Lesson Explainer: Volatilization Gravimetry Chemistry

In this explainer, we will learn how to use volatilization gravimetry to calculate the quantity of analyte in a sample or determine the formula of a hydrated compound.

Volatilization gravimetry describes gravimetric analysis methods that use thermal or chemical energy to separate the substances of a mixture or chemical compound. The thermal or chemical energy is used to convert some solid reactant molecules into gaseous molecules. The gaseous molecules are usually liberated from the solid quite easily because they are volatile and can be evaporated relatively easily. Volatilization gravimetry differs from other gravimetric analysis methods that use precipitation or electricity to separate the components of a mixture or chemical substance.

Definition: Volatilization Gravimetry

Volatilization gravimetry is a mass analysis method that uses thermal or chemical energy to separate substances in order to measure their masses.

Example 1: Defining Volatilization Gravimetry

Which of the following statements best describes the process of volatilization gravimetry?

  1. The measurement of the change in concentration after neutralizing a sample
  2. The measurement of the change in volume after mixing two samples together
  3. The measurement of the change in mass after removing volatile compounds from a sample
  4. The measurement of the change in mass after reacting a sample to produce a precipitate

Answer

This question is asking us to define volatilization gravimetry. “Volatilization” means that we evaporate or otherwise remove gases with low boiling points. “Gravimetry” means we are measuring a mass. These definitions fit answer C.

The other answers are incorrect. Neutralization is not directly relevant to volatilization gravimetry, so answer A is incorrect. We are concerned with the change in mass, not the change in volume, so choice B is incorrect. Choice D describes another type of gravimetry, precipitation gravimetry. We can use these statements to determine that option C is the correct answer for this question.

Volatilization gravimetry can involve the separation of many different volatile molecules, such as carbon dioxide, chlorine gas, or nitrogen gas.

For example, chemical energy can be used to produce carbon dioxide gas from sodium hydrogen carbonate, and the carbon dioxide can then be converted into different chemical products. The products can be chemical compounds that are easy to weigh and measure. The following chemical equation shows how an aqueous solution of sulfuric acid can be used to separate carbon dioxide gas molecules from sodium bicarbonate. 2NaHCO()+HSO()2CO()+2HO()+NaSO()3242224aqaqglaq

Example 2: Identifying the Volatile Compound in a Reaction for Volatilization Gravimetry

A student wants to determine the amount of sodium bicarbonate in a sample using volatilization gravimetry. The student reacts the sample with dilute sulfuric acid, resulting in the reaction 2NaHCO()+HSO()2CO()+2HO()+NaSO()3242224aqaqglaq

What volatile molecule will the student measure?

  1. CO()2g
  2. NaSO()24aq
  3. HSO()24aq
  4. HO()2l
  5. NaHCO()3s

Answer

This question is asking us to identify the volatile compound that is separated during volatilization gravimetry. Volatilization gravimetry is a process that uses chemical or thermal energy to separate volatile compounds.

A volatile compound is a compound with a low boiling point.

During volatilization gravimetry, volatile compounds can easily separate as a gas. In this reaction, the gaseous product that emerges is carbon dioxide (CO2). We can use these statements to determine that option A is the correct answer for this question.

However, here we will focus on volatilization gravimetry involving the loss of water.

The crystalline form of some salts can incorporate water molecules into their framework. These water molecules are known as the water of crystallization or water of hydration.

Definition: Water of Crystallization

Water of crystallization is the presence of water molecules within the structure of a crystal.

When a salt contains molecules of water within its structure, it is known as a hydrated salt or complex. We can write the chemical formula for a hydrated salt by writing the number of water molecules associated with a single unit of the salt. A common hydrated salt is copper(II) sulfate, CuSO·5HO42. We can see from the chemical formula that each unit of CuSO4 has 5 water molecules associated with it.

Definition: Hydrated Salt

A hydrated salt is a substance that contains water of crystallization.

If we were to remove all the water molecules from a hydrated salt, we would form the anhydrous salt. When water is being removed during volatilization gravimetry, we are going from the hydrated salt to the anhydrous salt.

Definition: Anhydrous Salt

An anhydrous substance is a substance that does not contain water.

Blue crystals of hydrated copper(II) sulfate can be transformed into white crystals of anhydrous copper(II) sulfate if they are heated with a Bunsen burner flame. The anhydrous copper(II) sulfate crystals can be converted back into blue hydrated copper(II) sulfate if they are saturated with enough water molecules, as illustrated below.

We can use the change in mass to determine properties such as the number of water molecules per formula unit in a sample, the mass of water within the sample, or even the mass percentage of a hydrated compound in a mixture.

