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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.

Sagot :

Kalahira
69,57% + 30,43% = 100%

100% ----------- 92,02g
69,57% oxygen ----- x
x = 64,004 ≈ 64,00g oxygen

100% ---------- 92,02g
30,43% nitrogen ----x
x = 27,99 ≈ 28g nitrogen

Mass of oxygen = 16g
Mass of nitrogen = 14g

moles of oxygen = 64g : 16g = 4
moless of oxygen = 28g : 14g = 2

N : O
4 : 2 ||:2
2 : 1

empirical formulas = N₄O₂
molecular formulas = N₂O

N₂ + ¹/₂O₂ ---> N2O
kenmyna

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


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