Question

The sulfur content of an ore is determined gravimetrically by reacting the ore with concentrated nitric acid and potassium chlorate, converting all sulfur to sulfate. The excess nitrate and chlorate is removed by reaction with concentrated hydrochloric acid and the sulfate is precipitated using barium cation.
Ba2+ (aq) + SO42- (aq) = BaSO4 (s)

Analysis of 12.3430 grams of a sulfur containing ore yielded 12.5221 grams of BaSO4. What is the percent by mass sulfur in the ore? (BaSO4 = 233.43 g/mol). Show all calculation.

Answer 1

Answer:

13.92 %

Explanation:

Mass of $BaSO_4$ = 12.5221 g

Molar mass of $BaSO_4$ = 233.43 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{12.5221\ g}{233.43\ g/mol}$

Moles of $BaSO_4$ = 0.0536 moles

According to the given reaction,

$Ba^{2+}_{(aq)}+SO_4^{2-}_{(aq)}\rightarrow BaSO_4_{(s)}$

1 mole of $BaSO_4$ is formed from 1 mole of $SO_4^{2-}$

Thus,

0.0536 moles of $BaSO_4$ is formed from 0.0536 moles of $SO_4^{2-}$

Moles of $SO_4^{2-}$ = 0.0536 moles

Moles of sulfur in 1 mole $SO_4^{2-}$ = 1 mole

Moles of sulfur in 0.0536 mole $SO_4^{2-}$ = 0.0536 mole

Molar mass of sulfur = 32.065 g/mol

Mass = Moles * Molar mass = 0.0536 * 32.065 g = 1.7187 g

Mass of ore = 12.3430 g

Mass % = $\frac{Mass\ of\ Sulfur}{Mass_{ore}}\times 100$ = $\frac{1.7187}{12.3430}\times 100$ = 13.92 %