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.

The empirical formula is  No2   and molecular formula is N2O4

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

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

i'm pretty sure it is letter c.) car