What is the molecular formula of a compound that contains 10.25 grams Nitrogen 23.35 grams oxygen and molecular mass of 92.0 grams per mole?

Answer 1

The question is missing something; I must assume that the compound is 23% N by mass and 77% O by mass. Starting with NO3 as an initial guess turns up 22.6% N, 77.4% O, which agrees with the above if we round off to two significant figures. It has a molar mass of 62, precisely half the stated molecular mass. One could presume a dimer of NO3, i.e., N2O6, which is hardly a common species but does appear in the literature.

Answer 2

Divide by the Atomic Mass of each

N -- 10.25 / 14 = 0.732

O -- 23.35 / 16 = 1.459

Now divide each answer by the smallest answer.

N -- 0.732 / 0.732 = 1

O -- 1.459 / 0.732 = 1.99 ... round to 2

Empirical formula NO2

Empirical mass = 14 + 16x2 = 46

molar mass / emp. mass = 92 / 46 = 2

multiply subscripts by this number

(NO2)2 = N2O4