PRACTICE MODULE CHAPTER 1 SK015

= ( ) , = ( )
( )
( )
12.6
= [2(1)+32.1+4(16)] = 0.1284 = 0.0642

2 /

= 0.1284 mol Volume of solution = 64.2 mL

7. Calculate the mass of sodium hydroxide that needs to be added to 150 g of water to produce 1.2
molal of solution? (7.2g)

, = ( ) = ( )
( )
( )

1.2 / = Mass of NaOH = 0.18 mol x [23+16+1] g/mol
0.15 = 7.2 g

Number of moles of solute = 0.18 mol

8. What is the molal concentration of a solution prepared by dissolving 0.30 mol of CuCl2 in 40.0 mol
of water? (0.42)

= ( ) , = ( )
( )
( )

Mass of water = 40.0 mol x 18 g/mol = 0.30
= 720 g → 0.72 kg 0.72

= 0.42 mol/kg

Calculate the molality of the following aqueous solutions, 0.840 M sugar (C12H22O11) solution

(density=1.12 g/mL) (1.01 molal)

▪ Assume volume of solution = 1000 mL Mass of C12H22O11 (solute)

= = 0.84 x [12(12)+22(1)+11(16)] g mol-1
= 287.28 g


1.12 / = Mass of solvent = mass of solution – mass of solute

1000 = 1120 g – 287.28 g =832.72 g

Mass of solution = 1120 g ( )

= ( )
( )
= 0.84
= ( ) 0.8327

0.840 M = = 1.01 mol/kg

1

12

PRACTICE MODULE CHAPTER 1 SK015

= 0.840 mol

9. A solution containing 121.8 g of Zn(NO3)2 per litre has a density of 1.107 g/mL. Calculate its molal
concentration. (0.653)

= Mass of solvent = mass of solution – mass of solute



= 1107 g – 121.8 g =985.2 g

1.107 / = 1000 = ( )

Mass of solution = 1107 g ( )

( ) = 0.6431
= ( )
0.9852
121.8
= [65.4 + 2(14 + 3 16)] ( ) = 0.653 mol/kg

= 0.6431 mol

Percentage by mass

13. Calculate the percent by mass of the solute in a aqueous solution containing 5.50 g of NaBr in 78.2
g of solution?

Solution:

= × % Mass of solute (NaBr) = 5.50g
Mass of solution = 78.2 g

= . × %

.

= 7.03%

14. Hydrochloric acid can be purchased as a solution of 37% HCl. What mass of this solution contains

7.5 g of HCl? (20.27 g)

Solution:
Mass of solute(HCl) = 7.5 g ⁄ % = 37%

= × %


% = . × %


Mass of solution = 20.27 g

15. A solution contains 66% H2SO4 by weight and has density of 1.58 g mL-1. How many moles of the

acid present in 1.0 L of the solution? (10.64)

Solution:
⁄ % = 66% , Density of solution = 1.58 g mL-1, Vsolution = 1.0 L ≡ 1000 mL

13

PRACTICE MODULE CHAPTER 1 SK015

= × % =



% = × % 1.58 / =

1000

Mass of solute = 1042.8 g Mass of solution = 1580 g
1042.8

2 4 = 2(1) + 32 + 4(16) /
= 10.64 mol

16. Calculate the amount of water(in gram) that must be added to 5.00 g of urea, (NH2)2CO in the

preparation of 16.2 percent by mass. (25.9)

Solution:
⁄ % = 16.2%, mass of urea, (NH2)2CO = 5.00g, mass of water =??

= × %



. % = . × %



Mass of solution = 30.8642 g.
Since Mass of solution = mass of solute + mass of solvent(H2O)

∴ .

17. The density of 20% by mass of nitric acid solution HNO3 is 1.11 g/mL. Calculate the volume acid
needed to prepare 2.0 L of 3.00 mol L-1 HNO3 solution.

Solution:

⁄ % = 20% ,Density of solution = 1.11 g mL-1, Vsolution = 2.0 L ≡ 1000 mL Molarity = 3 M, VHNO3 = ??

, = ( ) Mass of HNO3 = 3 × molar mass
( ) = 6 mol × 63 g mol-1 = 378 g

number of moles of HNO3 = 3.00 mol L-1 × 2L

= 6 mol

= × 100% =



20% = 378 × 100% 1890
1.11 / =

Mass of solution = 1890 g Volume of solution = 1702.7 mL

14

PRACTICE MODULE CHAPTER 1 SK015

18. A mixture containing benzene and toluene has 18.4 g of toluene and its percentage composition is
30%. Calculate the number of moles of benzene in this solution. (0.55)
Solution:

= × 100% ( )
= ( )


30% = 18.4 × 100%



Mass of solution = 61.33 g C6H6= 42.93
[6(12)+6(1)] ( )
Mass of solution = mbenzene + mtoluene

mbenzene = 61.33 -18.4 = 42.93g = 0.55 mol

Percentage by volume

19. Calculate the percentage by volume (% v/v) of ethanol in a solution containing 18.6 cm3 of ethanol
in 120 cm3 solution. (15.5)

Solution:

, ⁄ % = × %


= . × % = . %


20. Rubbing alcohol is commonly used as an antiseptic for a small cuts. It is sold as 70% by volume
solution of isopropyl alcohol in water. Calculate the volume of isopropyl alcohol use to make 750
mL of rubbing alcohol? (525 mL)

Solution:

, ⁄ % = × %


70% = × %


Volume of isopropyl alcohol = 525 mL

Mole fraction

21. What is the mole fraction of CuCl2 in a solution prepared by dissolving 0.30 mol of CuCl2 in 40.0
mol of H2O? (0.007)

Solution:

Mole fraction of CuCl2, =


= . = 0.007

( . + )

22. A solution is prepared by mixing 55g of toluene, C7H8 and 55 g of bromobenzene C6H5Br. What is
the mole fraction of each component? (0.63, 0.37)
Solution:

15

PRACTICE MODULE CHAPTER 1 SK015

( ) nT = 0.5978 + 0.3505 =0.9483 mol
= ( )
Mole fraction of C7H8 , =


C7H8= 55 = . = 0.63
7(12)+8(1) ( )
.

