An empirical formula experiment
The empirical formulaThe simplest whole number ratio of the atoms of each element in a compound. of a simple compoundA substance formed by the chemical union of two or more elements. can be found using experiments. This page outlines one common experiment. Eye protection should be worn when carrying out experiments like this.
Aim
To determine the empirical formula of magnesium oxide.
Method
- Measure and record the massThe amount of matter an object contains. Mass is measured in kilograms (kg) or grams (g). of an empty crucibleA small porcelain dish that can be heated very strongly over a Bunsen burner when it is supported in a pipeclay triangle. with its lid.
- Put a length of magnesium ribbon into the crucible.
- Measure and record the total mass of the crucible, its lid and contents.
- Place the crucible on a tripod with a pipe clay triangle. Strongly heat the crucible for several minutes using a Bunsen burner.
- When the magnesium has stopped glowing, turn off the Bunsen burner and allow the crucible to cool down.
- Repeat step 3.
Results
The table shows some example results.
Mass (g) | |
Mass of crucible and lid (step 1) | 50.00 |
Mass of crucible, lid and magnesium (step 3) | 50.24 |
Mass of crucible, lid and magnesium oxide (step 7) | 50.40 |
Mass of crucible and lid (step 1) | |
---|---|
Mass (g) | 50.00 |
Mass of crucible, lid and magnesium (step 3) | |
---|---|
Mass (g) | 50.24 |
Mass of crucible, lid and magnesium oxide (step 7) | |
---|---|
Mass (g) | 50.40 |
Analysis
- calculate the mass of magnesium used.
- mass = 50.24 g - 50.00 = 0.24 g
- calculate the mass of oxygen gained during heating
- mass = 50.40 g - 50.24 = 0.16 g
- calculate the empirical formula of magnesium oxide (Ar of Mg = 24 and Ar of O = 16)
Step | Action | Result | Result |
1 | Write the element symbols | Mg | O |
2 | Write the masses | 0.24 g | 0.16 g |
3 | Write the Ar values | 24 | 16 |
4 | Divide masses by Ar | 0.24 梅 24 = 0.01 | 0.16 梅 16 = 0.01 |
5 | Divide by the smallest number | 0.01 梅 0.01 = 1 | 0.01 梅 0.01 = 1 |
6 | Write the formula | MgO | MgO |
Step | 1 |
---|---|
Action | Write the element symbols |
Result | Mg |
Result | O |
Step | 2 |
---|---|
Action | Write the masses |
Result | 0.24 g |
Result | 0.16 g |
Step | 3 |
---|---|
Action | Write the Ar values |
Result | 24 |
Result | 16 |
Step | 4 |
---|---|
Action | Divide masses by Ar |
Result | 0.24 梅 24 = 0.01 |
Result | 0.16 梅 16 = 0.01 |
Step | 5 |
---|---|
Action | Divide by the smallest number |
Result | 0.01 梅 0.01 = 1 |
Result | 0.01 梅 0.01 = 1 |
Step | 6 |
---|---|
Action | Write the formula |
Result | MgO |
Result | MgO |
Empirical formula is MgO.
Question
In an experiment, 1.27 g of hot copper reacts with iodine vapour to form 3.81 g of copper iodide. Calculate the empirical formula of copper iodide. (Ar of Cu = 63.5, Ar of I = 127)
Step | Action | Result | Result |
1 | Write the element symbols | Cu | I |
2 | Write the masses | 1.27 g | 3.81 g - 1.27 g = 2.54 g |
3 | Write the Ar values | 63.5 | 127 |
4 | Divide masses by Ar | 1.27 梅 63.5 = 0.02 | 2.54 梅 127 = 0.02 |
5 | Divide by the smallest number | 0.02 梅 0.02 = 1 | 0.02 梅 0.02 = 1 |
6 | Multiply (if needed) to get whole numbers | (not needed: already have whole numbers) | (not needed: already have whole numbers) |
7 | Write the formula | Cul | Cul |
Step | 1 |
---|---|
Action | Write the element symbols |
Result | Cu |
Result | I |
Step | 2 |
---|---|
Action | Write the masses |
Result | 1.27 g |
Result | 3.81 g - 1.27 g = 2.54 g |
Step | 3 |
---|---|
Action | Write the Ar values |
Result | 63.5 |
Result | 127 |
Step | 4 |
---|---|
Action | Divide masses by Ar |
Result | 1.27 梅 63.5 = 0.02 |
Result | 2.54 梅 127 = 0.02 |
Step | 5 |
---|---|
Action | Divide by the smallest number |
Result | 0.02 梅 0.02 = 1 |
Result | 0.02 梅 0.02 = 1 |
Step | 6 |
---|---|
Action | Multiply (if needed) to get whole numbers |
Result | (not needed: already have whole numbers) |
Result | (not needed: already have whole numbers) |
Step | 7 |
---|---|
Action | Write the formula |
Result | Cul |
Result | Cul |
More guides on this topic
- Equations and formulae - Edexcel
- Hazards and risks - Edexcel
- Atomic structure - Edexcel
- The periodic table - Edexcel
- Ionic compounds - Edexcel
- Simple molecular substances - Edexcel
- Giant covalent substances - Edexcel
- Metals and non-metals - Edexcel
- Mole calculations (higher) - Edexcel
- Sample exam questions - key concepts in chemistry - Edexcel