QUESTIONS : EMPIRICAL / MOLECULAR FORMULA

Question 1 : 

A compound was analyzed and found to contain 13.5 g calcium, Ca; 10.8 g oxygen, O and 0.675 g H. What is the empirical formula of the compound?

[ RAM : Ca=40 , O=16 , H=1 ]

Solution:

Step 1 : Given ...

Mass of calcium, Ca = 13.5 g

Mass of oxygen, O = 10.8 g

Mass of hydrogen, H = 0.675 g

Step 2 : Find number of mole each elements.

Mole of Ca = 13.5 / 40 = 0.3375

Mole of O = 10.8 / 16 = 0.675

Mole of H = 0.675 / 1  = 0.675

Step 3 : Simplest mole ratio each elements.

Mole ratio of Ca = 0.3375 / 0.3375 = 1

Mole ratio of O = 0.675 / 0.3375 = 2

Mole ratio of H = 0.675 / 0.3375 = 2

Step 4 : Write the empirical formula
Empirical formula = CaO₂H₂ ≈ Ca(OH)₂

Question 2 : From table result for an experiment below. Find empirical formula of magnesium oxide?

Result of experiment:

Mass of crucible + lid	24.0 g
Mass of crucible + lid + magnesium ribbon	26.4 g
Mass of crucible + lid + magnesium oxide	28.0 g

[ RAM : Mg=24 ; O=16]

Solution:

	Magnesium, Mg	Oxygen, O
Mass (g)	26.4 - 24.0 = 2.4	28.0 - 26.4 = 1.6
Number of mole (mol)	2.4 / 24 = 0.1	1.6 / 16 = 0.1
Mole ratio	0.1/0.1	0.1/0.1
Simplest ratio	1	1
Empirical formula	MgO

Question 3 :
A hydrocarbon X, consists of 82.75% of carbon and 17.25% of hydrogen by mass. The relative molecular mass of X is 58. Determine the empirical formula and molecular formula of hydrocarbon X.
[RAM : C=12, H=1 ]

Solution :

∴ empirical formula hydrocarbon X.

	Carbon, C	Hydrogen , H
Assume percentage(%) as Mass (g)	82.75	17.25
Number of mole (mol)	82.75/12 = 6.9	17.25 / 1 = 17.25
Mole ratio	6.9 / 6.9 = 1	17.25 / 6.9 =  2.5
Simplest ratio	2	5
Empirical formula	C2H5

∴ molecular formula hydrocarbon X.

(empirical formula) n = molecular formula
Step 1
Relative molecular mass (empirical formula) n = Relative molecular mass molecular mass

Step 2
(C₂H₅) n = 58

Step 3
[2(12) + 5(1) ]n = 58

Step 4
29 n = 58

Step 5
n = 2


Step 6
Molecular formula = (C₂H₅)2 ≈ C₄H₁₀