= 0.5978 mol Mole fraction for C6H5Br,
= 1-0.63
C6H5Br =
= 0.37
55
[6(12)+5(1)+79.9] ( )

= 0.3505 mol

23. Calculate the mole fractions of each compound of the following solutions:
19.4 g of H2SO4 in 0.251 L of H2O (density of water is 1 g/mL) (H2SO4 =0.014, H2O=0.986)

( ) nT = 0.1978 + 13.9444 =14.1422 mol
= ( )
Mole fraction of C7H8 , =


H2SO4= 19.4 = . = 0.014
[2(1)+32.1+4(16)] ( ) .

= 0.1978 mol Mole fraction for H2O,
= 1-0.014
▪ =
= 0.986


1.00 / =

251

Mass of water = 251 g

H2O= 251
[2(1)+16] ( )

= 13.9444 mol

24. A solution contains 35% by mass HBr and has a density of 1.30 g mL-1. Calculate the molarity and
molality of this solution. (5.62 M, 4.32 molal)
Solution:

Assumes mass of solution is 100g , = ( )

= × 100% ( )


35% = × 100% = 0.4326
100
0.07692
Mass of solute, HBr = 35 g

( ) = 5.62 M
= ( / )

HBr= 35 Mass of solvent = msolution – msolute
[ 1+ 79.9] / = 100 – 35 g = 65g

= 0.4326 mol

= = ( )

( )

16

PRACTICE MODULE CHAPTER 1 SK015

100 = 0.4326
1.30 / =
Volume of solution = 76.9231 mL 0.065

= 0.0769 L = 6.656 mol/kg

25. An aqueous solution of hydrofluoric acid is 30.0% HF, by mass, and has a density of 1.101 g cm-3.
What are the molarity of HF in this solution?
Solution :

Assume 100.0 grams of solution. Therefore:

30.0 g is HF
70.0 g is H2O

Solution for molarity:
Determine moles of HF in 100.0 g of 30.0% solution:
30.0 g of this solution is HF
30.0 g / 20.0059 g/mol = 1.49956 mol
Determine volume of 100.0 g of solution:
density = mass / volume
1.101 g/mL = 100.0 g / x
x = 90.8265 mL
Determine molarity:

1.49956 mol / 0.0908265 L = 16.5 M
26. A bottle of commercial sulfuric acid (density 1.787 g/ml) is labelled as 86% by weight. What is the

molarity of acid?
Solution:
Determine moles of H2SO4 in 100.0 g of 86% solution:
86 g of this solution is H2SO4
86 g / 98.08 g/mol = 0.876835237 mol
Determine volume of 100.0 g of solution:
density = mass / volume
1.787 g/mL = 100.0 g / x
x = 55.96 mL

17

PRACTICE MODULE CHAPTER 1 SK015

Determine molarity:

0.876835237 mol / 0.05596 L = 15.67 M = 15.7 M (to three sf)

27. Concentrated phosphoric acid is 90% H3PO4 by mass and the remaining mass is water. The
molarity of H3PO4 in 90% H3PO4 is 12.2 M at room temperature.
i. What density of this solution at room temperature? (1.33 g/mL)
Solution:

Assumes mass of solution is 100g , = ( )
= × 100%
( )

12.2 M = 0.9184
90% = × 100%
( )
100
Volume of solution = 0.07528 L
Mass of solute, H3PO4 = 90 g =
( )

= ( / )
= 100
H3PO4= 90
[ 3(1)+ 31+4(16)] / 75.2787

= 0.9184 mol Density of solution = 1.33 g mL-1

ii. What volume (in mL) of this solution is needed to make a 1.00 L of a 1.00 M phosphoric acid? (82.0
mL)
Solution:

DILUTION, M1V1 = M2V2

= = ( . )( . ) = . = .

.

28. A 6.4 molal NaCl solution has a density of 1.2 g cm-3. Calculate
i. the percentage by mass of NaCl. (27.24%)

Solution: Mass of solution = msolvent + msolute
▪ Assume mass of solvent/water = 1kg = 1000 g + 374.4 g
= 1374.4g
, = ( )

( )

6.4 = ( ) = × 100%

1
= 374.4 × 100%
nNaCl = 6.4 mol
mNaCl = 6.4 mol x (23+35.5) g/mol 1374.4

= 374.4 g = 27.24%

ii. the molarity of the solution [Ar = Na=23, Cl=35.5]. (5.588 mol L-1)
Solution:

18

PRACTICE MODULE CHAPTER 1 SK015

= ( )



1374.4 = ( )

1.2 / 3 =

= . 6.4
= 1.1453 L = 1.1453

= 5.588 mol L-1

29. Calculate the molality of a 6.0 mol/L aqueous solution of an acid HA with a density of 0.878 g mL-1.
[Mr HA = 98.0]. (20.69 mol kg-1)

Assume volume of solution = 1000 cm3 = ( )

( )

= 6.0 / =



( ) 1.0
1000 3
0.878 / 3 = , = 6.0 mol

= Mass of solute, HA = 6.0 mol x 98.0 g mol-1
= 588 g

Mass of solvent = mass of solution – mass of solute
= 878 g – 588 g = 290 g

= ( )

( )

= 6.0
0.290

= 20.69 mol/kg

30. Calculate the mol fraction of hydrochloric acid which contain 35% HCl by mass.
Solution:

Assumes mass of solution is 100g Mass of solvent = 100-35 = 65g

= × 100% H2O = 65
[ 2(1)+ 16] /


35% = × 100% = 3.6111 mol
100

Mass of solute, HCl = 35 g

( ) nT = 0.9589 + 3.6111 =4.57 mol
= ( / )
Mole fraction of HCl , =

35
HCl = [ 1+ 35.5] / = . = 0.21

= 0.9589 mol .

19

PRACTICE MODULE CHAPTER 1 SK015

1.3 STOICHIOMETRY Circle Map

LIMITING REACTANT Limiting
1. Limiting reactant is the reactant that reactant
completely used in the chemical reaction
and limit the amount of product formed. By
using definition of limiting reactant,
complete your I-think map(circle map).

