Question

A 2.5600 gram sample of a sulfur-containing compound is analyzed by precipitating the sulfur as BaSO4. If 1.1756 g of BaSO4 are obtained, what is the percentage of sulfur in the sample?

Answer 1

Answer: The percent sulfur in the compound is 6.308 %.

Explanation:

We are given:

Mass of barium sulfate = 1.1756 g

We know that:

Molar mass of barium sulfate = 233.38 g/mol

Molar mass of sulfur atom = 32.07 g/mol

As, all the sulfur in the compound is converted to barium sulfate. So, the mass of sulfur in barium sulfate will be equal to the mass of sulfur present in the compound.

To calculate the mass of sulfur in given mass of barium sulfate, we apply unitary method:

In 233.38 g of barium sulfate, mass of sulfur present is 32.07 g

So, in 1.1756 g of barium sulfate, mass of sulfur present will be = $\frac{32.07}{233.38}\times 1.1756=0.1615g$

To calculate the percentage composition of sulfur in compound, we use the equation:

$\%\text{ composition of sulfur}=\frac{\text{Mass of sulfur}}{\text{Mass of compound}}\times 100$

Mass of compound = 2.5600 g

Mass of sulfur = 0.1615 g

Putting values in above equation, we get:

$\%\text{ composition of sulfur}=\frac{0.1615g}{2.5600g}\times 100=6.308\%$

Hence, the percent sulfur in the compound is 6.308 %.

Answer 2

Supposing all the sulfur in the BaSO4 came from the sample being analyzed: 

(1.1756 g)/(233.3909 g/mol) x (1 mol S/ 1 mol BaSO4) x (32.0655 g S/mol) / (2.5600 g) = 0.063092
                 = 6.3092% Sulfur