Lesson Explainer: Precipitation Gravimetry Chemistry

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

Precipitation gravimetry is a quantitative analytic technique that can be used to determine the mass of an analyte. The analyte can be removed from the solution by forming a precipitate with a known composition. By weighing the precipitate and working backward, the initial mass of the analyte present in the solution can be determined.

A precipitation reaction is a reaction that forms an insoluble solid product from the reaction between two soluble substances in solution.

The chemical responsible for causing a precipitate to form is known as the precipitating agent.

Definition: Precipitation Gravimetry

Precipitation gravimetry is an analytical technique that uses the formation and mass of a precipitate to determine the mass of an analyte.

It is important for the precipitate to be a pure substance with a definite and known composition.

How To: Setting Up and Performing Precipitation Gravimetry

  1. The sample of interest is dissolved in a solvent, commonly water, to give an aqueous solution.
  2. An excess of the precipitation agent is then added to the aqueous solution. A precipitate should form.
  3. The solution is then filtered using ashless filter paper to separate the precipitate from the solution.
  4. A few drops of the precipitating agent are then added to the filtered solution to ensure all the analyte has precipitated.
  5. The solution is then refiltered using the same ashless filter paper.
  6. The precipitate is washed with some deionized water to remove any impurities.
  7. The precipitate and filter paper are then placed into a crucible and ignited. This both dries the precipitate and removes the filter paper.
  8. Once cool, the precipitate and the crucible are then weighed. The final mass of the precipitate is then obtained by subtracting the mass of the empty crucible from the mass of the crucible and precipitate.

Ashless filter paper is used so that during the heating and drying phase, the precipitate is not contaminated with ash. If needed, the precipitate can be left to dry further in a desiccator.

There are other considerations when performing precipitation gravimetry such as particle size, solubility, and impurities. However, we will assume that such considerations are not important.

Example 1: Explaining Why Ashless Filter Paper Is Used in Gravimetric Analysis

In the precipitation method, why do we use ashless filter paper in chemical analysis?

  1. Because it filters the precipitate from the solution efficiently
  2. Because we know its mass, so we can calculate the mass of the precipitate easily
  3. Because it is affordable
  4. Because it is completely ignited without leaving ash

Answer

As the name suggests, the precipitation method involves reacting an aqueous solution of our analyte with another aqueous solution to produce a precipitate. The precipitate that is produced should have a definite and known composition.

Once the precipitate has been removed, it needs to be separated from the solution. The best technique for this separation is filtration, which will leave the solid precipitate on the filter paper and allow the remaining solution to pass through.

However, the precipitate will still be wet. The precipitate can then be dried in a crucible at very high temperatures. The intense heating in the crucible will completely burn up the filter paper. If the filter paper produces ash when being burned, it could contaminate the precipitate. Ashless filter paper does not produce ash when burnt and therefore does not contaminate the precipitate.

The correct answer is, therefore, D.

It is essential for the precipitate to be completely dry as any remaining solvent will give an incorrect mass reading. Any error will propagate through the calculation and give an inaccurate mass of the analyte.

Precipitation gravimetry can be used to determine the mass of sodium sulfate in an aqueous solution. A good precipitating agent would be barium chloride, as the sulfate and barium ions would react to form the insoluble barium sulfate.

The barium sulfate can then be isolated and dried. Once the mass of the precipitate has been determined, the first step is to write a balanced chemical equation for the full reaction: BaCl()+NaSO()BaSO()+2NaCl()2244aqaqsaq

The equation shows that the number of moles of barium sulfate is equal to the number of moles of sodium sulfate in the original solution. The number of moles of barium sulfate can be calculated using the formula 𝑛=𝑚𝑀, where 𝑛 is the amount in moles, 𝑚 is the mass in grams, and 𝑀 is the molar mass in grams per mole.

Having obtained the number of moles of barium sulfate, the number of moles of sodium sulfate can be determined.

The mass of sodium sulfate in the original solution can then be calculated using the formula 𝑛×𝑀=𝑚, where 𝑛 is the amount in moles, 𝑚 is the mass in grams, and 𝑀 is the molar mass in grams per mole.