Other important formula


= × %

WORKSHEET 1.3

Balancing chemical equation

1. Give oxidation numbers for the underlined atoms in these molecules and ions:

a. NaClO4 (+7) b. PtCl62- (+4) c. CaI2 (+2) d. SnF2 (+2)

e. Al2O3 (+3) f. ClF3 (+3) g. H3AsO3 (+3) h. SbF6 ̄ (+5)

i. KNO3 (+5) j. MnO2 (+4) k. MnO42- (+6) l. K2Cr2O7 (+6)

2. Balance the following chemical equation.
a) 2Al + 3/2O2 → Al2O3
b) Al(NO3)3 + 3NaOH → Al(OH)3 + 3NaNO3
c) 2KNO3 → 2KNO2 + O2
d) 3O2 + CS2 → CO2 + 2SO2
e) 2KClO3 → 2KCl + 3O2
f) 3BaF2 + 2K3PO4 → Ba3(PO4)2 + 6KF
g) H2SO4 + Mg(NO3)2 → MgSO4 + 2HNO3
h) 2HCN + CuSO4 → H2SO4 + Cu(CN)2
i) GaF3 + 3Cs → 3CsF + Ga
j) N2 + 3H2 → 2NH3

3. Balance the following chemical equation.
a) MgF2 + Li2CO3 → MgCO3 + 2LiF
b) 2AgNO3 + Cu → Cu(NO3)2 + 2Ag
c) 2AlBr3 + 3K2SO4 → 6KBr + Al2(SO4)3
d) C2H6 + 7/2O2 → 2CO2 + 3H2O
e) 2NH3 + H2SO4 → (NH4)2SO4
f) 6K + B2O3 → 3K2O + 2B
g) 2N2 + 5O2 → 2N2O5
h) NaOH + H2CO3 → Na2CO3 + H2O
i) 16Al + 3S8 → 8Al2S3

20

PRACTICE MODULE CHAPTER 1 SK015

j) Ca3(PO4)2 + SiO2 + C → 3CaSiO3 + CO + 2P
Balancing redox equation

MnO4 - + C2O4 2- → MnO2 + CO2

In Acidic Medium Reactant - MnO4- → Product - MnO2

Step 1: calculate oxidation x + 4(-2) = -1 x + 2(-2) = 0
number one of the species
to identify which reactant x = +7 x = +4
will undergoes reduction or
oxidation Oxidation number of Mn decrease from +7 to +4.

Thus, Mn in MnO4- (+7) underges reduction to MnO2 (+4)

Step 2: Write incomplete Oxidation: C2O42- → CO2
half equations for the Reduction: MnO4- → MnO2
oxidation and reduction
process

Step 3: Balance half Oxidation: C2O42- → 2CO2
equation in terms of the Reduction: MnO4- + 4H+ → MnO2 + 2H2O
number of atoms Oxidation: C2O42- → 2CO2 + 2e-

Step 4: Balance the half Reduction: MnO4- + 4H+ + 3e- → MnO2 + 2H2O
equation in terms of
number of charges by
adding electron.

Step 5: Multiply the half Oxidation: ( C2O42- → 2CO2 + 2e- ) x 3
equation with a suitable 3C2O42- → 6CO2 + 6e-
factor

*both equation should have same Reduction: ( MnO4- + 4H+ + 3e- → MnO2 + 2H2O ) x 2
number of electrons 2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O

Step 6: Combine the half 3C2O42- + 2MnO4- + 8H+ → 6CO2 + 2MnO2 + 4H2O
equation to give balanced
overall equation)

In Basic Medium 3C2O42- + 2MnO4- + 8H+ +8OH- → 6CO2 + 2MnO2 + 4H2O + 8OH-
4
Step 7: add 8OH- on both
side of chemical equation 3C2O42- + 2MnO4- + 8H2O → 6CO2 + 2MnO2 + 4H2O+ 8OH-
to eliminate 8H+. then,
cancel out H2O on both 3C2O42- + 2MnO4- + 4H2O → 6CO2 + 2MnO2 + 8OH-
sides of the equation.

Overall equation in basic
medium

21

PRACTICE MODULE CHAPTER 1 SK015

1. Write the balanced half reactions of the following reactions:

a. Cu + NO3– → Cu2+ + N2O4 in acidic solution

Oxidation: 2NO3- + 4H+ + 2e- → N2O4 + 2H2O
Reduction: Cu → Cu2+ + 2e-

b. CO2 + 2 NH2OH CO + N2 + 3 H2O in basic solution
Oxidation : 2NH2OH N2 + 2H2O + 2H+ + 2e-
Reduction: CO2 + 2H+ + 2e- CO + H2O

c. 2 H+ + H2O2 + 2 Fe2+ 2 Fe3+ + 2 H2O in acidic solution

Oxidation: Fe2+ Fe3+ + e-

Reduction: 2H+ + H2O2 + 2e- 2H2O

d. H+ + 2H2O + 2MnO4- + 5SO2 2 Mn2+ + 5 HSO4- in acidic solution
Oxidation: MnO4- + 8H+ + 5e- Mn2+ + 4H2O
Reduction: SO2 + 2H2O HSO4̄ + 3H+ + 2e-

e. NO2 NO3 - + NO in basic solution

Oxidation: MnO4- + 8H+ + 5e- Mn2+ + 4H2O
Reduction: NO2 + H2O NO3̄ + 2H+ + e-

f. Mn2+ + BiO3- → MnO4- + Bi3+ in acidic solution
Oxidation: Mn2+ + 4H2O → MnO4- + 8H+ + 5e
Reduction: BiO3- + 6H+ + 2e- → Bi3+ + 3H2O

2. Use the ion-electron method to complete and balance the following redox equations, occurring in either
acidic or basic aqueous solution, as indicated. Identify the oxidation and reduction half reactions in
each case.

a. In acidic aqueous solution: XeO3 + BrO3– → Xe + BrO4–
Oxidation : (BrO3– + H2O → BrO4– + 2H+ + 2e- ) x3
Reduction: XeO3 + 6H+ + 6e- → Xe + 3H2O
Overall : XeO3 + 3BrO3– → Xe + 3BrO4–

b. In acidic aqueous solution: MnO42- + CH3OH → Mn2+ + HCO2H
Oxidation : CH3OH + H2O → HCO2H + 6H+ + 6e-

Reduction: MnO42- + 8H+ + 6e- → Mn2+ + 4H2O

Overall : CH3OH + MnO42- + 2H+ → Mn2+ + HCO2H + 3H2O

c. In acidic aqueous solution: Cr2O72– + I- → Cr3+ + IO3–
Oxidation : Cr2O72– + 14H+ + 6e- → 2Cr3+ + 7H2O
Reduction: I- + 3H2O → IO3- + 6H+ + 6e-
Overall : Cr2O72– + I- + 8H+ → 2Cr3+ + IO3- + 7H2O

d. In basic aqueous solution: SO2 + MnO4– → SO 2– + MnO2
Oxidation : SO2 + 2H2O → SO42– + 4H+ + 2e- 4

