When organic compounds containing sulfur are burned, 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 \(\mathrm{NaOH}\) solution. A \(1.302-\mathrm{g}\) 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.

Step by step solution

01

Calculate the moles of NaOH used

We can calculate the moles of NaOH by multiplying its concentration by the volume in liters: moles of \(\mathrm{NaOH} = \mathrm{Molarity} \times \mathrm{Volume}\) moles of \(\mathrm{NaOH} = (0.1000\: \mathrm{mol/L}) \times (28.44 \times 10^{-3}\: \mathrm{L})\) moles of \(\mathrm{NaOH} = 0.002844 \: \mathrm{mol}\)

02

Determine the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\)

Since the balanced equation shows a 1:1:1 ratio between \(\mathrm{NaOH}\), \(\mathrm{SO}_{2}\), and \(\mathrm{H}_{2}\mathrm{SO}_{4}\) (keeping in mind that there are two acid hydrogens in \(\mathrm{H}_{2}\mathrm{SO}_{4}\)), we can determine that the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) and \(\mathrm{SO}_{2}\) are equal to the moles of \(\mathrm{NaOH}\). In other words: moles of \(\mathrm{H}_{2}\mathrm{SO}_{4} = \) moles of \(\mathrm{SO}_{2} = 0.002844 \: \mathrm{mol}\)

03

Calculate the moles of sulfur in the original coal sample

Each mole of \(\mathrm{SO}_{2}\) has one mole of sulfur. Thus, moles of S = moles of \(\mathrm{SO}_{2} = 0.002844 \: \mathrm{mol}\)

04

Calculate the mass of sulfur

To calculate the mass of sulfur, we need to multiply the moles of sulfur by its molar mass. The molar mass of sulfur is 32.06 g/mol: mass of S = moles of S × molar mass of S mass of S = (0.002844 \: \mathrm{mol}) \times (32.06 \: \mathrm{g/mol}) mass of S = 0.09114 \: \mathrm{g}$

05

Find the mass percent of sulfur in the coal sample

To find the mass percent of sulfur, we need to divide the mass of sulfur by the total mass of the coal sample and multiply the result by 100: mass percent of S = \(\frac{\text{mass of S}}{\text{mass of coal sample}} \times 100\) mass percent of S = \(\frac{0.09114 \: \mathrm{g}}{1.302 \: \mathrm{g}} \times 100\) mass percent of S = 7.00 % So, the mass percent of sulfur in the coal sample is 7.00 %.

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