What is the
A chlorine oxide is 59.7%% by mass CI What is the
empirical formula of the oxide?
A chlorine oxide is 59.7%% by mass CI What is the
empirical formula of the oxide?
A chlorine oxide is 59.7% by mass CI. What is the empirical formula of the oxide?
Solution:
w(Cl) = 59.7% or 0.597
w(O) = 100% − w(Cl) = 100% − 59.7% = 40.3% or 0.403
Calculate the mass of each element in 100 g of a chlorine oxide:
Mass of Cl = w(Cl) × Mass of sample = 0.597 × 100 g = 59.7 g Cl
Mass of O = w(O) × Mass of sample = 0.403 × 100 g = 40.3 g O
Calculate the moles of each element in a chlorine oxide:
Molar mass of chlorine (Cl) is 35.453 g/mol
Molar mass of oxygen (O) is 15.999 g/mol
Therefore
Moles of Cl = (59.7 g Cl) × (1 mol Cl / 35.453 g Cl) = 1.684 mol Cl
Moles of O = (40.3 g O) × (1 mol O / 15.999 g O) = 2.519 mol O
Calculate the mole ratio between the elements. Divide the moles of each element by the least number of moles.
Cl: 1.684 / 1.684 = 1.0
O: 2.519 / 1.684 = 1.5
Thus, the empirical formula of a chlorine oxide is ClO1.5 or Cl2O3
Answer: The empirical formula of a chlorine oxide is Cl2O3
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