Lesson Video: Precipitation Gravimetry Chemistry

In this video, we will learn how to determine the mass or composition of an analyte using precipitation gravimetry.


Video Transcript

In this video, we will learn how to determine the mass or composition of an analyte using precipitation gravimetry.

Some chemical reactions that take place in water or in aqueous conditions produce in soluble products. There are many colorful examples of this. For example, if a solution of lead nitrate is mixed with a solution of potassium iodide, a bright yellow solid is immediately formed. This is a completely unexpected result in this reaction, since the two starting solutions are colorless. Both lead nitrate and potassium iodide are soluble salts. Most common nitrates are soluble in water. Most salts of group one metals are also soluble.

Lead nitrate and potassium iodide are ionic compounds. And when dissolved, the ions are free to move around and interact with each other. A double decomposition reaction takes place here, and the ion swap over. The two products formed are potassium nitrate and lead iodide. Lead iodide does not dissolve in water. It has very limited solubility, and it’s responsible for the dense yellow solid observed in this reaction. The yellow lead iodide solid is known as a precipitate. And the reaction could be described as a precipitation reaction. A precipitate is simply a solid formed in a reaction from a mixture of solutions.

So, how could we collect a pure dry sample of this lead iodide precipitate? Firstly, we would need to separate it from the surrounding liquid that contains other dissolved ions. A piece of filter paper and a filter funnel would be used for this task. This procedure is called filtration, not funneling as is often believed. The yellow lead iodide solid will not pass through the filter paper, whereas any solution will drop into the flask that supports the funnel. The yellow lead iodide that remains on the filter paper will be contaminated with other dissolved ions. These solid filtrate should therefore be rinsed with distilled or deionized water whilst it’s in the filter paper.

Now that the solid precipitate has been separated from the solution containing other dissolved ions, it must be dried fully to remove traces of water. This can be achieved by placing the filter paper and the solid in a warm dry place. Ideally, a thermostatic drying oven is used set to 110 degrees Celsius. This temperature is plenty hot enough to remove the water but not too hot to risk burning the paper or melting or decomposing the product. To ensure that all traces of water are truly removed, the paper and the filtrate can be removed and its mass recorded on an accurate balance. The paper and the filtrate is returned to the drying oven for a few more hours, and then its mass is rerecorded. If the dry masses are the same, we can be convinced that all the water has been removed. This technique is known as drying to constant mass.

Now that we’ve seen how a pure dry sample of a precipitate can be collected, we can explore how this idea can be used to quantitatively analyze solutions of ions. A quantitative chemical analysis is one where the amount of one or more constituents in a sample are determined. As an example, the result of the chemical analysis may be expressed in concentration units or as a percentage by mass. Beyond simply confirming the presence of various ions, precipitates can be used to perform quantitative chemical analysis. The exact mass of a precipitate formed must be established. This can be achieved by taking mass measurements before and after the filtration and drying processes described earlier.

For example, the mass of a dry clean filter paper is recorded on a three-decimal-place balance. Balances that are accurate to the nearest 0.1 of a milligram are often referred to as analytical balances. They measure mass to four decimal places. As these balances are very sensitive, they usually incorporate a draft shield to stop airflow. The clean dry filter paper would then be used to collect all of the precipitate from a chemical reaction. The filter paper and the precipitate would be washed and dried, as described earlier. Having been dried to constant mass, the mass of the filter paper and the precipitate is then recorded. The difference in mass is equivalent to the mass of precipitate that was formed in the chemical reaction.

Sometimes the precipitate is dried in an ignition crucible using ashless filter paper. So we do not need to worry about preweighing the filter paper first. When a substance is analyzed based upon its mass, the method of analysis is known as gravimetry. The quantitative analytical method can be described as a gravimetric technique. In a chemical analysis, the substance being analyzed is often referred to as the analyte. In precipitation gravimetry, the analyte may be a soluble compound that is converted to an insoluble precipitate during the course of a simple reaction.

As an example, we could take a solution of sodium chloride which will be the analyte that is to be determined here. To this analyte solution we could add an excess of silver nitrate solution. The chemical reaction would produce a white precipitate of silver chloride and a solution of sodium nitrate. Using an analytical balance and a piece of filter paper, the mass of silver chloride precipitate can be established as described earlier. If we want to answer the question “What mass of sodium chloride was in the analyte solution?,” we need to look carefully at the balanced equation for the reaction that occurred.

From the balanced chemical equation, we can see that one mole of sodium chloride reacts with one mole of silver nitrate. Since the silver nitrate is in excess, all of the chloride ions are reacted. One mole of chloride ions, therefore, produces one mole of silver chloride as a solid precipitate. To convert the mass of silver chloride precipitated to the moles of silver chloride precipitated, we would take its mass and divide by the molar mass for silver chloride. The molar mass for silver chloride is found to be 143.4 grams per mole. The moles of silver chloride formed is therefore 0.139 grams divided by 143.4 grams per mole. This equates to 0.000969 moles of silver chloride.

The moles of silver chloride is now linked to the moles of sodium chloride in the analyte solution via the mole ratio in this balanced equation. This will enable us to find the mass of sodium chloride in this analyte solution. Since the mole ratio of silver chloride to sodium chloride in this balanced equation is one to one, there’s an equal number of moles of sodium chloride present in the original analyte solution. The molar mass of sodium chloride equates to 58.5 grams per mole. The mass of sodium chloride in the analyte solution is therefore equal to the moles of sodium chloride multiplied by its molar mass. This equates to 0.000969 moles multiplied by 58.5 grams per mole. The mass of sodium chloride in the analyte solution was therefore 0.0567 grams to three significant figures.

We will now look at a question to further develop the concept of precipitation gravimetry.

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.

Now it’s time to review the key points. Some chemical reactions produce precipitates. A precipitate is a solid product formed from a mixture of reacting solutions. A precipitate may be collected by filtration, washing, and then drying. Gravimetric analysis is a quantitative technique where the analysis is based upon mass measurements. The mass of a precipitate formed can be linked to the mass of a reacting analyte using a balanced equation.

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