Empirical And Molecular Formulae

Back to: CHEMISTRY SS1

 

Welcome to class! 

In today’s class, we will be talking about empirical and molecular formulae. Enjoy the class!

EMPIRICAL AND MOLECULAR FORMULAE

The empirical formula is the formula which shows the simplest whole-number ratios of atoms present in a compound while the 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.

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 / Atomic mass)    =               (2.04 / 12)    :       ( 0.34 / 1)   :      (2.73 / 16)

.                                                                         =                         0.17    :           0.34        :           0.17

Dividing through by the smallest value           =               (0.17 / 0.17)   :   (0.34 / 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 + 1×2 +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/ Atomic mass)      =                 (81.8 / 12)        :       (18.2 / 1)

.                                                                    =                             6.82      :           18.2

Dividing through by the smallest value       =                  (6.82 / 6.82)     :     18.2/6.82

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

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
5. Mercury (i) dioxonitrate (iii)
6. Sodium hydrogen trioxocarbonate (IV)
7. Oxochlorate (I) acid

READING ASSIGNMENT

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

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.  CH₄     C.    C₇H₁₂    D.   C₆H₆
4. An oxide of nitrogen contains 69.6% of oxygen by mass. Its empirical formula is   A.   N₂O₃   B.  N₂O₂   C.  N₂O    D.   NO₂
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.   MO₂   C.  M₂O    D.   M₂0₃

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:
3. The empirical formula of X and Y
4. The molecular formula of each compound if the molar mass of X is 26gmol^-1 and Y is 78gmol^-1.

 

In our next class, we will be talking about Identification and Types of Alloys.  We hope you enjoyed the class.

Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.

School Owner? Looking for ready-made content and tools to save time and grow easily? Book a free demo session now

Get more class notes, videos, homework help, exam practice on Android [DOWNLOAD]

Get more class notes, videos, homework help, exam practice on iPhone [DOWNLOAD]