From this process, the mass of sodium sulfate, the analyte, in the original solution can be determined.

Example 2: Calculating the Mass of Analyte from the Mass of a Precipitate

Consider the equation 2NaOH()+MgCl()2NaCl()+Mg(OH)()aqaqaqs22

From the equation, what is the mass of NaOH when 5 g of Mg(OH)2 precipitates? Give your answer to one decimal place. [Na=23g/mol, O=16g/mol, H=1g/mol, Mg=24g/mol]

Answer

From the balanced chemical equation, the reaction of aqueous sodium hydroxide with aqueous magnesium chloride produces a precipitate of Mg(OH)2. We need to use the final mass of the precipitate to determine the mass of NaOH in the solution.

The precipitate of Mg(OH)2 was found to have a mass of 5 g. We can use this to calculate the number of moles of Mg(OH)2 in the precipitate using the equation 𝑛=𝑚𝑀, where 𝑛 is the amount in moles, 𝑚 is the mass in grams, and 𝑀 is the molar mass in grams per mole.

The molar mass of magnesium hydroxide can be calculated by summing the average molar masses of the constituent atoms: 𝑀=𝑀+(2×𝑀)+(2×𝑀)𝑀=24/+(2×16/)+(2×1/)𝑀=58/.()()()()()()Mg(OH)MgOHMg(OH)Mg(OH)222gmolgmolgmolgmol

We can then substitute the amount in moles and the molar mass into the equation: 𝑛=558/𝑛=0.0862.ggmolmoles

From the balanced chemical equation, we can see that two moles of NaOH produce one mole of Mg(OH)2. Therefore, the number of moles of NaOH is double that of Mg(OH)2: 𝑛=2×𝑛𝑛=2×0.0862𝑛=0.1724.()()()()NaOHMg(OH)NaOHNaOH2molesmoles

We can now calculate the mass for NaOH using the equation 𝑛=𝑚𝑀.

And rearranging for 𝑚, 𝑛×𝑀=𝑚.

The molar mass of NaOH is 𝑀=𝑀+𝑀+𝑀𝑀=23/+16/+1/𝑀=40/.()()()()()()NaOHNaOHNaOHNaOHgmolgmolgmolgmol

We can now substitute in the values for 𝑀 and 𝑛 to calculate the mass of NaOH: 𝑛×𝑀=𝑚0.1724×40/=𝑚6.8965=𝑚.molesgmolg

Rounding to one decimal place gives the mass of NaOH as 6.9 g.

Example 3: Balancing a Chemical Equation and Calculating the Mass of an Analyte from Precipitation Gravimetry

A chemist wants to determine the mass of calcium chloride present in an aqueous solution. Upon the addition of excess silver nitrate, a precipitate was seen to form. After filtration and drying, the mass of the precipitate was found to be 0.85 g.

  1. What is the equation for the reaction between calcium chloride and silver nitrate?
    1. CaCl()+2AgNO()2AgCl()+Ca(NO)()2332aqaqaqs
    2. CaCl()+AgNO()AgCl()+CaNO()aqaqss33
    3. CaCl()+AgNO()AgCl()+CaNO()aqaqsaq33
    4. CaCl()+2AgNO()2AgCl()+Ca(NO)()2332aqaqsaq
    5. 2CaCl()+Ag(NO)()AgCl()+2CaNO()aqaqsaq3223
  2. What is the mass of calcium chloride in the original solution? Give your answer to 2 decimal places. [Ca=40g/mol, Cl=35.5g/mol, Ag=108g/mol]

Answer

Part 1

The formulas for calcium chloride and silver nitrate are CaCl2 and AgNO3. From this, we can instantly conclude that options B, C, and E are incorrect.

Aqueous solutions of calcium chloride and silver nitrate will react in a double substitution reaction. The cations of each ionic compound will swap to form silver chloride and calcium nitrate. The question states that a precipitate is formed. In this reaction, the precipitate formed is silver chloride (AgCl), which will have the state symbol (s). From the remaining options, only D has silver chloride as a solid. The equation is also balanced and so the correct answer is D.

