Objectives:

To calculate the percent water by mass in several hydrated compounds and to dehydrate an

unknown solid sample and identify it by comparing its percent water with known hydrated

compounds.

Material: Hydrated salts (BaCl2∙2H2O, CaSO4∙2H2O, CuSO4∙5H2O, NiCl2∙6H2O, MgSO4∙7H2O).

Note: the magnesium sulfate crystals should be clear and transparent; the presence of white

crystals indicates a mixture of 6-hydrate and lower.

Equipment: A smart device and internet connection.

Safety: Handle the crucible, lid, wire triangle, and iron ring carefully—objects remain hot long

after heating is discontinued. Solid samples may tend to spatter during heating. Wear safety

goggles at all times in the laboratory.

Waste disposal: All solid waste must be disposed of in inorganic waste.

Virtual Lab Instructions

:

1. Read the introduction.

2. Watch the lecture and experiment’s videos on Labflow.

3. Answer prelab questions. (20 points)

4. Open report and answer YES to “Are you completing this experiment online?”

5. Request “Provisional Data”

and confirm.

6. Use the provided data to complete the report and answer post lab questions. (70 points)

7. Answer the quiz questions. (10 points)

Lab 5: Water of Hydration

Chemistry 111 Lab 5 Water of Hydration

Many solids, especially inorganic salts, occur naturally as hydrates. This means that water molecules are

incorporated into the crystal lattice of the compound in such a way that the water is chemically bound.

Some of these materials may spontaneously lose water molecules when placed in contact with dry air, a

process known as efflorescence. Other hydrates are hygroscopic, meaning that they absorb water from

humid air. These materials are often used as desiccants, or drying agents. For example, many electronic

products are shipped with small packets of silica, which act as a desiccant to protect the electronics from

the effects of humidity. Some solids absorb so much humidity from the air that they then start to dissolve

in the large quantity of absorbed water, a process known as deliquescence.

Some hydrates exist in equilibrium with moisture in the air, absorbing or releasing water molecules

depending on the relative humidity. One example is cobalt (II) chloride, which absorbs or releases water in

a reversible process:

CoCl2∙6H2O ⇆ CoCl2 + 6H2O
(pink) (blue)

Cobalt chloride is used in many novelty items, such as in weather strips to predict the weather. When the

chance of rain increases, the relative humidity is high, and the cobalt chloride absorbs water to form the

hydrated salt and the weather strip turns pink. When the relative humidity drops during fair weather, the

compound releases water and reverts to the anhydrous form, and the weather strip turns blue.

Hydrated salts are not simply solids that are wet or that have moisture absorbed on their surface. They are

compounds in which water molecules are incorporated into the crystal structure of the salt, chemically

combined with the cations and anions. This is reflected in the chemical formula of the salt, which includes

the specific number of water molecules in the unit cell (or moles of water per mole of salt). Thus, the

hydrated cobalt chloride (CoCl2∙6H2O) includes 6 water molecules in its unit cell, or 6 moles of water per

mole of salt. The common notation for hydrated salts is to use a raised dot (∙) between the formula for the

salt and the number of water molecules in the compound. Alternatively, the formula for hydrated salts can

be written with the water molecules in parentheses, with the number of water molecules indicated using a

subscript. In other words,

CoCl2∙6H2O = CoCl2(H2O)6

Because water is part of the chemical structure of a hydrated salt, it is important to specify the amount of

water when naming hydrates. The rule for naming hydrated salts is to start with the name of the salt and

add the term “hydrate” with a Greek numerical prefix to indicate the molar ratio of water in the compound.

These prefixes are listed in Table 5.1 below. Thus,

CuSO4∙5H2O would be called “copper (II) sulfate pentahydrate,” and

BaCl2∙2H2O would be called “barium chloride dihydrate.”

In this exercise you will identify an unknown hydrated salt from a list of potential unknowns. Water of

hydration is easily removed from the compound by heating to above 100 oC (the boiling point of water).

By measuring the mass of water lost upon heating from a given amount of hydrated salt, the % water by

mass can be calculated and compared to the theoretical values based on the formulas of known compounds.

INTRODUCTION

Chemistry 111 Lab 5 Water of Hydration

Table 5.1. Greek Numerical Prefixes

Number of Water

Molecules

Greek

Prefix

1 mono

2 di

3 tri

4 tetra

5 penta

6 hexa

7 hepta

:
:
:
:
:
:

Chemistry 111 Lab 5 Water of Hydration

Pre-Lab Questions

Define the following terms:

– hydrate:

– desiccant:

– anhydrous:

– efflorescence:

– deliquescence:

2. Name these compounds:

(a) Ca(NO3)2∙4H2O:

(b) FeF3∙3H2O:

3. A sample of a hydrated salt weighing 1.167 g is heated until no more water is given off. The mass of

the anhydrous salt was 0.629 g. What is the mass of water lost by the sample? Calculate the mass %

of water in the hydrated salt.

4. Calculate the mass % water in Na2S2O3∙5H2O. Report your answer to four significant figures.

5. Two of the possible unknowns in this lab are also commonly known as “gypsum” and “Epsom salts”,

respectively. Do a little research (library, web) to identify these compounds.

Chemistry 111 Lab 5 Water of Hydration

Watch the experiment’s video which demonstrates the procedure for Water of Hydration lab. Open

“Report” and answer “Yes” to “Are you Completing this experiment online?”. Request provisional data

and confirm.

Use the provided data to calculate % water in the hydrated sample. Compare %water in the sample with

the %water values in Data Sheet, Part 1 to determine the identity of the hydrated sample.

A summary of the procedure is provided below.

1. Obtain a clean crucible and lid. Heat gently with a Bunsen burner for one minute, and then heat strongly

to redness for about three minutes. Allow to cool completely.

