A compound containing only nitrogen and oxygen is
63.64% N by mass the empirical formula of the
compound is
Solution:
w(N) = 63.64% or 0.6364
w(O) = 100% − w(N) = 100% − 63.64% = 36.36% or 0.3636
Calculate the mass of each element in 100 g of the compound:
Mass of N = w(N) × Mass of samole = 0.6364 × 100 g = 63.64 g N
Mass of O = w(O) × Mass of sample = 0.3636 × 100 g = 36.36 g O
Calculate the moles of each element in the compound:
Molar mass of nitrogen (N) is 14.0067 g/mol
Molar mass of oxygen (O) is 15.999 g/mol
Therefore
Moles of N = (63.64 g N) × (1 mol N / 14.0067 g N) = 4.5435 mol N
Moles of O = (36.36 g O) × (1 mol O / 15.999 g O) = 2.2726 mol O
Calculate the mole ratio between the elements. Divide the moles of each element by the least number of moles.
N: 4.5435 / 2.2726 = 1.999
O: 2.2726 / 2.2726 = 1.000
Thus, the empirical formula of the compound is N2O
Answer: The empirical formula of the compound is N2O
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