Reduction: MnO4– + 4H+ + 3e- → MnO2 + 2H2O

22

PRACTICE MODULE CHAPTER 1 SK015

: 3SO2 + 2MnO4– + 2H2O → 3SO42– + 2MnO2 + 4H+
Overall (basic) : 3SO2 + 2MnO4– + 4OH- → 3SO42– + 2MnO2 + 2H2O

3. Balance each redox reaction in acid solution.

a. NO3¯ + Fe2+ → HNO2 + Fe3+
Ans: NO3¯ + 2Fe2+ + 3H+ → HNO2 + 2Fe3+ + H2O

b. MnO4 - + S2O3 2- → S4O62- + Mn 2+-

Ans: 16 H + + 2 MnO4 - + 10 2- →2 Mn2+ + 8 H2O + 5 S4O6 2-

S2O3

c. ClO3 - + Cl - → Cl2 + ClO2
Ans: 4 H + + 2 ClO3 - + 2 Cl - →2 ClO2 + 2 H2O + Cl2

4. Balance each redox reaction in Basic Solutions
a. Zn + NO3- → Zn(OH)42- + NH3
Ans: 4Zn + 6H2O + NO3- + 7OH- → 4Zn(OH)42- + NH3

b. Cu(NH3)42+ + S2O42- → SO32- + Cu + NH3
Ans: Cu(NH3)42+ + S2O42- + 4OH- → 2SO32- + Cu + 4NH3 +2H2O

5. Write balanced equations for the following reactions:

a. Cr(OH)3 + Br2 CrO42- + Br- in basic solution
Oxidation : Cr(OH)3 + H2O → CrO42- + 5H+ + 3e-
Reduction: Br2 + 2e- → 2Br ̄
Overall Basic: 2Cr(OH)3 + 3Br2 + 10OH- → 2CrO42- + 8H2O + 6Br¯

b. O2 + Sb H2O2 + SbO2- in basic solution
Oxidation : Sb + 2H2O → SbO2- + 4H+ + 3e-

Reduction: O2 + 2H+ + 2e- → H2O2
Overall: 2Sb + 2H2O + 3O2 + 2OH- → 2SbO2- + 3H2O2

c. ClO2- ClO2 + Cl- in acidic solution

Oxidation: ClO2- ClO2 + e-

Reduction: ClO2- + 4H+ + 4e- Cl- + 2H2O

Overall: 5ClO2- + 4H+ Cl- + 4ClO2 + 2H2O

23

PRACTICE MODULE CHAPTER 1 SK015

LIMITING REACTANT

1. Suppose 316.0 g aluminum sulfide reacts with 493.0 g of water. Determine limiting reactant. What mass of the
excess reactant remains? Calculate percentage yield of H2S if actual mass of H2S is 197.0 g

STEP 1 Al2S3 + 6H2O → 2Al(OH)3 + 3H2S
Determine limiting reactant

Calculate given moles for both reactant = ( )
( )

2 3 = 316
[2(27)+3(32.1)] ( )

= 2.1025 mol

493
2 = [2(1) + 16] ( )

= 27.3889 mol

STEP 2 Write relationship between reactants 1 mol of Al2S3 react with 6 mol of H2O
from the balance chemical equation

STEP 3 Determine mole of H2O needed using 1 mol of Al2S3 react with 6 mol of H2O
given mole of Al2S3 (or vise versa)

2.1025 mol of Al2S3 react with x mol of H2O

= 2.1025 × 6 2 = 12.615 mol of H2O(needed)
1

STEP 4 Compare mole H2O needed with mole Number of moles of H2O needed is less than number of
H2O given moles of H2O available. H2O is an excess reactant. Hence,
Al2S3 is the limiting reactant.

STEP 5 By using limiting reactant, determine 1 mol of Al2S3 react with 6 mol of H2O
mole/mass of excess reactant used in the 2.1025 mol of Al2S3 react with x mol of H2O
reaction

= 2.1025 × 6 2 = 12.615 mol of H2O(needed)
1

STEP 6 Calculate mass of excess reactant remain Number of moles of excess reactant = navailable- nneeded
=27.3889 – 12.615 = 14.7739 mol H2O

Mass of excess H2O = 14.7739 x 18 g/mol
= 265.93 g

Calculate percentange yield of H2S if actual mass of H2S is 197.0 g

24

PRACTICE MODULE CHAPTER 1 SK015

STEP 7 Write formula of percentage yield
= × %
STEP 8 Write relationship between limiting
reactant and H2S Use limiting reactant to calculate the number of moles
of product formed.

1 mol of Al2S3 produces 3 mol of H2S

STEP 9 Determine mole/mass of H2S based on Use limiting reactant to calculate the number of moles
mole limiting reactant (theoritical yield) of product formed.

1 mol of Al2S3 produces 3 mol of H2S

2.1025 mol of Al2S3 produces x mol of H2S

= 2.1025 × 3 H2S = 6.3075 mol of H2S
1

Mass of H2S = 6.3075 mol × [2(1)+32.1] g/mol

STEP Determine % yield of H2S = 215.09 g (theoretical yield)
10 Given actual yield = 197.0 g


= . × %

= 91.6%

2. Nitric acid react with oxygen gas to give nitrogen dioxide (NO2), a dark brown gas:
2NO + O2 → 2NO2

In one experiment 0.886 mol of NO is mixed with 0.503 mol of O2.
i. Calculate which of the reactant is the limiting reactant.

Solution:

= 0.886 ( ) and 2 = 0.503 ( )

From the equation:

2 mol of NO react with 1 mol of O2

0.886 mol of NO react with x mol of O2

= 0.8860 × 1 2 = 0.4430 mol of 2(needed)
2

Number of moles of O2 needed is less than number of moles of O2 available.
O2 is an excess reactant. Thus, NO is the limiting reactant.

ii. Calculate the number of mol of NO2 produced.
Use limiting reactant to calculate the number of moles of product formed.
2 mol of NO produces 2 mol of NO2
0.886 mol mol of NO produces 0.886 mol of NO2

25

PRACTICE MODULE CHAPTER 1 SK015

3. Consider the reaction:
MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O

If 0.86 mol of MnO2 and 48.2 g HCl react, which reagent will used up first? How many grams of Cl2 will
be produced? (23.4)

Solution:

= ( ) number of moles of MnO2 = 0.86 mol (available)
( )

= 48.2
1+35.5 ( )

= 1.3205 mol (available)

From the equation:
1 mol of MnO2 react with 4 mol of HCl
0.86 mol of MnO2 react with x mol of HCl
= 0.860 × 4 = 3.44 mol of HCl(needed)

1

Number of moles of HCl needed is more than number of moles of HCl available.
HCl is the limiting reactant. Hence HCL will be used up first in this reaction.

