Question

A student performs a gravimetric analysis experiment and is given 1.94 g 1.94 g of a contaminated mixture containing anhydrous magnesium chloride and potassium nitrate. to determine the percentage by mass of magnesium chloride in the mixture, excess silver nitrate is added to the mixture to precipitate the chloride ion as silver chloride. the mass of the silver chloride precipitate is found to be 1.43 g 1.43 g. which of the following is the mass percent of magnesium chloride in this sample?

a.25%
b.24%
c. 47%
d. 74%

Answer 1

To determine the mass percent of magnesium chloride in the mixture, calculate the mass of magnesium chloride and divide it by the total mass of the mixture, then multiply by 100. The mass percent is not one of the given answer choices. None of the options is correct

Step-by-step explanation:

To determine the mass percent of magnesium chloride in the mixture, we need to calculate the mass of magnesium chloride and divide it by the total mass of the mixture, then multiply by 100. The mass of the silver chloride precipitate is 1.43 g. Since silver chloride is AgCl and contains one chloride ion (Cl-) for every Ag+, the molar mass of chloride is equal to the molar mass of AgCl. The molar mass of AgCl is 143.32 g/mol (107.87 g/mol for Ag + 35.45 g/mol for Cl).

Moles of Cl- = Mass of AgCl / Molar mass of AgCl = 1.43 g / 143.32 g/mol = 0.009968 mol

Moles of MgCl2 = 2 x Moles of Cl- (from the balanced equation) = 2 x 0.009968 mol = 0.019936 mol

Mass of MgCl2 = Moles of MgCl2 x Molar mass of MgCl2 = 0.019936 mol x (95.21 g/mol for Mg + 2 x 35.45 g/mol for Cl) = 3.3 g

Mass percent of MgCl2 = (Mass of MgCl2 / Total mass of mixture) x 100 = (3.3 g / 1.94 g) x 100 = 170.1%

Therefore, the mass percent of magnesium chloride in this sample is 170.1%, which is not one of the given answer choices. Please check the question or options provided.