CHAPTER 1 - MATTER

Example 2

1.00 g sample of compound A was burnt
in excess oxygen yield 2.52 g of CO2
and 0.443 g of H2O. Determine the
empirical formula of the compound.

50

Answer:

Element C H O

Mass 12.01 x 2.52 2.02 x 0.443 = 1- 0.6877 - 0.04922
(g) = 0.2631
44.01 18.02

=0.6877 =0.04922

51

Answer:

Element C H O

Mass (g) 0.6877 0.04922 0.2631

Number 0.6877 0.04922 0.2631
of moles 12.01 1.01 16.00
=0.0573 =0.01644
(mol) =0.04873
3.5 3 1
Ratio of
moles 7 6 2

Simplest
ratio

∴ Empirical formula = C7H6O2

52

Exercise

A combustion of 0.202 g of an organic
sample that contains carbon, hydrogen
and oxygen produce 0.361g carbon
dioxide and 0.147 g water. If the relative
molecular mass of the sample is 148, what
is the molecular formula.

Ans : C6H12O4 53

Answer:

Element C H O

Mass 12.01 x 0.361 2.02 x 0.147 = 0.202- 0.0985 -
(g) 0.01648
44.01 18.02
= 0.08702
=0.0985 =0.01648

54

Answer:

Element C H O

Mass (g) 0.0985 0.01648 0.08702

Number 0.0985 0.01648 0.08702
of moles 12.01 1.01 16.00

(mol) =0.008201 =0.01632 =0.005439
3 1
Ratio of 1.5
moles 6 2

Simplest 3
ratio

∴ Empirical formula = C3H6O2

55

Molecular formula = n (C3H6O2)
Mr n (C3H6O2) = 148

n[3(12.01)+6(1.01)+2(16.00)] =148
74.09n = 176
n=2

∴ Molecular formula = C6H12O4

56

Concentration of Solutions
• A solution is a homogenous mixture of two

or more substances.

• Example: 57
sugar + water = solution
sugar = solute
water = solvent

• The concentration of solution is the
amount of solute present in a given
quantity of solvent or solution.

• Concentration of a solution can be
expressed in various ways:

A. Molarity
B. Molality
C. Mole Fraction
D. Percentage by Mass
E. Percentage by Volume

58

A. Molarity (M)

• The number of moles of solute per cubic
decimetre (dm3) or litre (L) of solution.

Molarity, M moles of solute
volume of solution (dm3)

The unit of molarity : molL-1 or moldm-3 or molar

59

Example 1

Determine the molarity of 85.0 ml of
ethanol solution contains 1.77 g of
ethanol, C2H5OH.

60

Answer:

moles of C2H5OH  1.77 g
46.08 gmol

 0.0384 mol

moles of solute
Molarity,M of C2H5OH = volume of solution(L)

= 0.0384 mol
85 L
1000

= 0.452 molL-

61

Example 2
A solution is prepared by dissolving 0.586 g
of sodium carbonate, Na2CO3 in 250 cm3 of
water. Calculate its molarity.

62

Answer:

moles of Na 2CO3  0.586 g
105.99 gmol

 5.53 x 10-3 mol

moles of solute
Molarity,M of Na2CO3 = volume of solution(dm-3)

= 5.53 x 10-3 mol
250 dm3
1000

= 0.0221 moldm-3

63

Exercise: Molarity

1. Calculate the molarity of a solution of
1.71 g sucrose, C12H22O11 dissolved in
500 ml of water.
[0.01 molL-]

2. How many gram of potassium

dichromate, K2Cr2O7 required to prepare
a solution of 250 ml with concentration of

2.16 M

[158.9 g] 64

Answer:

1. nsucrose  1.71
342.34

 4.995 x103 mol

M  4.995 x103 mol
0.5 L
sucrose

 0.01molL 1

65

Answer:

2. Molarity  mol solute

volume of solution (L)

2.16  mol solute
0.25

 0.54 mol

0.54  mass K 2Cr2O7
294 .2

mass K 2Cr2O7 158.9 g

66

B. Molality (m)

• Molality is the number of moles of
solute dissolved in a given of solvent
in kg.
Molality, m  moles of solute
mass of solvent (kg)

• The unit for molality is molkg-1 or molal
(m) 67

Example 1

• Calculate the molality of a sulphuric acid
solution containing 16.8 g of sulphuric acid
in 200 g of water.

