Question

A sample of bismuth weighing 0.687 g was converted to bismuth chloride by reacting it first with HNO3, and then with HCI, 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?
(b) What is the theoretical percent of Bi in the bismuth chloride, based on this formula?

Answer 1

Answer:

The empirical formula is BiCl3

% Bi = 66.27 %

Explanation:

Step 1: Data given

Mass of bismuth = 0.687 grams

Mass of bismuth chloride produced = 1.032 grams

Molar mass of bismuth = 208.98 g/mol

Molar mass of bismuth chloride = 315.33 g/mol

Step 2: The balanced equation

Step 3: Calculate moles of Bi

Moles Bi = mass Bi / molar mass Bi

Moles Bi = 0.687 grams / 208.98 g/mol

Moles Bi =  0.00329 moles

Step 4: Calculate moles of Cl

Mass of Cl = 1.032 - 0.687  = 0.345 moles

Moles Cl = 0.345 moles / 35.45 g/mol

Moles Cl = 0.00973 moles Cl

Step 5: Calculate mol ratio

We divide by the smaller number of moles:

Bi: 0.00329 / 0.00329 = 1

Cl: 0.0097The empirical formula is BiCl33/0.00329 = 3

Step 6: Calculate molar mass of BiCl3

Molar mass = 208.98 + 3*35.45 = 315.33 g/mol

Step 7: Calculate percent of Bi

% Bi = (208.98 / 315.33) * 100%

% Bi = 66.27 %