Question

1.     Define the terms molecular formula and empirical formula and give an example to illustrate.
 

 

2.     You are given a magnesium ribbon weighing 0.042 g. You perform a combustion reaction to synthesize magnesium oxide yielding a final product mass of 0.070 g.  What is the empirical formula of the metal oxide?  (Show all work.)

Answer 1

An empirical formula gives the simplest whole number ratio of atoms of different elements present in a compound

The molecular formula shows the exact number of atoms of different types present in a molecule of a compound

Examples of empirical and molecular formula

If carbon and hydrogen are present in a compound in the ratio of 1:2 the empirical formula for the compound is CH₂

the empirical formula mass = 14

Molar mass = n × emperical mass

If we know the molar mass of compound is 42 g mol^-1 then n=3

Molecular formula = n × emperical formula mass

= 3 × (CH₂) = C₃H₆ that is propene

2.

Mass of oxygen = mass of magnesium - mass of magnesium oxide

= 0.07 - 0.042 = 0.028 grams

Moles of Mg = mass/mol. Wt. = 0.042/ 24.3 = 0.001769

Moles of O = mass/mol.wt. = 0.028/16 = 0.00175

Moles og Mg/Moles of O = 0.001769/0.00175 = 1.01

Therefore Mg : O = 1.01: 1.0

Empericsl formula = MgO