Question

Question 10 chap 4

A sample of 0.7960 g of an unknown compound containing barium ions (Ba2+) is dissolved in water and treated with an excess of Na2SO4. If the mass of the BaSO4 precipitate formed is 0.7651 g, what is the percent by mass of Ba in the original unknown compound?

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 = 0.7651 g

we have below equation to be used:

number of mol of BaSO4,

n = mass of BaSO4/molar mass of BaSO4

=(0.7651 g)/(233.37 g/mol)

= 3.278*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 * 3.278*10^-3

= 3.278*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

= 3.278*10^-3 mol * 137.3 g/mol

= 0.4501 g

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

= 0.4501 * 100 / 0.7960

= 56.5 %

Answer: 56.5 %