If the weight of an oxide of nitrogen is 108g and 4.02g of nitrogen combine with 11.48g of oxygen to form it, what is the molecular formula of the compound

Molecular formula is N₄O₁₀

We are given;

We are required to calculate the molecular formula of the compound.

Atomic mass of Nitrogen as 14.0 g/mol

Moles = Mass ÷ Atomic mass

Therefore;

Moles of Nitrogen = 4.02 g ÷ 14.0 g/mol

                             = 0.287 moles

Moles of Oxygen

Atomic mass of oxygen as 16.0 g/mol

Moles of Oxygen = 11.48 g ÷ 16.0 g/mol

                            = 0.7175 moles

The mole ratio of N : O will be;

O.287 moles : 0.7175 moles

                 1 : 2.5

Multiplying by 2 , we get

                2 : 5

Therefore, the empirical formula of the nitrogen oxide is N₂O₅

Molecular formula = (empirical formula)n

Mass = 108 g

(N₂O₅)n = 108 g

((14×2) + (16 ×2))n = 108

        60n = 108

             n = 1.8

                = 2

Thus, Molecular formula = (N₂O₅)2

                                        = N₄O₁₀

Therefore, the molecular formula is N₄O₁₀

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3