Question

To analyze for barium in an unknown white powder, an excess of sodium sulfate solution was added to a 0.624 g sample of the powder dissolved in water. A white precipitate of BaSO4 was isolated, dried, and found to weight 0.438 g. What is the mass percent of Ba in the white powder sample?

a. 0.267%

b. 41.3%

c. 58.9%

d. 70.2%

e. none of the above Write the balanced chemical equation: Show all calculation:

Answer 1

Molar mass of BaSO4 = 1*MM(Ba) + 1*MM(S) + 4*MM(O)

= 1*137.3 + 1*32.07 + 4*16.0

= 233.37 g/mol

mass of BaSO4 = 0.438 g

we have below equation to be used:

number of mol of BaSO4,

n = mass of BaSO4/molar mass of BaSO4

=(0.438 g)/(233.37 g/mol)

= 1.877*10^-3 mol

This is number of moles of BaSO4

one mole of BaSO4 contains 1 moles of Ba

we have below equation to be used:

number of moles of Ba = 1 * number of moles of BaSO4

= 1 * 1.877*10^-3

= 1.877*10^-3

Molar mass of Ba = 137.3 g/mol

we have below equation to be used:

mass of Ba,

m = number of mol * molar mass

= 1.877*10^-3 mol * 137.3 g/mol

= 0.2577 g

Mass % Ba = mass of Ba * 100 / mass of sample

= 0.2577*100/0.624

= 41.3 %

Answer: b