Question

A 0.4958 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 1.0569 g. What is the mass percentage of chlorine in the original compound? %

A student determines the copper(II) content of a solution by first precipitating it as copper(II) hydroxide, and then decomposing the hydroxide to copper(II) oxide by heating. How many grams of copper(II) oxide should the student obtain if his solution contains 46.0 mL of 0.447 M copper(II) nitrate?

Answer 1

1)

m = 0.4958 g of XCl

all Cl- --> AgCl

AgCl =1.0569 g

% of Cl in sample

MW Cl = 35.5

MW AgCl = 143.32

fraction of Cl = 35.5/143.32 = 0.2477

then

mass of Cl = 1.0569 *0.2477 = 0.26179413 g of Cl

% mass of Cl in sample = mass of Cl / sample mass * 100 = 0.2618 / 0.4958 *100 = 52.80354 % of smaple is Cl

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