## Friday, March 18, 2011

### 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 = CaO2H2Ca(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