Volatilization gravimetry can be performed in the laboratory with relatively simple or incredibly sophisticated scientific equipment. Some of the simplest volatilization gravimetry experiments are conducted with little more than a clamp stand, a silicon or porcelain crucible, and a Bunsen burner.

How To: Determining the Amount of Water in a Substance by Volatilization Gravimetry

A simple and commonly used setup for performing volatilization gravimetry is shown below:

This relatively simple scientific setup is ideal for determining the amount of water that is liberated from a hydrated chemical complex such as hydrated salts. The steps to perform this analysis are as follows:

  1. Place a weighed sample of the substance into a crucible.
  2. Weigh the crucible. This gives the mass of the crucible plus the mass of the sample.
  3. Secure the crucible to a point above a Bunsen burner using a ring stand and a clay triangle.
  4. Turn on the Bunsen burner and heat the crucible.
  5. Measure the mass of the crucible, and keep heating until a constant mass is obtained.
  6. Measure the final mass of the crucible. This gives the mass of the crucible plus the mass of the sample with the volatile molecules removed.

During the heating process, the water molecules are liberated from the hydrated complex to form an anhydrous product. We can then use the masses of the sample and the crucible before and after heating to determine how much water has been lost.

It should be somewhat clear by now that hydrated chemical complexes can be weighed and heated to determine the relative abundance of volatile compounds such as water molecules. Hydrated chemical complexes can be heated with a Bunsen burner to determine the mass of water molecules that are liberated from hydrated chemical complexes, and this mass can then be used to determine the degree of hydration.

Consider a hydrated cobalt(II) sulfate complex that has the molecular formula CoSO·HO42𝑥, where 𝑥 is some unknown integer. The mass of a sample of the complex was recorded as 4.97 grams. After heating to an anhydrous state, a constant mass of 2.74 grams was obtained. The data is summarized in the following table.

Mass of Sample before HeatingMass of Sample after Heating
4.97 g2.74 g

The mass of the liberated water molecules is simply calculated as the difference between the mass of the sample before heating and the mass of the sample after heating. The mass of the water lost is therefore calculated according to the following equation.

Equation: Mass of Water Lost

The mass of the lost water can be calculated with the following equation: Massofsamplebeforeheatingmassofsampleafterheatingmassofwaterlostgggmassofwaterlost=4.972.74=2.23=.

We now have to determine how many moles of water molecules are contained in 2.23 grams of water. The number of moles of water can be calculated using the formula 𝑛=𝑚𝑀, where 𝑛 is the number of moles, 𝑚 is the mass in grams, and 𝑀 is the molar mass in grams per mole. The molar mass of water is 18 g/mol. 𝑛=2.2318/𝑛=0.124.ggmolmoles

Next, we can calculate the number of moles of cobalt(II) sulfate molecules in the 2.74 grams of the anhydrous cobalt(II) sulfate complex. The molar mass of cobalt(II) sulfate can be calculated by summing the average molar masses of the constituent atoms: 𝑀=𝑀+𝑀+4×𝑀𝑀=59/+32/+(4×16/)𝑀=155/.()()()()()()CoSOCoSOCoSOCoSO444gmolgmolgmolgmol

Using the formula, we get 𝑛=𝑚𝑀𝑛=2.74155/𝑛=0.0177.ggmolmolesofCoSO4

The calculations show that there were 0.124 moles of water molecules and 0.0177 moles of cobalt(II) sulfate contained within the hydrated cobalt(II) sulfate complex.

The molar values can be divided to determine the ratio between the number of water molecules and cobalt(II) sulfate molecules in the original hydrated cobalt(II) sulfate complex. 0.1240.0177=7.0056,molHOmolCoSO24 which, rounded to the nearest integer, equals 7molHOmolCoSO24.

The results from the volatilization gravimetry show that there are seven molecules of water per unit of cobalt(II) sulfate. Therefore, the molecule formula is CoSO·7HO42.

Whenever we use volatilization gravimetry for hydrated substances, certain assumptions are made. One of these is that water is the only molecule lost. However, the loss of any other volatile compound will also cause the mass to decrease.

Example 3: Finding the Number of Molecules in a Hydrate from the Mass of the Hydrate and Water

A sample of cobalt(II) chloride hydrate (CoCl·HO)22𝑥 is heated until its mass remains constant. For every 1.00 g of cobalt(II) chloride produced, 0.831 g of water is liberated. What is the value of 𝑥, where 𝑥 is an integer? [Co=59g/mol, Cl=35.5g/mol, H=1g/mol, O=16g/mol]

Answer

This question is asking us to find the water of crystallization per unit for a cobalt(II) chloride hydrate. We are told that for every 1.00 g of anhydrous cobalt(II) chloride produced, 0.831 g of water is liberated.

