Question

An analysis of an oxide of nitrogen with a molecular weight of 92.02 amu gave 69.57% oxygen and 30.43% nitrogen. What are the empirical and molecular formulas for this nitrogen oxide? Complete and balance the equation for its formation from the elements nitrogen and oxygen.

Answer 1

The empirical formula is No2 and molecular formula is N2O4

empirical formula calculation

find the moles of each reactant

=% composition/molar mass

molar mass of oxygen=16 g/mol ,while hat of nitrogen= 14 g/mol from periodic table

moles is therefore=

nitrogen= 30.43/14=2.174 moles

oxygen= 69.57/1 6=4.348 moles

find the mole ratio by dividing each moles with smallest number of mole( 2.174)

nitrogen = 2.174/2.174= 1

oxygen = 4.348/2.174=2

The balanced equation for its formation from the element nitrogen and oxygen is as below

N2 +2 O2 → N2O4

therefore the empirical formula= NO2

molecular formula calculation

[NO2] n= 92.02 amu

[(14) + (16x2)]n = 92.02 amu

46n= 92.02

divide both side by 46

n=2

therefore the molecular formula is gotten by multiplying the empirical formula by 2= [NO2]2= N2O4