Determination Of Water In A Hydrate Lab Report
Percent Water in a Hydrate
PURPOSE
To determine the percent water in a hydrate sample.
INTRODUCTION
Many substances contain water molecules as a part of their crystal structure. We call such solids hydrates, and we call the bound water the water of hydration.
A hydrate has a definite number of water molecules bound to each anhydrous salt unit. The formula of the hydrate copper(II) sulfate pentahydrate is
CuSO4 · 5 H2O
The dot indicates that the molecules of water are attached to the ions in CuSO4 by weak bonds.
We can drive off the water of hydration by heating the hydrate. If blue CuSO4 · 5 H2O is heated, the water of hydration is released as water vapor, and solid white anhydrous CuSO4 remains. …show more content…
The reverse reaction of Equation 1 may also occur. Anhydrous CuSO4 is white, but upon exposure to air, the anhydrous salt absorbs water. This reaction produces blue CuSO4·5 H2O.
In some cases, compounds can actually dissolve in their water of hydration. Salts such as calcium chloride (CaCl2), calcium sulfate (CaSO4), and magnesium sulfate (MgSO4), can absorb so much water that they form solutions. These salts are said to be deliquescent. We often use anhydrous salts of this type as drying agents.
Some hydrated salts such as Na2SO4·10 H2O tend to lose water even without heating, just from exposure to dry air under normal conditions of temperature and pressure. Such salts are said to be efflorescent.
In this experiment, you will determine the mass of the salt before and after heating. The difference in mass between the hydrated and the dehydrated (anhydrous) salt is equal to the mass of water in the salt. From these data, you will calculate the formula of the hydrate and the percent by mass of water in a hydrated …show more content…
Caution: Wear your safety goggles.
1. Support a clean crucible on a clay triangle and heat with an intense flame for 5 minutes. Handle the crucible with the crucible tongs for the rest of the experiment; do not use your fingers. Determine the mass of the cooled crucible.
2. Add at most 3.00 g of an unknown hydrated salt to the crucible and determine the mass of the salt and crucible.
3. Return the crucible with sample to the clay triangle.
4. At first, heat the sample slowly and then gradually intensify the heat. Do not allow the crucible to become red-hot. This can cause the anhydrous salt to decompose. Heat the sample for 15 minutes. Determine the mass of the anhydrous salt and crucible.
5. Reheat the sample for 5 more minutes. Again measure the mass. Continue this step until a constant mass is achieved.
6. Discard the anhydrous salt in the trash.
Data has been provided within the following data table for completion of this
PURPOSE
To determine the percent water in a hydrate sample.
INTRODUCTION
Many substances contain water molecules as a part of their crystal structure. We call such solids hydrates, and we call the bound water the water of hydration.
A hydrate has a definite number of water molecules bound to each anhydrous salt unit. The formula of the hydrate copper(II) sulfate pentahydrate is
CuSO4 · 5 H2O
The dot indicates that the molecules of water are attached to the ions in CuSO4 by weak bonds.
We can drive off the water of hydration by heating the hydrate. If blue CuSO4 · 5 H2O is heated, the water of hydration is released as water vapor, and solid white anhydrous CuSO4 remains. …show more content…
The reverse reaction of Equation 1 may also occur. Anhydrous CuSO4 is white, but upon exposure to air, the anhydrous salt absorbs water. This reaction produces blue CuSO4·5 H2O.
In some cases, compounds can actually dissolve in their water of hydration. Salts such as calcium chloride (CaCl2), calcium sulfate (CaSO4), and magnesium sulfate (MgSO4), can absorb so much water that they form solutions. These salts are said to be deliquescent. We often use anhydrous salts of this type as drying agents.
Some hydrated salts such as Na2SO4·10 H2O tend to lose water even without heating, just from exposure to dry air under normal conditions of temperature and pressure. Such salts are said to be efflorescent.
In this experiment, you will determine the mass of the salt before and after heating. The difference in mass between the hydrated and the dehydrated (anhydrous) salt is equal to the mass of water in the salt. From these data, you will calculate the formula of the hydrate and the percent by mass of water in a hydrated …show more content…
Caution: Wear your safety goggles.
1. Support a clean crucible on a clay triangle and heat with an intense flame for 5 minutes. Handle the crucible with the crucible tongs for the rest of the experiment; do not use your fingers. Determine the mass of the cooled crucible.
2. Add at most 3.00 g of an unknown hydrated salt to the crucible and determine the mass of the salt and crucible.
3. Return the crucible with sample to the clay triangle.
4. At first, heat the sample slowly and then gradually intensify the heat. Do not allow the crucible to become red-hot. This can cause the anhydrous salt to decompose. Heat the sample for 15 minutes. Determine the mass of the anhydrous salt and crucible.
5. Reheat the sample for 5 more minutes. Again measure the mass. Continue this step until a constant mass is achieved.
6. Discard the anhydrous salt in the trash.
Data has been provided within the following data table for completion of this