Question

A series of sulfate samples is to be analyzed by precipitation as BaSO₄. if it is known that the sulfate content in these samples ranges between 20% and 55%,

what minimum sample mass should be taken to ensure that a precipitate mass no smaller than 0.200 g is produced?

Answer 1

A minimum sample mass of 0.5155 g should be taken to ensure that the produced precipitate of BaSO4 is no less than 0.200 g, assuming a sulfate content of no less than 20% in the samples.

Step-by-step explanation:

To find the minimum sample mass that should be taken to ensure that a precipitate mass no smaller than 0.200 g of BaSO4 is produced, we must consider the sample with the lowest possible sulfate concentration of 20%. First, we'll use the molar mass of BaSO4 to find how many moles 0.200 g represents.

0.200 g BaSO4 × (1 mol BaSO4 / 233.43 g BaSO4) = 0.000857 mol BaSO4

Since the stoichiometry between BaSO4 and MgSO4 in the reaction is 1:1, the moles of MgSO4 needed for BaSO4 to precipitate would be the same.

0.000857 mol MgSO4 × (120.37 g MgSO4 / 1 mol MgSO4) = 0.1031 g MgSO4

This is the amount of MgSO4 needed to produce 0.200 g of BaSO4. However, in the question, MgSO4 represents only 20% of the mixture at minimum. To find the total sample mass, divide by the percentage composition:

0.1031 g / 0.20 = 0.5155 g

This means a sample of at least 0.5155 g should be taken to ensure a precipitate of BaSO4 is no less than 0.200 g.