Back to: CHEMISTRY SS1
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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
- An organic compound on analysis yielded 2.04g carbon, 0.34g hydrogen and 2.73g oxygen.
- Calculate the empirical formula.
- 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
- An organic compound has the empirical formula CH2. If its molecular mass is 42gmol-1, what is the molecular formula?
- Determine the relative molecular mass of calcium trioxocarbonate (v).
- Define the term radical.
- Write the formula of the following compounds
- Mercury (i) dioxonitrate (iii)
- Sodium hydrogen trioxocarbonate (IV)
- Oxochlorate (I) acid
READING ASSIGNMENT
New School Chemistry for Senior Secondary School by O. Y. Ababio, Pg 31-32
WEEKEND ASSIGNMENT
- The % by mass of carbon in CO2 is A. 37% B. 27% C. 48% D. 52%
- What is the molar mass of Na2SO4? A. 172 B. 168 C. 142 D. 133
- The empirical formula of a compound is CH, the molecular formula could be A. C2H4 B. CH4 C. C7H12 D. C6H6
- An oxide of nitrogen contains 69.6% of oxygen by mass. Its empirical formula is A. N2O3 B. N2O2 C. N2O D. NO2
- 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
- Calculate the % by mass of water of crystallization in Al2(SO4)3.9H2O
- Two compounds X and Y have the same % composition by mass 92.3% carbon and 7.7% hydrogen. Calculate the:
- The empirical formula of X and Y
- 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.
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Awesome post, thanks for the useful information.
what a wonderful site but please can i know how to get atomic mass
wounderfull class