Part 2

From the balanced chemical equation, the reaction of aqueous silver nitrate with aqueous calcium chloride produces a precipitate of AgCl. We need to use the final mass of the precipitate to determine the mass of CaCl2 in the solution.

The precipitate of AgCl was found to have a mass of 0.85 g. We can use this to calculate the number of moles of AgCl in the precipitate, using the equation 𝑛=𝑚𝑀, where 𝑛 is the amount in moles, 𝑚 is the mass in grams, and 𝑀 is the molar mass in grams per mole.

The molar mass of silver chloride can be calculated by summing the average molar masses of the constituent atoms: 𝑀=𝑀+𝑀𝑀=108/+35.5/𝑀=143.5/.()()()()()AgClAgClAgClAgClgmolgmolgmol

We can then substitute the amount in moles and the molar mass into the equation: 𝑛=0.85143.5/𝑛=0.0059.ggmolmoles

From the balanced chemical equation, we can see that one mole of CaCl2 produces two moles of AgCl. Therefore, the number of moles of CaCl2 is half that of AgCl: 𝑛=0.5×𝑛𝑛=0.5×0.0059𝑛=0.0029.()()()()CaClAgClCaClCaCl222molesmoles

We can now calculate the mass of CaCl2 using the equation 𝑛=𝑚𝑀.

And rearranging for 𝑚, 𝑛×𝑀=𝑚.

The molar mass of CaCl2 is 𝑀=𝑀+(2×𝑀)𝑀=40/+(2×35.5/)𝑀=111/.()()()()()CaClCaClCaClCaCl222gmolgmolgmol

We can now substitute in the values for 𝑀 and 𝑛 to calculate the mass of CaCl2: 𝑛×𝑀=𝑚0.0029×111/=𝑚0.3287=𝑚.molesgmolg

Rounding to two decimal places gives us that the mass of CaCl2 in the solution is 0.33 g.

The precipitation gravimetry method can also be used to determine the composition of a salt or the percentage mass of a mixture of salts.

Example 4: Using Precipitation Gravimetry to Identify the Halogen in a Compound

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? [Mg=24g/mol, O=16g/mol, H=1g/mol, F=19g/mol, Br=80g/mol, Cl=35.5g/mol, I=127g/mol]

Answer

In this question, we are given a salt with the chemical formula MgX2 and need to identify X. All we know is that X is one of the halogens.

In the experiment, a precipitate of Mg(OH)2 is formed by reacting an aqueous solution of the unknown salt with an aqueous solution of NaOH. We can write the balanced chemical equation for this reaction: MgX+2NaOHMg(OH)+2NaX22

Even though X is unknown, as a halogen it has a valency of 1 and will therefore form an ionic compound with sodium of the form NaX.

The precipitate of Mg(OH)2 was found to have a mass of 0.187 g. We can use this to calculate the number of moles of Mg(OH)2 in the precipitate, using the equation 𝑛=𝑚𝑀, where 𝑛 is the amount in moles, 𝑚 is the mass in grams, and 𝑀 is the molar mass in grams per mole.

The molar mass of magnesium hydroxide can be calculated by summing the average molar masses of the constituent atoms: 𝑀=𝑀+(2×𝑀)+(2×𝑀)𝑀=24/+(2×16/)+(2×1/)𝑀=58/.()()()()()()Mg(OH)MgOHMg(OH)Mg(OH)222gmolgmolgmolgmol

We can then substitute the amount in moles and the molar mass into the equation: 𝑛=0.18758/𝑛=0.00322.ggmolmoles

From the balanced chemical equation, we can see that one mole of MgX2 produces one mole of Mg(OH)2. Therefore, the number of moles of MgX2 is 0.00322 moles.

We can now calculate the molar mass for MgX2 using the equation 𝑛=𝑚𝑀.

