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When a charge is passed through Boron Trichloride, a new compound forms with 0.0878g Cl for every 0.0131g B...
What is the empirical formula for the new compound? (All numbers should be subscript)
a) BCl
b) BCl2
c) BCl3
d) B2Cl3
e) B3Cl
2 Answers
- 1 decade agoFavorite Answer
The ratio of the masses in the new compound is .0878/.0131, which is approximately equal to 6.702. If you then divide this by the ratio of the atomic mass of chlorine to the atomic mass of boron, you end up with the ratio of the number of chlorine atoms to the number of boron atoms, which determines the emperical formula.
6.702/(35.45/10.81)=2.04, which is approximately equal to 2, so there's two atoms of chlorine for every atom of boron.
b) Bcl2
- phoenixshadeLv 51 decade ago
Convert the masses of each element to moles (look up the atomic weights)
Chlorine
(0.0878 g) / (35.453 g/mol) = 0.00248 mol
Boron
(0.0131 g) / (10.811 g/mol) = 0.00121 mol
Calculate the molar ratio by dividing by the smallest.
Cl: 0.00248 / 0.00121 = 2.05
This is very close to exactly two chlorines per boron, so the empirical formula as BCl2. The answer is (b).
