Calculating Empirical and Molecular Formula Flashcards
(25 cards)
What is the empirical formula?
Gives you the smallest whole number ratio of atoms in a compound
What’s the step-by-step method for calculating an empirical formula?
- List all the elements
- Write down their experimental masses (or %).
- Divide each mass by its atomic mass (Ar) to get moles.
- Divide all mole values by the smallest one.
- Multiply to get whole numbers (if needed).
- Use the whole number ratio to write the formula.
How is empirical formula different to molecular formula?
- EF = the simplest formula that tells you the ratio of different elements in the compound
- MF = the actual number of atoms of each element in a single molecule
What are molecular formulaes?
The whole-number multiples of empirical formulae
When does combustion happen?
When a substance reacts with oxygen when its burned in air
What is the purpose of heating the crucible before use in a combustion experiment?
To clean it by removing any traces of oil or water.
How do you find the mass of magnesium used in the combustion experiment?
Subtract the mass of the crucible and lid from the mass of the crucible, lid, and magnesium.
Why is a lid used during combustion but not fully closed?
To prevent solids from escaping and to allow oxygen to enter.
What indicates the reaction has completed in the magnesium combustion?
All magnesium ribbon turns white (into magnesium oxide).
How is the mass of magnesium oxide found in this experiment?
Subtract the initial mass of crucible and lid from the final mass after reaction.
Method: Finding EF using combustion
- Heat a clean crucible until red hot, then let it cool.
- Weigh the crucible and lid.
- Add magnesium ribbon, then reweigh.
- Heat the crucible with the lid slightly open to allow oxygen in.
- Heat strongly for ~10 minutes until the magnesium turns white.
- Let it cool, then reweigh.
- Mass of magnesium oxide - initial reading for the mass of crucible and lid
Method: Finding EF using reduction
- Weigh a test tube with a rubber bung.
- Spread a small amount of copper(II) oxide in the test tube.
- Re-insert the bung and reweigh.
- Pass gas through the tube to displace air, then light it.
- Heat the tube strongly for ~10 minutes until copper turns brownish-pink.
- Let it cool, then turn off the gas.
- Reweigh the tube with the product.
- Use the mass difference to calculate the moles and empirical formula.
What is reduction?
The loss of oxygen from a substance
What happens to copper(II) oxide during heating with gas?
It changes from black to a brownish-pink color (copper metal).
Why is gas passed through the test tube before lighting?
To expel air and avoid an explosive mixture.
How is the empirical formula determined in this experiment?
By calculating the moles of oxygen removed and copper formed, then finding the simplest ratio.
What are solid salts made of?
-A lattice of positive and negative ions
-Some salts also include water molecules in the lattice
What is “water of crystallisation”?
-Water molecules trapped inside the salt’s crystal lattice
-These water molecules are chemically bound in a fixed ratio
What does “hydrated” mean in terms of salts?
-The salt contains water of crystallisation
-E.g. CuSO₄·5H₂O is hydrated copper sulfate
What does “anhydrous” mean?
-The salt has no water of crystallisation
-It has been heated or dried to remove the water
Why do water molecules stay in the salt lattice?
- Water is polar:
- Hydrogen atoms = slightly positive (δ⁺)
- Oxygen atoms = slightly negative (δ⁻)
- They are attracted to ions and held in place
What does the formula of a hydrated salt show?
- The number of moles of water per mole of salt
- E.g. CuSO₄·5H₂O → 5 moles of water per mole of salt
What happens when a hydrated salt is heated?
- It loses water of crystallisation
- Becomes anhydrous
What is the method to calculate the formula of a hydrated salt?
- Measure mass of hydrated salt
2, Measure mass after heating (anhydrous salt) - Find mass of water lost
- Calculate moles of water and anhydrous salt
- Find the ratio → gives X in formula: Salt·XH₂O