Use limiting reactant to calculate the number of moles of product formed.

4 mol of HCl produces 1 mol of Cl2

1.3205 mol of HCl produces x mol of Cl2

= 1.3205 × 1 Cl2 = 0.3301 mol of Cl2
4

Mass of Cl2 = 0.3301 mol × 35.5(2)
= 23.44 g

4. 2.0 g of aluminum was reacted with solution containing 11.5 g H2SO4 to produce Al2(SO4)3 and
hydrogen gas.
i. Balance the chemical equation.
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
ii. Determine the limiting reactant

Solution:

= ( ) 2 4
( ) 11.5

= 2 = 2(1) + 32.1 + 4(16) ( )
27.0 ( )
= 0.1172 mol (available)
= 0.0741 mol (available)

From the equation:

2 mol of Al react with 3 mol of H2SO4

0.0741 mol of Al react with x mol of H2SO4

= 0.0741 × 3 H2SO4 = 0.1112 mol of H2SO4(needed)
2

Number of moles of H2SO4 needed is less than number of moles of H2SO4 available.
H2SO4 is an excesss reactant. Hence Al is the limiting reactant.

26

PRACTICE MODULE CHAPTER 1 SK015

iii. What is the number of mol of hydrogen released?
Solution:

Use limiting reactant to calculate the number of moles of product formed.

2 mol of Al produces 3 mol of H2

0.0741 mol of Al produces x mol of H2

= 0.0741 × 3 H2 = 0.1112 mol of H2
2

iv. Calculate then mass of Al2(SO4)3 produced
Solution:

Use limiting reactant to calculate the number of moles of product formed.

2 mol of Al produces 1 mol of Al2(SO4)3

0.0741 mol of Al produces x mol of Al2(SO4)3

= 0.0741 × 1 Al2(SO4)3= 0.0371 mol of Al2(SO4)3
2

Mass of Al2(SO4)3 = number of moles x molar mass
= 0.0317 x [2(27)+ 3(32.1+4(16))] = 10.85 g

v. Calculate the amount of excess reactant after the reaction completed
Solution:

Amount of excess reactant = number of moles available – number of moles available
= 0.1172-0.1112
= 6 x 10-3 @ 0.59g

5. Write the balanced equation for the reaction that occurs when iron (II)chloride is mixed with sodium
phosphate forming iron (II) phosphate and sodium chloride.
i. If 23 grams of iron (II) chloride reacts with 41 grams of sodium phosphate, what is the
limiting reagent? How much sodium chloride can be formed?
Solution:

( ) Na3PO4 = 41
= ( ) [3(23)+31+4(16)] ( )

23 = 0.25 mol(available)
55.9 + 2(35.5) ( )
FeCl2=

= 0.1812 mol(available)

From the equation:

3FeCl2 + 2Na3PO4 → Fe3(PO4)2 + 6NaCl

3 mol of FeCl2 react with 2 mol of Na3PO4

0.1812 mol of FeCl2 react with x mol of Na3PO4

x= 0.1812 × 2 mol Na3PO4 = 0.1208 mol of Na3PO4 (needed)
3

Number of moles of Na3PO4 needed is less than number of moles of Na3PO4 available.
Na3PO4 is the excess reactant. FeCl2 is limiting reactant.

Use limiting reactant to calculate the number of moles of product formed.
3 mol of FeCl2 produces 6 mol of NaCl

27

PRACTICE MODULE CHAPTER 1 SK015

0.1812 mol of FeCl2 produces x mol of NaCl

x= 0.1812 × 6 mol NaCl = 0.3625 mol of NaCl
3

Mass of NaCl = 0.3625 mol × (23+35.5)
= 21.21 g

ii. How much of the excess reagent remains when this reaction has gone to completion?
Solution:

3 mol of FeCl2 react with 2 mol of Na3PO4

0.1812 mol of FeCl2 react with x mol of Na3PO4

x= 0.1812 × 2 mol of Na3PO4 = 0.1208 mol of Na3PO4 (reacted)
3

Number of moles of excess reagent remains = navailabe - nneeded
= 0.25 mol – 0.1208 mol = 0.1292 mol

Mass of excess reagent remains = 0.1292 mol x [3(23) + 30.97 + 4(16)] g/mol = 21.18 g

iii. If 16.1 grams of sodium chloride are formed in the reaction, what is the percent yield of this
reaction?

Solution:


= × %

= . × % = . %
.

6. Zinc and sulphur react to form zinc sulphide according to the equation.
Zn + S ---------> ZnS

If 25.0 g of zinc and 30.0 g of sulphur are mixed,

i. Which chemical is the limiting reactant?
Solution:

( ) = 30
= ( ) 32.1 ( )

25 = 0.9346 mol(available)
= 65.4 ( )

= 0.3823 mol(available)

From the equation:

1 mol of Zn react with 1 mol of S
0.3823 mol of Zn react with 0.3823 mol of S(needed)

Number of moles of S needed is less than number of moles of S available.
S is an excess reactant. Thus, Zn is the limiting reactant.

28

PRACTICE MODULE CHAPTER 1 SK015

ii. How many grams of ZnS will be formed?
Use limiting reactant to calculate the number of moles of product formed.
1 mol of Zn produces 1 mol of ZnS
0.3823 mol of Zn produces 0.3823 mol of ZnS

Mass of ZnS formed = 0.3823 mol × (65.4+32.1) g/mol
= 37.27 g

iii. How many grams of the excess reactant will remain after the reaction is over?

Number of moles of excess reagent remains = navailabe - nneeded
= 0.9346 mol – 0.3823 = 0.5523 mol

Mass of excess reagent remains = 0.5523 mol x [32.1] g/mol = 17.73 g

7. Which element is in excess when 3.00 grams of Mg is ignited in 2.20 grams of pure oxygen? What
mass is in excess? What mass of MgO is formed?
Solution:

( ) 2 = 2.20
= ( / ) 2(16) ( −1)

3.00 = 0.06875 mol(available)
= 24.3 ( / )

= 0.1235 mol(available)

From the equation:

2Mg + O2 → 2MgO

2 mol of Mg react with 1 mol of O2

0.1235 mol of Mg react with x mol of O2

x= 0.1235 × 1 mol O2 = 0.06175 mol of O2 (needed)
2

Number of moles of O2 needed is less than number of moles of O2 available.
O2 is an excess reactant. Thus, Mg is the limiting reactant.

