Question Video: Using Precipitation Gravimetry to Identify the Halogen in a Compound Chemistry

A magnesium halide salt has the formula MgX₂. A 0.593-g sample of MgX₂ was dissolved in 100 mL of deionized water, followed by the addition of excess NaOH. The precipitate of Mg(OH)₂ was filtered, washed, and dried. The precipitate was found to have a mass of 0.187 g. What is the identity of X? [A] I [B] F [C] Br [D] Cl

05:43

Video Transcript

A magnesium halide salt has the formula MgX2. A 0.593-gram sample of MgX2 was dissolved in 100 milliliters of deionized water, followed by the addition of excess NaOH. The precipitate of MgOH2 was filtered, washed, and dried. The precipitate was found to have a mass of 0.187 grams. What is the identity of X? (A) I, (B) F, (C) Br, (D) Cl.

In this question, we’re being asked to find the identity of X which will be the halogen that’s responsible for the halide ion in the ionic salt MgX2. The halogen present as the halide ion in MgX2 could be iodine, fluorine, bromine, or chlorine. With this in mind, the actual formula of MgX2 could be MgI2, MgF2, MgBr2, or MgCl2. Firstly, we need to look at the reaction that occurs between the magnesium halide salt and NaOH, which is sodium hydroxide. Both are dissolved in water, so these are aqueous solutions. We are told in this question that one of the products is MgOH2. This is magnesium hydroxide, which is an insoluble precipitate. The other product of this reaction will be a combination of the sodium ions and the halide ions from the two reactants concerned. Sodium halides are soluble, so this product is an aqueous solution.

We now need to balance this equation. We can see from the formula for magnesium hydroxide that one mole of magnesium hydroxide contains two moles of hydroxide ions. This makes sense as the magnesium ion is present as a two plus ion and the hydroxide ions are one minus ions. The charges must balance up, so we need two hydroxide ions for every magnesium two plus ion. We therefore need two moles of sodium hydroxide. This will provide two moles of sodium ions and two moles of hydroxide ions on the left side of our equation, which will balance with the hydroxide ions on the right side. We therefore need to form two moles of NaX, our sodium halide, which will contain two moles of sodium ions. Two moles of NaX also contains two moles of X−, or halide ions, which balance with the two moles of halide ions in the magnesium halide salt we started with.

This equation is now balanced, and we can use it with the quantitative data provided in the question. We know in this question that the mass of magnesium hydroxide precipitate collected was 0.187 grams. Since we know the precise composition of magnesium hydroxide from its formula, we can find the molar mass of magnesium hydroxide. This equates to 58.3 grams per mole. The moles of magnesium hydroxide produced can then be found by taking its mass in grams and dividing it by the molar mass in grams per mole. The moles of magnesium hydroxide is therefore 0.0032.

We can link the moles of magnesium hydroxide formed to the moles of magnesium halide salt in the analyte using the mole ratio in the balanced equation. One mole of magnesium hydroxide originates from one mole of magnesium halide salt. There must therefore have been 0.0032 moles of the magnesium halide salt in the sample. Since the 0.0032 moles of magnesium halide salt was found in 0.593 grams of sample, we can use this information to find the molar mass.

Since we now have the moles and the mass for the magnesium halide salt, we can use this information to find the molar mass of the magnesium halide salt itself. The molar mass of MgX2 is the mass of the sample divided by the number of moles that it contains. The molar mass of MgX2 is therefore 185.3 grams per mole. Using the atomic mass from magnesium, we can find the atomic mass for the element X in this magnesium halide salt. By subtracting the atomic mass for magnesium from the molar mass of the magnesium halide salt, we have the mass of twice the halogen present. Twice the atomic mass of the halogen present equates to 161. The atomic mass of the halogen X is therefore 161 divided by two, which equals 80.5.

We need to bear in mind that this will not be the precise atomic mass of our halogen X, as it’s derived from experimental data where there are errors. The only halogen with an atomic mass close to 80.5 is bromine. The atomic mass of bromine is 79.9. The correct answer is therefore bromine.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.