68

Answer:

Molality, m  moles of solute
mass of solvent (kg)

16.8

molality of H2SO4  98
200

1000

 0.1714
0.2

 0.86 molkg1

69

Example 2

Calculate the mass of NaOH to be added to
150 g of water for the preparation of a 1.2 m
of NaOH solution.

70

Answer:

Molality, m  moles of solute
mass of solvent (kg)

molality of NaOH  n NaOH
150
1000

1.2  n NaOH
0.15

n NaOH  0.18 mol

71

Exercise: Molality

1. What is the molal concentration of a
solution when 0.30 mol of CuCl2 is
dissolved in 40.0 mol of water?
[0.416 m]

2. What is the molality of a solution made by
dissolving 36.5 g of naphthalene, C10H8 in
425 g of toluene, C7H8?
[0.670 m]

72

Answer:

1. mass of H2O40.0 x18.02
 720.8 g
 0.7208 kg

molality of CuCl 2  0.30
0.7208

 0.416 m

73

Answer:

2. mass of toloune (solvent )  0.425 kg

mol napthalene  128 36.5 g 1
.08 gmol

 0.28498 mol

molality of C10H8  0.28498 mol
0.425

 0.67 m

74

C. Mole Fraction (X)

• Mole fraction is the ratio of the number of

moles of one component to the total

number of moles of all components

present.

mole fraction of A, XA = total moles of A
moles of all components

XA = nA
n total

75

 Mol fraction is always smaller than 1

 The total mol fraction in a mixture
(solution) is equal to one.
XA + XB + XC + X….. = 1

 Mole fraction has no unit (dimensionless)
since it is a ratio of two similar quantities.

76

Example:
A sample of ethanol, C2H5OH contains
200.0 g of ethanol and 150.0 g of water.
Calculate the mole fraction of
(a) ethanol
(b) water
in the solution.

77

Solution: 200.0 g
a) n ethanol
-1
n water
45.03 gmol
X ethanol  4.4375 mol

150.0 g
18.02 gmol1
 8.324 mol

4.4375 mol
(4.4375  8.324) mol
 0.3477

78

b) X water = 1 - 0.3477
= 0.6523

79

Exercise: Mole Fraction

A solution is prepared by mixing 55 g of toluene,
C7H8 and 55 g of bromobenzene C6H5Br.
What is the mole fraction of each component?
[Ar C=12.01, H=1.01, Br=79.9]

80

Solution:

n C7H8  55
92.15

 0.5969 mol

n C 6H5 Br  55
157.01

 0.3503 mol

X C7H8  n n C7H8
 nC
C 7H8 6H5Br

 0.5969
0.5969  0.3503

 0.63

X C6H5Br  1  0.63 81
 0.37

D. Percentage by Mass (%w/w)

• Percentage by mass is defined as the
percentage of the mass of solute per mass
of solution.

%ww  mass of solute x100
mass of solution

note : mass of solution  mass of solute  mass of solvent

% w/w has no unit.

82

Example 1:
A sample of 0.892 g of potassium chloride,
KCl is dissolved in 54.362 g of water.
What is the percent by mass of KCl in the
solution?

Solution:
%00 mmaassss 0.08.98290g2.809g.52849g.53246g2.3g6x21g00 01000%

 1.6100

83

Example 2:

A solution is made by dissolving 4.2 g of
sodium chloride, NaCl in 100.00 mL of
water. Calculate the mass percent of
sodium chloride in the solution.