Number of moles of excess reagent remains = navailabe - nneeded
= 0.06875 – 0.06175 = 7x10-3 mol

Mass of excess reagent remains = 7 x 10-3 mol x [2(16)] g/mol = 0.224 g

Use limiting reactant to calculate the number of moles of product formed.
2 mol of Mg react with 1 mol of 2MgO
0.1235 mol of Mg react with 0.1235 mol of MgO

Mass of MgO formed = 0.1235 mol × (24.3+16) g/mol = 4.98 g

8. When MoO3 and Zn are heated together they react
3Zn(s) + 2MoO3(s) ----------> Mo2O3(s) + 3ZnO(s)

29

PRACTICE MODULE CHAPTER 1 SK015

What mass of ZnO is formed when 20.0 grams of MoO3 is reacted with 10.0 grams of Zn? Consider
the reaction.
Solution:

( ) 3 = 20.0
= ( / ) 96+3(16) ( −1)

10.0 = 0.1389 mol(available)
= 65.4 ( / )

= 0.1529 mol(available)

From the equation:

3 mol of Zn react with 2 mol of MoO3

0.1529 mol of Zn react with x mol of MoO3

x= 0.1529 × 2 mol MoO3 = 0.1019 mol of MoO3 (needed)
3

Number of moles of MoO3 needed is less than number of moles of MoO3 available.
MoO3 is an excess reactant. Thus, Zn is the limiting reactant.

Use limiting reactant to calculate the number of moles of product formed.
3 mol of Zn produce 3 mol of ZnO
0.1529 mol of Zn react with 0.1529 mol of ZnO
Mass of ZnO formed = 0.1529 mol × (65.4+16) g/mol = 12.45 g

9. Given reaction; I2O5(g) + 5CO(g) → 5CO2(g) + I2(g)
a) 80.0 grams of iodine(V) oxide, I2O5, reacts with 28.0 grams of carbon monoxide, CO.
Determine the mass of iodine I2, which could be produced?

Solution:

( ) = 28.0
= ( / ) [12+16] ( −1)

80.0 = 1 mol(available)
2 5 = [2(126.9) + 5(16)]( / )

= 0.2397 mol(available)

From the equation:

1 mol of I2O5 react with 5 mol of CO
0.2397 mol of I2O5 react with x mol of CO
x = 0.2397 × 5 mol CO = 1.1985 mol of CO (needed)

1

Number of moles of CO needed is more than number of moles of CO available.
Thus, CO is the limiting reactant.

Use limiting reactant to calculate the number of moles of product formed.
5 mol of CO produce 1 mol of I2
1 mol of CO react with 0.2 mol of I2

Mass of I2 formed = 0.2 mol × [2(126.9] g/mol = 50.76 g

30

PRACTICE MODULE CHAPTER 1 SK015

b) If, in the above situation, only 0.160 moles, of iodine, I2 was produced.
i) what mass of iodine was produced?
Solution:
Mass of I2 produced = 0.160 mol x [2(126.9] g/mol = 40.61g
ii) what percentage yield of iodine was produced.
Solution:

= × %

= . × % = . %

.

10. Silver nitrate, AgNO3, reacts with ferric chloride, FeCl3, to give silver chloride, AgCl, and ferric

nitrate, Fe(NO3)3. In a particular experiment, it was planned to mix a solution containing 25.0 g of

AgNO3 with another solution containing 45.0 grams of FeCl3.

i. Write the chemical equation for the reaction.

3AgNO3 + FeCl3 → Fe(NO3)3 + 3AgCl

ii. Which reactant is the limiting reactant?

Solution:

( ) 3
= ( / ) 45.0

3 = [55.9 + 3(35.5)] ( −1)
25.0
= 0.2771 mol(available)
= [107.9 + [(14 + 3(16)]( / )

= 0.1471 mol(available)

From the equation:

3 mol of AgNO3 react with 1 mol of FeCl3

0.1471 mol of AgNO3 react with x mol of FeCl3

x= 0.1471 × 1 mol FeCl3 = 0.049 mol FeCl3 (needed)
3

Number of moles of FeCl3 needed is less than number of moles of FeCl3 available.
FeCl3 is an excess reactant. Thus, AgNO3 is the limiting reactant.

iii. What is the maximum number of moles of AgCl that could be obtained from this mixture?

Solution:
Use limiting reactant to calculate the number of moles of product formed.
3 mol of AgNO3 produced 3 mol AgCl
0.1471 mol of AgNO3 produced 0.1471 mol AgCl

iv. What is the maximum number of grams of AgCl that could be obtained?

Mass of AgCl formed = 0.1471 mol × [107.9 + 35.5] g/mol = 21.09 g

v. How many grams of the reactant in excess will remain after the reaction is over?

Number of moles of excess reagent remains = navailabe - nneeded
= 0.2771 – 0.049 = 0.2281 mol

Mass of excess reagent remains = 0.2281 mol x [55.9 + 3(35.5)] g/mol = 37.04 g

31

PRACTICE MODULE CHAPTER 1 SK015

11. Solid calcium carbonate, CaCO3, is able to remove sulphur dioxide from waste gases by the
reaction:

CaCO3+ SO2 + other reactants ------> CaSO3 + other products

In a particular experiment, 255 g of CaCO3 was exposed to 135 g of SO2 in the presence of an
excess amount of the other chemicals required for the reaction.

a) What is the theoretical yield of CaSO3?
Solution:

( ) 2
= ( / ) 135.0

3 = [32.1 + 2(16)] ( −1)
255.0
= 2.1061 mol(available)
= [40.1 + [(12 + 3(16)]( / )
= 2.5475 mol(available)

From the equation:

1 mol of CaCO3 react with 1 mol of SO2
2.5475 mol of CaCO3 react with 2.5475 mol of SO2(needed)

Number of moles of SO2 needed is more than number of moles of SO2 available.
SO2 is the limiting reactant.

Use limiting reactant to calculate the number of moles of product formed.
1 mol of SO2 produced 1 mol CaSO3
2.1061 mol of SO2 produced 2.1061 mol CaSO3

Theoritical yield of CaSO3 = 2.1061 mol x [40.1+32.1+3(16)] g mol-1
= 253.15g

b) If only 198 g of CaSO3 was isolated from the products, what was the percentage yield of CaSO3
in this experiment?(78.21%)
Solution:

= × %

= × % = . %
.