From the masses given, we can calculate the number of moles of water that is liberated and the number of moles of anhydrous cobalt(II) chloride. We can calculate the number of moles using the formula 𝑛=𝑚𝑀, where 𝑛 is the number of moles, 𝑚 is the mass in grams, and 𝑀 is the molar mass in grams per mole.

The molar mass of CoCl2 can be calculated as follows: 𝑀=𝑀+2×𝑀𝑀=59/+(2×35.5/)𝑀=130/.()()()()()CoClCoClCoClCoCl222gmolgmolgmol

The molar mass of HO2 can be calculated as follows: 𝑀=𝑀+2×𝑀𝑀=16/+(2×1/)𝑀=18/.()()()()()HOOHHOHO222gmolgmolgmol

We can now calculate the number of moles of CoCl2 and HO2: 𝑛=𝑚𝑀𝑛=𝑚𝑀𝑛=1.00130/𝑛=0.83118/𝑛=0.00769.𝑛=0.04617.ggmolggmolmolesofCoClmolesofHO22

Finally, we can take the ratio of these two amounts in moles, making sure we divide the more abundant amount of water by the less abundant amount of CoCl2 in order to end up with an integer value. 0.046170.00769=6.00166.molHOmolCoClmolHOmolCoCl2222

Rounding to the nearest integer, the value equals 6molHOmolCoCl22.

The chemical formula of hydrated cobalt(II) chloride is (CoCl·6HO)22. The value of 𝑥 is 6.

We can also use the results from volatilization gravimetry analysis to determine the percentage of water of crystallization. The percentage of water of crystallization tells us the amount of water, by mass, that is present in a hydrated crystal.

Equation: Percentage of Water of Crystallization

%=×100%waterofcrystallizationmassofthewaterofcrystallizationmassofhydratedcompound

We can calculate the percentage of water of crystallization for our hydrated cobalt(II) sulfate salt. We determined the chemical formula of this salt to be CoSO·7HO42.

Taking the molar mass of water as 18 g/mol and that of CoSO4 as 155 g/mol, we can calculate the molar mass of the hydrated complex CoSO·7HO42. 𝑀=𝑀+7×𝑀𝑀=155+(7×18)/𝑀=281/.()()()()()CoSO·7HOCoSOHOCoSO·7HOCoSO·7HO42424242gmolgmol

The formula for calculating the percentage of water of crystallization is: %=(×𝑀)()×100%%=(7×18)/281/×100%%=44.84%.waterofcrystallizationno.ofmolesofwatermoleculesofcrystallizationofwatermassofthehydratedcompoundwaterofcrystallizationgmolgmolwaterofcrystallization

We have determined that the percentage of water of crystallization is 44.84% for the heptahydrate cobalt(II) sulfate complex. This number implies that the hydrated cobalt(II) sulfate salt complex contains 45% water by mass.

Example 4: Calculating the Percentage of Water of Crystallization Given the Mass of the Hydrated and Dehydrated Salt

A 0.3548 g sample of a hydrated salt was strongly heated until a constant mass of 0.3015 g was obtained. What is the percentage of water of crystallization in this hydrated salt? Give your answer to 2 decimal places.

Answer

This question is asking us to find the percentage of water of crystallization, or the percent by mass of water in the hydrated salt. To find this number, we will need to compare the mass of the evaporated water to the overall mass of the hydrated salt.

To find the mass of evaporated water, we can subtract the starting mass from the end mass, assuming that all mass lost is due to water evaporation. Massofhydratedsaltmassofanhydroussaltmassofwaterlostggg=0.35480.3015=0.0533.

To find the percentage of water of crystallization, we divide this number by the mass of the hydrated salt to get a percentage. 0.05330.3548×100%=15.0225%.gg

Rounding to two decimal places gives a final value of 15.02%.

This number means that 15.02% of the mass of the hydrated salt is the mass of water of crystallization.

Key Points

  • Volatilization gravimetry involves separating out volatile compounds and measuring the change in mass.
  • Using volatilization gravimetry, we can determine the mass of the volatile compound and the mass of the remainder and use these values to calculate the number of moles of each.
  • Water of crystallization are water molecules that are incorporated into the crystal structure of a substance.
  • Salts that contain water are called hydrated, while salts without water are called anhydrous.
  • The degree of hydration of a hydrated salt is how many water molecules are incorporated per formula unit of salt.
  • Volatilization gravimetry lets us determine the degree of hydration of a salt.
  • Volatilization gravimetry also lets us determine the percentage of water of crystallization, by mass, in a salt.

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