2. Using crucible tongs transfer the crucible and lid to

the balance and weigh. Never touch the crucible or

cover with your bare hands as oils and moisture

from your skin will add mass to the crucible. The

proper technique for using tongs to carry the

crucible is illustrated in Figure 5.1.

Figure 5.1. Proper technique for using

crucible tongs.

3. Transfer about 1 gram of hydrated salt to the crucible and reweigh. The mass of the sample can

be determined by difference.

4. Support the crucible and cover on the clay triangle as

shown in Figure 5.2. The lid should be slightly ajar to

allow steam to escape during heating. Heat gently for

about 5 minutes—vigorous heating of the sample may

cause it to splatter or to undergo decomposition. After

several minutes, lift the lid to observe any changes in the

material. When heating is finished, replace the lid to

completely cover the sample to prevent re-adsorption of

water, and allow the crucible to cool.

5. When the crucible and lid have cooled completely, weigh

the crucible and anhydrous salt sample. Record the mass

to the nearest 0.001 g on your data sheet.

Figure 5.2. Using a ring stand.

VIRTUAL PROCEDURE

Chemistry 111 Lab 5 Water of Hydration

6. Return the crucible and lid to the clay triangle and heat gently for another 5 minutes. Cool completely

and weigh, recording the mass on your data sheet. If the weight after the second heating agrees with

the mass after the first heating (within 0.005 g) then no further heating is necessary. If the second

heating appeared to drive off additional water, then heat a third time, cool, and re-weigh. Continue until

you obtain agreement between successive weighings.

7. Scrape out the anhydrous salt into an evaporating dish or small beaker. Record any observations

regarding the appearance of the anhydrous salt.

8. Repeat steps 2–6 using a second sample of the same unknown. Calculate the % water by mass in each

sample. Average your values for the two trials.

9. Add the anhydrous salt from the second trial to the evaporating dish or beaker, and add a few drops of

water. Record any changes in the appearance of the salt on your data sheet. Dissolve any remaining

solid and rinse down the sink with plenty of water. Clean the crucible and cover and return to the lab

cart or lab drawer as instructed.

Chemistry 111 Lab 5 Water of Hydration

Part 1

1. Calculate the theoretical % water by mass for each of the possible unknowns listed on your data
sheet. The simplest way to calculate the mass % of water is to begin by assuming one mole of the

compound. The % water by mass can then be determined as the mass of water divided by the mass

of compound, times 100%. For example, the % water by mass in FeCl3∙6H2O would be calculated

as follows:

% water =
mass of 6 moles H2O

mass of one mole FeCl3∙6H2O
× 100%

=
(6 moles)(18.02 g/mole)

(55.85 g Fe)+(3 x 35.45 g Cl)+( 6 x 18.02 g H2O)
× 100%

=
108.12 g H2O

270.32 g FeCl3∙6H2O
× 100% = 40.00%

Part 2

2. Calculate the mass of each hydrated salt sample by difference.

3. Calculate the mass of water lost as the difference between the mass of the hydrated salt and the
mass of the anhydrous salt.

4. Calculate % water by mass in each sample as:

% water =
weight of water lost

????ℎ? ?? ℎ??????? ????
× 100%

Record the % water by mass for each trial on your data sheet. Calculate the average % water for your

trials

5. Identify your unknown hydrated salt by comparing the average % water (by mass) for your unknown

with the calculated theoretical % water for the salts listed on Part 1 of your data sheet.

CALCULATIONS

Chemistry 111 Lab 5 Water of Hydration

Data Sheet

Part 1. Calculate mass % H2O in the hydrated salts below. You will use these values to identify the

unknow salt in the next question. (Show calculaciones)

BaCl2∙2H2O:

CaSO4∙2H2O:

CuSO4∙5H2O:

NiCl2∙6H2O:

MgSO4∙7H2O:

Chemistry 111 Lab 5 Water of Hydration

Data Sheet

Part 2. Two sets of data are shown below for an unknown sample. Use the provided data to calculate the

average %water and to identify the Unknown.

HINT: Use the %water calculated in Part 1 to identify the sample.

Unknown Sample: ___________

Trial 1 Trial 2

Mass of crucible + lid (g) ____________ ___________

Mass of crucible, lid + hydrated salt (g) ____________ ___________

Mass of crucible, lid + anhydrous salt (g) ____________ ___________

(weight 1)

Mass of crucible, lid+ anhydrous salt (g) ____________ ___________

(weight 2)

Mass of crucible, lid + anhydrous salt (g) ____________ ___________

(weight 3)

Calculations

Mass of hydrated salt (g) __________ __________

Mass of anhydrous salt (g) _________ _________

Mass of H2O lost (g) _________ _________

% water in Unknown _________ _________

AVERAGE % WATER : ____________

IDENTITY of SAMPLE = ______________

Chemistry 111 Lab 5 Water of Hydration

Post Lab Questions

1. You have identified the unknown in Part 2 of datasheet. Compare %water in the sample with the actual
%water for the salt in Part 1 and calculate the % error and indicate if it is a positive or negative error.

Discuss possible sources of error, paying attention to the direction of the error in each case.

2. Write the formula for your sample and give its correct chemical name.

3. Did the appearance of your unknown change when it was dehydrated and re-hydrated? Why do you

think this is so?

4. Heating some hydrates too strongly may cause them to decompose. For example, some sulfates can

emit SO2 or SO3 gas upon strong heating. If this were to occur in this experiment would the

calculated mass % water be too low or too high? Explain.

5. A student takes 1.857 g of hydrated iron (III) nitrate and heats it in a crucible. The anhydrous material

weighed 1.112 g. What is the formula for this hydrated salt? Show your work.