Question

1. In sulphur estimation 0.157g of an organic compound gave 0.4813 g of BaSo4. What is the percentage of sulphur in the organic compund?

Answer 1

137+32+4x16= 233.

And it's 0.157g.

Because of these mistakes I do think ur answer is wrong and thereby I provide u the right one.

Thank you! :=)

Answer 2

Answer : The percentage of sulfur in the organic compound is, 42.10 %

Solution : Given,

Mass of organic compound = 0.157 g

Mass of $BaSO_24$ = 0.4813 g

Molar mass of $BaSO_24$ = 233 g/mole

First we have to calculate the mass of sulfur.

As, 233 g of $BaSO_24$ contains 32 g of sulfur

So, 0.4813 g of $BaSO_24$ contains $\frac{32}{233}\times 0.4813=0.066g$ of sulfur

Now we have to calculate the percentage of sulfur in the organic compound.

$\text{Percentage of sulfur}=\frac{\text{Mass of sulfur}}{\text{Mass of organic compound}}\times 100=\frac{0.066}{0.157}\times 100=42.10\%$

Therefore, the percentage of sulfur in the organic compound is, 42.10 %