A sample of bismuth weighing at 0.687 g was converted to bismuth chloride by reacting it first with HNO3 and then with HCl followed by careful evaporation to dryness. The mass of the bismuth chloride obtained was 1.032 g
A) what is the empirical formula of bismuth chloride? ( I already solved this but I want to see if I got the right answer. I got BiCl3)
B) what is the theoretical percent of Bi in the bismuth chloride based in this formula? (What the heck. I have no idea what theoretical percent is)
C) what is the percent error in the above experimental determination?
Chemistry! Empirical formula?
2015-05-26 6:31 pm
回答 (1)
2015-05-26 9:37 pm
A)
(0.687 g Bi) / (208.9804 g Bi/mol) = 0.0032874 mol Bi
(1.032 g - 0.687 g) / (35.4532 g Cl/mol) = 0.0097311 mol Cl
Divide by the smaller number of moles:
(0.0032874 mol Bi) / 0.0032874 mol = 1.00
(0.0097311 mol Cl) / 0.0032874 mol = 2.96
Round to the nearest whole numbers to find the empirical formula:
BiCl3
So you are correct.
B) The question is asking: What percent of the mass of BiCl3 is contributed by the Bi?
If you are going to figure the percent mass in theory you use the atomic masses, instead of the masses given in the question, which would be the empirical or experimental percent.
(208.9804 g Bi/mol ) / (208.9804 g Bi/mol + 3 x (35.4532 g Cl/mol)) = 0.662715 = 66.2715% Bi
C) It depends on what was being experimentally determined. I thought the question was asking about the empirical formula, in which case there is no error, the empirical formula is "BiCl3" exactly.
If you are determining the amount of BiCl3 that should have been produced in the experiment, it's called percent yield, in which case:
(0.687 g) / (0.662715) = 1.0366 g BiCl3 in theory
(1.032 g) / (1.0366 g) = 0.9956 = 99.56% yield
If there's an error (and I am reluctant to call it that) it would be:
(1.0366 g - 1.032 g) / (1.0366 g) = 0.0044 = 0.44%
(0.687 g Bi) / (208.9804 g Bi/mol) = 0.0032874 mol Bi
(1.032 g - 0.687 g) / (35.4532 g Cl/mol) = 0.0097311 mol Cl
Divide by the smaller number of moles:
(0.0032874 mol Bi) / 0.0032874 mol = 1.00
(0.0097311 mol Cl) / 0.0032874 mol = 2.96
Round to the nearest whole numbers to find the empirical formula:
BiCl3
So you are correct.
B) The question is asking: What percent of the mass of BiCl3 is contributed by the Bi?
If you are going to figure the percent mass in theory you use the atomic masses, instead of the masses given in the question, which would be the empirical or experimental percent.
(208.9804 g Bi/mol ) / (208.9804 g Bi/mol + 3 x (35.4532 g Cl/mol)) = 0.662715 = 66.2715% Bi
C) It depends on what was being experimentally determined. I thought the question was asking about the empirical formula, in which case there is no error, the empirical formula is "BiCl3" exactly.
If you are determining the amount of BiCl3 that should have been produced in the experiment, it's called percent yield, in which case:
(0.687 g) / (0.662715) = 1.0366 g BiCl3 in theory
(1.032 g) / (1.0366 g) = 0.9956 = 99.56% yield
If there's an error (and I am reluctant to call it that) it would be:
(1.0366 g - 1.032 g) / (1.0366 g) = 0.0044 = 0.44%
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