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 ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

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