Lesson Notes By Weeks and Term - Senior Secondary School 1

SUBJECT: CHEMISTRY

CLASS:� SS 1

DATE:

TERM: 2nd TERM

REFERENCE BOOKS

• New Chemistry for Senior Secondary School by Osei Yaw Ababio; U.T.M.E Past Questions and Answers.
• Practical Chemistry for Senior Secondary Schools by Godwin Ojokuku
• Outline Chemistry for Schools & Colleges by Ojiodu C.C.
• Chemistry Pass Questions for S.S.C.E and UTME.

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WEEK FOUR��� ��� ��� ��� ��� ��� ��� ��� ��� ��� DATE-------------

TOPIC: EMPIRICAL AND MOLECULAR FORMULAE

Empirical formula is the formula which shows the simplest whole number ratios of atoms present in a compound while molecular formula is the formula which shows the actual number of each kind of atoms present in the molecule. The molecular formula of a compound is a whole number multiple of its empirical formulae.

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CALCULATIONS

1. An organic compound on analysis yielded 2.04g carbon, 0.34g hydrogen and 2.73g oxygen.
2. Calculate the empirical formulA.
3. If the relative molecular mass of the compound is 60. Calculate its molecular formulA.

����Solution:

����Elements ��� ��� ��� ��� ��� C��� ��� H��� ��� O

����Reacting mass��� ��� ��� ��� ��� 2.04��� ��� 0.34��� ��� 2.73

���Mole ratio = Reacting mass��� =��� ��� 2.04��� :��� 0.34��� :��� 2.73

�����Atomic mass��� ��� 12��� ��� � 1��� ��� � 16�

��� ��� ��� ��� =��� 0.17��� :��� 0.34��� :��� 0.17

��Dividing through by the ��� ��� 0.17��� :��� 0.34��� :��� 0.17

���Smallest value��� ��� ��� ��� 0.17��� ��� 0.17��� ��� 0.17

����Whole number ratio��� ��� ��� 1��� :��� � 2��� :��� � 1

������The empirical formula = CH2O

Relative molecular mass of the compound = 60

Let the molecular formula = (CH2O)n

��� (CH2O)n �= 60

��� (12 + 1x2 +16)n = 60

��� ��� ��� 30n � � = 60

��� ��� ��� � � n � � = 60/30 = 2

Therefore, the molecular formula is (CH2O)2 �= C2H4O2

Calculate the empirical formula of an organic compound containing 81.8% carbon and 18.2% hydrogen

Solution:

Element ��� ��� ��� ��� ��� C��� ��� H

% Composition by mass��� ��� ��� 81.8��� ��� 18.2

Mole ratio = % by mass� �=��� ��� ��� 81.8��� :��� 18.2

�������������������Atomic mass� � � � � � � ��� 12��� ��� � 1

��� ��� � � � � =��� ��� ��� 6.82��� :��� 18.2

Dividing through by the��� ��� ��� 6.82��� :��� 18.2

smallest value��� ��� ��� ��� ��� 6.82��� ��� 6.82

Whole number ratio��� ��� ��� ��� � 1��� :��� 2.67

Since the ratio is not completely whole, we continue to multiple to obtain the lowest multiple that is close to a whole number i.e.

1:2.67, 2:5.34, 3:8.01, 4:10.65, 5:13.35, etC. 3:8.01 is close to whole number.

Therefore, the empirical formula is C3H8

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GENERAL EVALUATION/REVISION

1. An organic compound has the empirical formula CH2. If its molecular mass is 42gmol-1, �����what is the molecular formula?
2. Determine the relative molecular mass of calcium trioxocarbonate (v).
3. Define the term radical.
4. Write the formula of the following compounds

1. Mercury (i) dioxonitrate (iii)

��� ��� B.Sodium hydrogen trioxocarbonate (IV)

1. Oxochlorate (I) acid

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READING ASSIGNMENT

New School Chemistry for Senior Secondary School by O. Y. Ababio, Pg 31-32

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WEEKEND ASSIGNMENT

1. The % by mass of carbon in CO2 is A.37% B. 27% C. 48% D. 52%
2. What is the molar mass of Na2SO4? A. 172 B.168 C.142 D.133
3. The empirical formula of a compound is CH, the molecular formula could be A. C2H4

B.CH4C. C7H12D. C6H6

1. An oxide of nitrogen contains 69.6% of oxygen by mass. Its empirical formula is A. N2O3������B. N2O2C. N2O D. NO2
2. 5.0g of an oxide of a metal (M) gave 4.0g of the metal when reduced with hydrogen. What is the empirical formula of the oxide? [M=64, O=16] A. MO B. MO2C. M2O D.�M203

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THEORY

1. Calculate the % by mass of water of crystallization in Al2(SO4)3.9H2O
2. Two compounds X and Y have the same % composition by mass 92.3% carbon and 7.7% hydrogen. Calculate the:

1. Empirical� formula of X and Y
2. Molecular formula of each compound if the molar mass of X is 26gmol-1 and Y is 78gmol-1.