12. A research supervisor told a chemist to make 100 g of chlorobenzene from the reaction of benzene
with chlorine and to expect a yield no higher that 65%. What is the minimum quantity of benzene
that can give 100 g of chlorobenzene if the yield is 65%? The equation for the reaction is:

C6H6 + Cl2 -----------> C6H5Cl + HCl

benzene chlorobenzene

32

PRACTICE MODULE CHAPTER 1 SK015

Solution:

( )
= × % = ( / )

65% = × % ℎ

153.8462
Theoritical yield of C6H5Cl = 153.8462 g = [6(12) + 5(1) + 35.5] ( )
= 1.3675 mol

From the equation; 1 mol of C6H5 produced 1 mol C6H5Cl
1.3675 mol of C6H5 produced 1.3675 mol C6H5Cl

Mass of benzene = 1.3675 mol x [6(12) + 6(1)] g/mol
= 106.67 g (minimum quantity)

13. Certain salts of benzoic acid have been used as food additives for decades. The potassium salt of
benzoic acid, potassium benzoate, can be made by the action of potassium permanganate on
toluene.

C7H8 + 2KMnO4 → KC7H5O2 + 2MnO2 + KOH + H2O

toluene potassium benzoate

If the yield of potassium benzoate cannot realistically be expected to be more than 68%, what is the

minimum number of grams of toluene needed to achieve this yield while producing 10.0 g of KC7H5O2?

Solution:

( )
= × % = ( / )

68% = × % KC7H5O2

14.7059
= [39.1 + 7(12) + 5(1) + 2(16)] ( )
Theoritical yield of KC7H5O2= 14.7059 g
= 0.0918 mol

From the equation; 1 mol of C7H8 produced 1 mol KC7H5O2
0.0918 mol of C7H8 produced 0.0918 mol KC7H5O2

Mass of toluene = 0.0918 mol x [7(12) + 8(1)] g/mol
= 8.45 g (minimum quantity)

33

PRACTICE MODULE CHAPTER 1 SK015

WORKSHEET 1.4

MULTIPLE CHOICE QUESTIONS

1. Isotopes differ from each other in what way?
A they have different numbers of neutrons in the nucleus.
B they have different numbers of protons in the nucleus.
C the have different numbers of electrons outside the nucleus.
D more than one response is correct.

2. How many protons, electrons and neutrons are there in an ion of chlorine isotope, 1375 −?
A 35 p, 35 e, 17 n
B 35 p, 17 e, 18 n
C 17 p, 35 e, 17 n
D 17 p, 18 e, 18 n

3. Naturally occurring carbon consists of four isotopes: 11C, 12C, 13C, and 14C. The proton number of
carbon is 6. Which of the following is true about carbon isotopes?
A Attraction forces of 14C nucleus is stronger than of 11C.
B 14C has the most number of neutrons.
C All carbon isotopes are having different atomic size.
D All carbon isotopes contain the same number of neutron and electrons.

4. Which of the following is a pair of isotopes?
A 13588 and 13680
B 7324 and 7353
C8336 and 8364
D 23925 and 23949

5. Which of the following contains only empirical formula?
A C3H8, N2O4, H2O
B C5H12, NO2, H2O
C C5H10, N2O4, H2O2
D C5H12, NO2, H2O2

6. Potassium oxalate has the chemical formula K2C2O4. Based on this information, the formula of
iron(III) oxalate is
A FeC2O4
B Fe(C2O4)2
C Fe2(C2O4)3
D Fe3(C2O4)2

7. An organic compound contains carbon, hydrogen and oxygen. 1.84 g of the organic compound

gives 3.52 g of carbon dioxide and 2.16 g of H2O on complete combustion. Determine the empirical

formula of the organic compound,

A C2H5O C CH4O

B C2H6O D C2H4O

34

PRACTICE MODULE CHAPTER 1 SK015

8. What is the stoichiometric coefficient of H2O when the following equation is properly balanced with
the smallest set of whole numbers?
NH3 + O2 → NO + H2O
A3
B4
C5
D6

9. The redox reaction between Fe3+ and MnO4- is shown as two half equations below:
Fe2+(aq) → Fe3+(aq) + e-

MnO4- (aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
Calculate the number of moles of MnO4- needed to oxidize 1 mol of Fe2+.

A 0.20 C. 0.50

B 0.40 D 1.00

10. Give S2O82-(aq) + 2e → 2SO42- (aq)
Mn2+(aq) + 4H2O(l) → MnO4- (aq) + 8H+(aq) + 5e
How many moles of S2O82- are needed to oxidise 1 mol of Mn2+?
A 0.4 mol C 2.0 mol

B 1.0 mol D 2.5 mol

11. Calculate the number of hydrogen atoms in 9.60 g of (NH4)2CO3 compound.

A 6.02 x 1022 C 4.82 x 1023

B 3.01 x 1023 D 6.02 x 1023

12. Borax, Na2[B4O5(OH)4], is used as a fireproof insulation material and as a washing powder.
Calculate the mass of boron present in 30 g of borax.
A 4.30 g
B 5.46 g
C 6.00 g
D 6.44 g

13. What volume of water in cm3 should be added to 10.0 cm3 NaOH 6.0 M to produce a solution of

NaOH 0.3 M?

A 10 cm3 C 200 cm3

B 190 cm3 D 500 cm3

14. A student wish to neutralize phosphoric acid with 0.2 M solution of sodium hydroxide. The reaction

equation is as follows:

H3PO4 + 3NaOH → Na3PO4 + 3H2O

What volume of 0.2 M sodium hydroxide is required to neutralize 50 mL phosphoric acid 0.1 M?

A 25 mL C 50 mL

B 75 mL D 150 Ml

35

PRACTICE MODULE CHAPTER 1 SK015

15. Hydrogen gas can be prepared through the reaction between zinc and hydrochloric acid as follows
Zn(s) + 2HCl (aq) → H2(g) + ZnCl2(aq)
If 13.0 g zinc react with excess HCl, calculate the volume of hydrogen gas in mL produced at
standard temperature and pressure,
A 4.86 C 4450
B 4.45 D 4863

16. Nitrogen monoxide reacts spontaneously with oxygen as represented by the equation: 2NO(g) +
O2(g) → 2NO2(g). in an experiment, 75.0 g of nitrogen monoxide is allowed to react with 64.0 g of
oxygen. Which of the following statement is/are correct?

