When organic compounds containing sulfur are bumed, sulfur dioxide is produced. The amount of \(\mathrm{SO}_{2}\) formed can be determined by the reaction with hydrogen peroxide: $$\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q)$$ The resulting sulfuric acid is then titrated with a standard NaOH solution. A 1.302 -g sample of coal is burned and the \(\mathrm{SO}_{2}\) is collected in a solution of hydrogen peroxide. It took 28.44 \(\mathrm{mL}\) of a $0.1000-M \mathrm{NaOH}$ solution to titrate the resulting sulfuric acid. Calculate the mass percent of sulfur in the coal sample. Sulfuric acid has two acidic hydrogens.

The balanced chemical equation for the reaction between sulfuric acid and sodium hydroxide is: $$\mathrm{H}_{2}\mathrm{SO}_{4}(a q) +2\mathrm{NaOH}(a q)\longrightarrow \mathrm{Na}_{2}\mathrm{SO}_{4}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l)$$ In this equation, one mole of sulfuric acid reacts with two moles of sodium hydroxide to produce one mole of sodium sulfate and two moles of water.

It took 28.44 mL of a 0.1000-M NaOH solution to titrate the resulting sulfuric acid. First, calculate the moles of NaOH used: $$moles \,of\, NaOH = Molarity \times Volume$$ where volume is converted from mL to L. $$moles \,of\, NaOH = 0.1000 \frac{moles}{L} \times 0.02844 L = 0.002844 \,moles \,of\, NaOH$$ From the stoichiometry of the balanced chemical equation, two moles of NaOH react with one mole of sulfuric acid. We use this information to determine the moles of sulfuric acid produced: $$moles \,of \,H_{2}SO_{4} = \frac{1}{2} \times moles \,of\, NaOH = \frac{1}{2} \times 0.002844 = 0.001422 \,moles \,of \,H_{2}SO_{4}$$

The molar mass of sulfuric acid (H2SO4) is 98 g/mol, and the molar mass of sulfur (S) is 32 g/mol. To determine the mass of sulfur in sulfuric acid, we can set up the following relationship: $$\frac{mass \,of\, S}{mass \,of\, H_{2}SO_{4}} = \frac{molar \,mass \,of \,S}{molar \,mass \,of \,H_{2}SO_{4}}$$ Solve for the mass of S: $$mass \,of\, S =\frac{molar \,mass\, of\, S}{molar \,mass\, of\, H_{2}SO_{4}} \times moles \,of\, H_{2}SO_{4} \times molar\, mass\, of\, H_{2}SO_{4}$$ $$mass\, of\, S =\frac{32 g/mol}{98 g/mol} \times 0.001422 moles \times 98 g/mol = 0.0466 \,g \,of \,S$$ Finally, calculate the mass percent of sulfur: $$Mass \,percent\, of\, S = \frac{mass \,of \,S}{mass\, of\, coal \,sample} \times 100$$ $$Mass\, percent \,of\, S = \frac{0.0466 \,g}{1.302 \,g} \times 100 = 3.58 \%$$ The mass percent of sulfur in the coal sample is approximately 3.58%.