Question

In order to determine the purity of ammonium sulfate, a sample with a mass of \(0.850 \mathrm{~g}\) is dissolved in \(\mathrm{KOH}\). The equation for the reaction that takes place is $$ \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O} $$ The ammonia liberated is distilled into a flask that contains \(50.00 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HCl}\). Not all the \(\mathrm{HCl}\) is consumed. The excess \(\mathrm{HCl}\) reacts with \(17.3 \mathrm{~mL}\) of \(0.120 \mathrm{M} \mathrm{NaOH}\). What is the mass percent of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in the sample?

Short answer

Answer: The mass percent of ammonium sulfate in the given sample is 81.01%.

Step-by-step solution

Calculate the moles of excess HCl

We will use the relationship between moles, volume, and concentration to calculate the moles of excess HCl. Moles of HCl = Volume of HCl x Concentration of HCl Here, we have the volume of NaOH that reacted with the excess HCl, and its concentration. So, first, we need to find the moles of NaOH and then we can find the moles of excess HCl. Moles of NaOH = 17.3 mL x 0.120 M = 0.002076 mol Since 1 mol of NaOH reacts with 1 mol of HCl, therefore, the moles of excess HCl are equal to the moles of NaOH. Moles of excess HCl = 0.002076 mol

Find the moles of reacted HCl with ammonia

We know the volume and concentration of the initial HCl solution. Total moles of HCl = Volume of HCl x Concentration of HCl Total moles of HCl = 50.00 mL x 0.250 M = 0.01250 mol As we have the moles of excess HCl, we can find out the moles of HCl that reacted with ammonia. Moles of reacted HCl with ammonia = Total moles of HCl - Moles of excess HCl Moles of reacted HCl with ammonia = 0.01250 mol - 0.002076 mol = 0.01042 mol

Find the moles of ammonia that reacted

In the balanced equation, each mole of NH4+ reacts with 1 mole of OH- to form 1 mole of NH3. So the moles of NH3 formed are equal to the moles of NH4+ which reacted. Moles of NH4+ = Moles of reacted HCl with ammonia Moles of NH4+ = 0.01042 mol

Calculate the mass of ammonium sulfate

We know the moles of NH4+ and the molar mass of ammonium sulfate ((NH4)2SO4). We can calculate the mass of ammonium sulfate: Molar mass of ammonium sulfate = (2 x (14.01+4 x 1.008) + 32.07 + 4 x 16.00) g/mol = 132.14 g/mol Mass of ammonium sulfate = Moles of NH4+ x (Molar mass of ammonium sulfate/2) Mass of ammonium sulfate = 0.01042 mol x (132.14 g/mol / 2) = 0.6886 g

Calculate the mass percent of ammonium sulfate

Now we have the mass of ammonium sulfate and the mass of the sample. We will use the following formula: Mass percent of ammonium sulfate = (Mass of ammonium sulfate / Mass of sample) x 100 Mass percent of ammonium sulfate = (0.6886 g / 0.850 g) x 100 = 81.01% The mass percent of ammonium sulfate in the sample is 81.01%.