Question

What mass of metal is present in a 45.8g mass of barium sulphate?

Answer 1

Hint: This problem can be easily solved by using the formula to find the no. of moles. Once the no. of moles of Barium are known, the mass of barium (in grams) can be found out by multiplying the moles of barium with the molar mass of barium metal.

Complete answer:
The formula for Barium Sulphate can be given as $BaS{O_4}$ .
To find the no. of moles present in 45.8 g of $BaS{O_4}$ we need to find the molar mass of $BaS{O_4}$ .
The molar mass of $BaS{O_4}$ = molar mass of Ba + molar mass of S + $4 \times $ molar mass of oxygen.
The molar mass of Barium $ = 137g/mol$
Molar mass of Sulphur $ = 32g/mol$
Molar mass of Oxygen $ = 16g/mol$
The molar mass of Barium Sulphate $ = 137 + 32 + 4 \times 16 = 233g/mol$
The no. of moles of a substance can be given by the formula: $Moles = \dfrac{{mass(g)}}{{MolarMass{\text{ }}(g/mol)}}$
No. of moles of $BaS{O_4}$ in 45.8 g $ = \dfrac{{45.8}}{{233}} = 0.0196mol$
Barium sulphate has one mole of barium, one mole of sulphur and 4 moles of oxygen. We are concerned with only the barium atom. Therefore, the no. of moles of Barium Metal in $BaS{O_4}$ is 0.0196 moles.
The mass (in grams) of Barium in 0.0196 moles $ = moles \times Molar{\text{ }}Mas{s_{Barium}} = 0.0196 \times 137$
Mass of barium present in 45.8 grams $ \cong 27g$
This is the required answer.

Note:
Barium Sulphate is the sulphate salt of barium. Barium sulphate has two components; one is the barium and second is the sulphate anion. The central atom sulphur is attached to four oxygen atoms. The colour of $BaS{O_4}$ is crystalline white. It is insoluble in water and also in alcohol. It is soluble in acids and has no odour.