Find the empirical formula for a compound if a 2.50 g sample contains 0.90 g of Calcium and 1.60 g of Chlorine? What is the molecular formula if the molar mass is 330 g per mol?
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A compound is made of calcium and chlorine. A 2.5 g sample contains 0.9 g of calcium and 1.6 g of chlorine. The percentage by molar mass of calcium in the compound is (0.9/40)/2.5 = 0.009 and the percentage by molar mass of chlorine is 0.018
The ratio of the two is 1:2. The empirical formula of the compound is CaCl2
As the molar mass of the compound is 330 g/ mole and the mass of the empirical formula is 110, the molecular formula is Ca3Cl6
The molecular formula of the compound is Ca3Cl6.
2.5 g of the compound contains 0.9 g Calcium and 1.6 g Chlorine.
now consider atoms mass ratio atom ratio(ratio of mass/at.mass)
Ca :Cl 0.9:1.6 0.9/40:1.6/35.5
=0.0225:0.045 = 1:2
Hence empirical formula of the compound is CaCl2
Now suppose the molecular formula be (CaCl2)f, where f is a numerical factor.
molar mass = (1x40 + 2x35.5)f = 111f
by condition 111f = 330
or f= 2.97 or 3 (approximated to the nearest integer).
Hence the molecular formula of the compound is (CaCl2)3 or, Ca3Cl6.
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