How to Verify the Law of Constant Composition
In your chemistry classes, you will work with the law of constant composition, which states that the proportion of elements within a compound is always equal. Verifying this law through hands-on experimentation is a useful way of connecting the laws printed in your textbook to real-life phenomena. When carrying out this laboratory experiment, always take precautions by wearing gloves, a lab coat, and goggles to protect yourself.
Other People Are Reading
Things You'll Need
- Goggles
- Lab coat
- Gloves
- Crucible
- Magnesium turnings
- Spatula
- Heat source
Instructions
-
-
1
Weigh a dry crucible, along with its cover, on a mass balance.
-
2
Weigh out 0.7 grams of magnesium turnings on the mass balance, then transfer them to your crucible with your spatula. Place your crucible on a heat source approved by your instructor, and switch on the heat. Heat your magnesium, covered, for eight minutes.
-
-
3
Turn off the heat source. Using heat-resistant gloves, remove your crucible from the source. Allow the contents of your crucible to cool for several minutes.
-
4
Add 20 milliliters of distilled water to your crucible, using a pipette. Place your crucible back on the heat source, cover it, switch the heat on, and allow all of the water to evaporate. Then, continue to heat the crucible for the next six minutes.
-
5
Remove the crucible from the source of heat and allow it to cool. Use your mass balance to find the total mass of the crucible, crucible cover and magnesium oxide.
-
6
Subtract the mass that you found in Step 1 from the mass that you found in Step 5 to find the mass of the magnesium oxide.
-
7
Write the equation for magnesium oxide formation: "2Mg+ O2=2MgO." Calculate the initial molar amount of magnesium. Since you started with 0.7 grams of Mg and the molar mass of Mg is 24.305 grams, you have 0.7/24.305 moles of Mg, or 0.029 moles of magnesium, to start with. Since the ratio of Mg to MgO is 1:1 (as the equation states, there are two molecules of Mg for every two molecules of MgO), the equation creates 0.029 moles of MgO.
-
8
Calculate the amount of MgO that the reaction should have produced. Since there should be 0.029 moles of MgO, which is equal, according to the periodic table, to 0.029*(24.3+16 grams), or 1.168 grams of MgO.
-
9
Make sure that the number that you calculated in Step 8 is roughly equal to the number that you calculated in Step 6. This verifies the law of constant composition.
-
1