Solution:

mass NaCl  4.2 g

mass H2O 100 g

0 mass of NaCl  4.2 g x 100 0 0
0 g 100
4.2 g

 4.03 0 84
0

Exercise: % w/w

1. How many grams of NaOH and water are needed
to prepare 250.0 g of 1.00% NaOH solution?
Ans : 2.50g; 247.5 g

2. Hydrochloric acid can be purchased as a solution
of 37% HCl. What is the mass of this solution contains
7.5 g of HCl?
Ans : 20.27 g

85

Solution:

1. mass NaOH  x g

mass H2O  y g
mass solution  x  y  250 g

00 mass of NaOH  xg x 100 0 0  1.00%
250 g

mass NaOH  x  2.5 g

mass H2O  250 g  2.5 g
 247.5 g

86

Solution:

2. 00 mass of HCl  7.5 g x 100 0 0  37%
xg

mass of solution  x  20.27 g

87

E. Percentage By Volume (%V / V)

• Percentage by volume is defined as the
percentage of volume of solute in milliliter
per volume of solution in milliliter.

% V  volume of solute (mL) x 100
V volume of solution (mL)

note :
Density of solution  mass of solution
volume of solution

88

Example 1:
A sample of 250.00 mL ethanol is labeled
as 35.5% (v/v) ethanol. How many
milliliters of ethanol does the solution
contain?

89

Solution:

00 volume of ethanol  Vethanol x100 00
Vsolution
VethVaentohalnol 35.53050.1x502%050010.020050%0m.0L0 mL

 88.8 mL

90

Conversion of Concentration Units

Example 1:
A 6.25 m of sodium hydroxide, NaOH
solution has has a density of 1.33 g mL-1
at 20 ºC. Calculate the concentration
NaOH in:
(a) molarity
(b) mole fraction
(c) percent by mass

91

Solution:

(a) M = nNaOH
Vsolution

6.25 m of NaOH
 there is 6.25 mol of NaOH in 1 kg of water

for a solution consists of 6.25 mol of NaOH and 1
kg of water;

Vsolution = mass solution
solution

92

masssolution = massNaOH + masswater
massNaOH = nNaOH  molar mass of NaOH

= 6.25 mol  (22.99 + 16.00 + 1.01) g mol 1
= 250 g

masssolution = 250 g + 1000 g

= 1250 g

Vsolution = 1250 g
1.33 g mL1
93

MNaOH = 6.25 mol
 112.3530 103 L 
 

= 6.65 mol L1

94

(b) XNaOH = nNaOH
nNaOH  nwater

1 kg of water contains 6.25 mol of NaOH

nwater = mass water
molar mass of water

= 1000 g mol 1
 16.00)
(2(1.01) g

XNaOH =  6.25 6.25 mol mol 
mol  1000

18.02

= 0.101 95

(c) %(w/w) of NaOH = massNaOH  100%

massNaOH  mass water

= 250 g  100%
250 g  1000 g

= 20.0%

96

Example 2:

An 8.00%(w/w) aqueous solution of
ammonia has a density of 0.9651 g mL-1.
Calculate the

(a) molality

(b) molarity

(c) mole fraction

of the NH3 solution

Answer: a) 5.10 mol kg-1
b) 4.53 mol L-1
c) 0.0842

97

Solution:

a) % by mass of NH3 Mass of NH3 x100%  8.00 %
Mass of solution

assume that ,

mass of NH3  8 g,
mass of solution 100g

mass solvent  100 g  8 g

 92 g  0.092 kg

mole of NH 3  8g 1
17.03 gmol

 0.4698 mol

molality NH 3  0.4698 mol
0.092 kg

 5.11m 98

Solution:

b) mass of solution 100g

mole of NH 3  8g 1
17.03 gmol

 0.4698 mol

from density of NH 3  mass solution  0.9651
Vsolution

V solution  100 g
0.9651

103.61ml  0.10361L

molarity NH 3  0.4698 mol
0.1036 L

 4.53 M 99