Page 28 - Q & A STPM 2022 Chemistry

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
Question 2

Compound X has the following composition by mass.
C, 60.0%; O, 26.8%; H, 13.2%
Which of the following compound is definitely not X?
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A CH 3 CH(OH)CH 3
B HOCH 2CH 2CH 2OH
C CH 3OCH 2CH 3
D CH 3 CH 2 CH 2 OH
Answer: B
Exam Tips
Exam Tips
Compounds A, C and D are isomeric with one oxygen atom in their structures,
whereas compound B has two oxygen atoms. So, Compound B is the odd one
out.
An alternative is to calculate the empirical formula of X.
C : O : H = 60.0 : 26.8 : 13.2
12 16
= 5 : 1.7 : 13.2
= 3 : 1 : 8
Empirical formula is C 3 H 8 O.
So, the simplest molecular formula is (C 3 H 8 O) 2 = C 6 H 16 O 2 , which does not
match compound B.

Section B Structured Questions
Question 3
An organic compound, Y, contains carbon, hydrogen and oxygen only. When
6.0 g of Y is burned completely in excess oxygen, 13.2 g of carbon dioxide and
7.2 g of water is produced. Determine the empirical formula of Y.

Answer:
44 g of CO 2 contains 12 g of C.
\ Mass of C in 13.2 g of CO 2 = 12 × 13.2 g = 3.6 g
44
18 g of H 2 O contains 2 g of H.
Term
2
3 \ Mass of H in 7.2 g H 2 O = 18 × 7.2 = 0.8 g
\ Mass of oxygen in 6.0 g of Y = 6.0 – 3.6 – 0.8 = 1.6 g

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