Answer :
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
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
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
