Question

A sample of 0.7360 g of an unknown compound containing barium ions (Ba2+) is dissolved in water and treated with an excess of Na2CO3. If the mass of the BaCO3 precipitate formed is 0.7578 g, what is the percent by mass of Ba in the original unknown compound?

Answer 1

mass of Compound = 0.7360 g

mass of BaCO3 = 0.7578 g

The balanced equation is

Ba+2(aq) + Na2CO3(aq) --------------------- BaCO3(s) + 2 Na+(aq)

1 mole                                                        1 mole

Na2CO3 is excess . So, Ba+2 ions are limiting reagent

molar mass of BaCO3 = 197.34 g/mole

molar mass of Ba+2 ions = 137.33 g

mass of Ba+2 ions present in 197.34 g of BaCO3 = 137.33 grams

mass of Ba+2 present in 0.7578 g of BaCO3 = ?

                                                                        = 137.33 x 0.7578 / 197.34 = 0.527 g

mass of Ba+2 ions = 0.527 g

% by mass of Ba+2 ions = mass of solute / mass of solution x100

% by mass of Ba+2 ions = 0.527 x100/0.7360 = 71.6%

percent by mass of Ba+2 ions = 71.6%