And rearranging for 𝑀, 𝑀=𝑚𝑛, where 𝑛 is the number of moles of MgX2, which we have calculated, and 𝑚 is the mass of MgX2, which is given in the question as 0.593 g. Substituting these values into the equation gives 𝑀=0.5930.00322=183.925/=184/.gmolesgmolgmol

The molar mass of MgX2 can be calculated as 𝑀=𝑀+(2×𝑀).()()()MgXMgX2

We have calculated 𝑀=184/()MgX2gmol, and know that 𝑀=24()Mg. We can therefore rearrange and solve for 𝑀()X: 184=24+(2×𝑀)18424=(2×𝑀)(18424)2=𝑀,()()()XXX which gives us a value for 𝑀()X of 80 g/mol.

We can then use the values given in the question, or a periodic table, to determine that the element with an atomic mass closest to 80 is bromine. Therefore, X is bromine.

Example 5: Using Precipitation Gravimetry to Calculate the Percentage Mass of NaCl in a Mixed Sample

A 2.45 g sample of white crystals is known to contain NaCl and KNO3. The sample is fully dissolved in deionized water and an excess of AgNO3 is then added, forming a precipitate of AgCl. Upon being filtered, washed, and dried, the precipitate is found to have a mass of 1.54 g. To the nearest integer, what is the percentage by mass of NaCl in the mixture? [Na=23g/mol, Cl=35.5g/mol, Ag=108g/mol]

Answer

In this question, we have a sample of two salts, NaCl and KNO3. However, we need to determine what percentage, by mass, of NaCl there is in the sample.

Both salts will dissolve in water to form an aqueous solution. When the precipitating agent AgNO3 is added to the solution, a precipitate of AgCl is formed. We can write a balanced chemical equation for the reaction between NaCl and AgNO3: NaCl()+AgNO()AgCl()+NaNO()aqaqsaq33

However, there will be no reaction between KNO3 and AgNO3. In aqueous solution, these two compounds will produce K+, Ag+, and NO3 ions.

The question tells us that the final mass of the AgCl precipitate is 1.54 g. From this, we can calculate the number of moles of AgCl using the equation 𝑛=𝑚𝑀, where 𝑛 is the amount in moles, 𝑚 is the mass in grams, and 𝑀 is the molar mass in grams per mole.

The molar mass of silver chloride can be calculated by summing the average molar masses of the constituent atoms: 𝑀=𝑀+𝑀𝑀=108/+35.5/𝑀=143.5/.()()()()()AgClAgClAgClAgClgmolgmolgmol

We can then substitute the amount in moles and the molar mass into the equation: 𝑛=1.54143.5/𝑛=0.010.ggmolmoles

From the balanced chemical equation, we can see that one mole of NaCl produces one mole of AgCl. Therefore, the number of moles of NaCl equals that of AgCl: 𝑛=𝑛𝑛=0.010.()()()NaClAgClNaClmoles

We can now calculate the mass of NaCl using the equation 𝑛=𝑚𝑀.

And rearranging for 𝑚, 𝑛×𝑀=𝑚.

The molar mass of NaCl is 𝑀=𝑀+𝑀𝑀=23/+35.5/𝑀=58.5/.()()()()()NaClNaClNaClNaClgmolgmolgmol

We can now substitute the values for 𝑀 and 𝑛 to calculate the mass of NaCl: 𝑛×𝑀=𝑚0.010×58.5/=𝑚0.627=𝑚.molesgmolg

Now that we know the mass of NaCl in the sample, we can calculate this as a percentage of the total mass: %=×100%%=0.6272.45×100%%=25.62%.bymassofNaClmassofNaClinsampletotalmassofsamplebymassofNaClggbymassofNaCl

Rounding to the nearest integer gives us a percentage by mass for NaCl of 26%.

Key Points

  • Precipitation gravimetry is an analytical technique that uses the formation and mass of a precipitate to determine the mass of an analyte.
  • The substance used to react and form the precipitate is known as the precipitating reagent.
  • The precipitate should be a pure substance with a definite and known composition.
  • Ashless filter paper is used experimentally to prevent the sample from being contaminated during the heating phase.
  • From the mass of the precipitate, the number of moles of the precipitate, the number of moles of the analyte, and finally the mass of the analyte can be determined.

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