Question

A 0.5895-g sample of impure magnesium hydroxide is dissolved in \(100.0 \mathrm{~mL}\) of \(0.2050 \mathrm{M} \mathrm{HCl}\) solution. The excess acid then needs \(19.85 \mathrm{~mL}\) of \(0.1020 \mathrm{M}\) \(\mathrm{NaOH}\) for neutralization. Calculate the percent by mass of magnesium hydroxide in the sample, assuming that it is the only substance reacting with the HCl solution.

Short answer

The percentage by mass of magnesium hydroxide in the impure sample is found to be 100.36%. However, this value is greater than 100%, indicating a possible error in measurement, assumption, or provided data. It is necessary to recheck the given data and calculations.

Step-by-step solution

Calculate moles of \(\mathrm{HCl}\) reacted with magnesium hydroxide

The first thing we need to do is find how many moles of HCl reacted with magnesium hydroxide. To do this, we can use the information given about the volume and concentration of NaOH. The balanced equation for the reaction between NaOH and HCl is: \(\mathrm{NaOH + HCl \rightarrow NaCl + H_2O}\) From the stoichiometry, there is a 1:1 relation between NaOH and HCl. Lets find the moles of NaOH used for the neutralization: moles(NaOH) = M(\(\mathrm{NaOH}\)) * V(\(\mathrm{NaOH}\)) where M(\(\mathrm{NaOH}\)) is the molarity of NaOH and V(\(\mathrm{NaOH}\)) is the volume of NaOH. moles(NaOH) = \(0.1020 \mathrm{M} \times \frac{19.85\,\mathrm{mL}}{1000\,\mathrm{mL}}\) moles(NaOH) = \(2.0307 \times 10^{-3}\) moles Since there is a 1:1 ratio between moles of NaOH and HCl, there are also \(2.0307 \times 10^{-3}\) moles of HCl that reacted with the magnesium hydroxide.

Calculate moles of magnesium hydroxide

Now we can find the moles of \(\mathrm{Mg(OH)_2}\) by looking at the balanced chemical equation of the reaction between magnesium hydroxide and the remaining hydrochloric acid. \(\mathrm{Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O}\) From the stoichiometry, 1 mole of magnesium hydroxide reacts with 2 moles of HCl. We can use the calculated moles of HCl to find the moles of magnesium hydroxide in the given sample: moles(\(\mathrm{Mg(OH)_2}\)) = \(\frac{1}{2}\) * moles(\(\mathrm{HCl}\)) moles(\(\mathrm{Mg(OH)_2}\)) = \(\frac{1}{2} \times (2.0307 \times 10^{-3}) \, \mathrm{moles}\) moles(\(\mathrm{Mg(OH)_2}\)) = \(1.0154 \times 10^{-3}\) moles

Calculate the mass of magnesium hydroxide

Now we can find the mass of magnesium hydroxide in the sample by using the molar mass of \(\mathrm{Mg(OH)_2}\): M(Mg(OH)\(_2\)) = 24.31 g/mol (Mg) + 2 * (16.00 g/mol (O) + 1.01 g/mol (H)) M(Mg(OH)\(_2\)) = 58.33 g/mol Now we can find the mass: mass (Mg(OH)\(_2\)) = moles (Mg(OH)\(_2\)) * M(Mg(OH)\(_2\)) mass (Mg(OH)\(_2\)) = \(1.0154 \times 10^{-3}\, \mathrm{moles} \times 58.33 \, \mathrm{g/mol}\) mass (Mg(OH)\(_2\)) = 0.5916 g

Calculate the percentage by mass of magnesium hydroxide in the sample

We can now calculate the percentage by mass of magnesium hydroxide in the sample as follows: percentage by mass of \(\mathrm{Mg(OH)_2}\) = \(\frac{mass \, (\mathrm{Mg(OH)_2})}{total \, mass \, (sample)}\times 100\) percentage by mass of \(\mathrm{Mg(OH)_2}\) = \(\frac{0.5916 \, \mathrm{g}}{0.5895 \, \mathrm{g}}\times 100\) percentage by mass of \(\mathrm{Mg(OH)_2}\) = 100.36% Since the percentage of magnesium hydroxide calculated is greater than 100%, it means that either an error occurred in the measurement or assumption, or the provided data is incorrect. We need to recheck the given data and the calculations.

Key concepts

Molar Mass

Understanding molar mass is essential when diving into the realm of chemistry. It is the weight of one mole of a substance, essentially a counting unit representing 6.02 x 10^23 particles of the substance, and is expressed in grams per mole (g/mol). When calculating the molar mass of a compound, such as magnesium hydroxide (\r\(\mathrm{Mg(OH)_2}\)), each element's molar mass is multiplied by its occurrence in the formula and then added together.

For magnesium hydroxide, the calculation would involve the atomic mass of magnesium (Mg), oxygen (O), and hydrogen (H). Each element’s molar mass – 24.31 g/mol for Mg, 16.00 g/mol for each O, and 1.01 g/mol for each H – contributes to the compound’s total molar mass. Hence, the molar mass of \r\(\mathrm{Mg(OH)_2}\) is calculated as:\r\[M(\mathrm{Mg(OH)_2}) = 24.31\,\mathrm{g/mol}\,(\mathrm{Mg}) + 2 \times (16.00\,\mathrm{g/mol}\,(\mathrm{O}) + 1.01\,\mathrm{g/mol}\,(\mathrm{H}))\], which equals 58.33 g/mol.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It's essentially the math behind chemistry. By using stoichiometry, we can predict the amount of product that will form during a reaction, as well as how much of the reactants are needed to produce a desired amount of product.

In the given exercise, stoichiometry helps us determine the relationship between magnesium hydroxide and hydrochloric acid (HCl). The balanced equation \r\(\mathrm{Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O}\) indicates that one mole of \r\(\mathrm{Mg(OH)_2}\) reacts with two moles of HCl. With this stoichiometric ratio, we can calculate the number of moles of magnesium hydroxide from the moles of HCl that have not reacted with the sample.

Acid-Base Titration

Acid-base titration is a practical application of stoichiometry used to determine the concentration of an unknown solution. In this process, a known concentration of one solution is gradually added to a known volume of another solution until the chemical reaction between the two is complete. The point at which the reaction is neutralized is often indicated by a color change with the help of an indicator.

For instance, in the textbook problem, the acid-base titration involves adding sodium hydroxide (NaOH) to the excess hydrochloric acid (HCl). The stoichiometric 1:1 ratio between NaOH and HCl allows the determination of the unreacted moles of HCl, which then can be used to figure out the amount of \r\(\mathrm{Mg(OH)_2}\) that reacted initially. It’s crucial to carry out the titration precisely, as an incorrect amount of NaOH added can lead to an inaccurate calculation of the percent composition of a substance in a sample.