I Nitrogen oxide is the limiting reactant.

II 115.0g of NO2 is produced.

III 24.0 g of O2 remained unreacted

WORKSHEET 1.5

MODEL OF PSPM QUESTIONS

1. a) Define nucleon number and isotope.

Give the number of protons, neutrons and electrons in each of the following species.

79

i. 35 Br –

ii. 130 Ba 2+
56

ANSWER Nucleon number :
b)
Total number of protons and neutrons in the nucleus of atom of an element.

Isotope :

Two or more atoms of the same element having same proton number but different

number of neutrons.

No. of protons No. of neutrons No. of electrons

i. Br – 35 44 36

ii. Ba 2+ 56 74 54

A 50 mL of a saturated NaOH solution containing 52% NaOH by weight with a density of

1.48 g mL-1 is used to prepare a 0.1 M NaOH solution. Determine the initial concentration

of NaOH solution and the amount of water required to prepare the 0.1 M NaOH solution.

ANSWER Mass of solution = density of solution × volume of solution
= 1.48 g  50mL
mL
= 74.0 g

% w/w = mass of NaOH  100 = 52 %
mass of solution

Mass of NaOH = 52  74.0g
100

= 38.48 g

36

PRACTICE MODULE CHAPTER 1 SK015

Moles of NaOH = 38.48g = 0.962 mol
40g/mol

Molarity = Mole of NaOH
Volume of solution (L)

= 0.962 mol = 19.24 M
50  10 −3 L

Prepare 0.1M solution:

M1V1 = M2V2

V2 = 50 mL  19.24 M
0.1 M

= 9620 mL
The amount of water required = 9620 – 50 mL

= 9570 mL @ 9.57 L

2. a) Compound A consists of the element C, H and O. The complete combustion of 4.624g of
the compound A yielded 6.557g of CO2 and 4.026g of H2O.Determine the empirical
formula of compound A.

ANSWER No of moles CO2 = 6.557 g
44 gmol-1

=0.1490 mol

1 mole of CO2 contain 1 mol of C
0.1490 mol of CO2 contain 0.1490 mol C

1 mole of C equivalent to 12 g of C
0.1490 mol of C equivalent to 12 g  0.149 mol

1 mol
= 1.788 g

No of moles H2O = 4.026 g
18 gmol-1

=0.2237 mol

1 mole of H2O contain 2 mol of H

0.2237 mol of H2O contain 2 mol  0.2237 mol
1 mol

= 0.4474 mol

1 mol of H2O equivalent to 1 g of H

0.4474 mol of H2O contain 1g  0.4474 mol
1 mol

= 0.4474 g

Mass of O = Mass of sample – Mass of C – Mass of H
= 4.624 – 1.788 – 0.4474
= 2.3886 g

Element CHO

37

PRACTICE MODULE CHAPTER 1 SK015

Mass (g) 1.788 0.4474 2.3883
0.4474 =0.4474 2.3883 =0.1493
No of mole (mol) 1.788 = 0.149 16.0
Mol ratio 12.0 1.0
0.4473 0.1493
0.149
0.149 0.149
Simplest Mole ratio 0.149 3 1
1

Empirical formula = CH3O

b) The reaction between acetic acid, CH3COOH and barium hydroxide, Ba(OH)2 produces a
salt. Determine the maximum mass of the salt obtained if 17.13g of barium hydroxide is
used.

ANSWER 2CH3COOH(aq) + Ba(OH)2(aq) → Ba(CH3COO)2(s) +2H2O(l)

Mole of Ba(OH)2 = 17.13 g = 0.100 mol
171.3 g/mol

From the stoichmetry equation:

1 mol of Ba(OH)2 produce 1 mol of Ba(CH3COO)2
So, 0.1 mol of Ba(OH)2 produce 0.1 mol of Ba(CH3COO)2

Molar mass of Ba(CH3COO)2 = 255.3 g/mol
Mass of Ba(CH3COO)2 = 0.1 mol × 255.3 g

= 25.53 g
3. a) FIGURE 1 shows the mass spectrum for copper, Cu, atom which has been identified to

have two isotopes.

FIGURE 1

38

PRACTICE MODULE CHAPTER 1 SK015

Relative 62.929 64.927
abundan 8 8

2693Cu 2695Cu

62 63 64 65 66
i. Define isotope. Mass/charge

ANSWER Two or more atoms of the same element having the same proton number but
different nucleon numbers.

ii. Using the information given in FIGURE 1, calculate the abundance of each
isotope for Cu.

ANSWER Lets,
x = % of 63Cu

X – 100 = % of 65Cu
Average atomic mass

= (atomic mass 63Cu x % abundance ) + (atomic mass 65 Cu x % abundance)

63.6 = ( x x 62.9298 ) + ( 100 - x x 64.9278 )
100 100

X = 66.46%
So,

% abundance of 63Cu = 66.46%
% abundance of 65Cu = 100 – 66.46

= 33.54%

b) Urea, (NH2)2CO is used as fertilizer and in animal feed. It is prepared by reacting ammonia

and carbon dioxide as shown:

2NH3(g) + CO2(g) (NH2)2CO(s) + H2O(l)

In a process, 637.2 g of ammonia is allowed to react with 1142 g of carbon dioxide.

i. Determine the limiting reagent in the reaction.

ANSWER Moles of NH3 = 637.2 g Moles of CO2 = 114.2 g
17.0 g/mol 44.0 g/mol
= 25.95 mol
= 37.48 mol

39

PRACTICE MODULE CHAPTER 1 SK015

From the equation,
2 mol of NH3 react with 1 mol of CO2
37.48 mol of NH3 react with 37.48 mol of NH3 x 1 mol CO2

2 mol NH3
= 18.74 mol CO2
Mole of CO2 needed (18.74 mol) less than given (25.95 mol) so CO2 is excess reactant and
the limiting reactant is NH3
ii. Calculate the mass of urea formed.
ANSWER From the equation,
2 mol of NH3 produce 1 mol of (NH2)2CO

37.48 mol of NH3 react with 37.48 mol of NH3 x 1 mol of (NH2 )2 CO
2 mol NH3

= 18.74 mol (NH2)2CO
Mass of (NH2)2CO = 18.74 mol x 60 g/mol

= 1124.4 g

Prepared by : Mdm Siti Fatimah & Mdm Nur Zarifah Syazana

40