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Find the empirical formula for a compound if a 2.50 g sample contains 0.90 g of Calcium...
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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.
Posted by justaguide on July 12, 2012 at 8:36 PM (Answer #1)
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.
Posted by llltkl on July 22, 2012 at 4:26 PM (Answer #2)
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