Question

Determine the empirical formula of a compound containing 22.6% nitrogen and 77.4% oxygen

Answer 1

empirical formula is the simplest ratio of whole numbers of components in a compound
for 100 g of the compound
N O
mass 22.6 g 77.4 g
number of moles 22.6 g / 14 g/mol 77.4 g / 16 g/mol
= 1.61 mol = 4.84 mol
divide by the least number of moles
1.61 / 1.61 = 1.00 4.84 / 1.61 = 3.01
rounded off to nearest whole numbers
N - 1
O - 3

therefore empirical formula - NO₃

Answer 2

The empirical formula of a compound that contain 22.6% nitrogen and 77.4% oxygen is calculated as below

calculate the moles of each element
= % composition/molar mass
that is
N(nitrogen) = 22.6/14= 1.61 moles
O(oxygen) = 77.4/16 = 4.84 mole

find the mole ratio by dividing each mole by the smallest mole ( 1.61 moles)

that is
N = 1.61/1.61 = 1
O = 4.84/1.61 = 3

therefore the empirical formula = NO3