Welcome to Organic Chemistry for Veterinary Science

CHEM1405

 Room 224, School of Chemistry Email: george@chem.usyd.edu.au

Lecture Topics

The following lecture notes may be downloaded in Adobe Acrobat format (PDF files). If you do not have Acrobat Reader click here.

Week 1 - Atomic Structure and the Periodic Table (these slides cover all three lectures)

If you want to brush up on your stoichiometry and mole calculations, have a look at the Bridging Course notes, particularly 'Handbook' topics 6, 7, 8 & 9. There are explanations of terms, examples, questions and full worked answers there. The ChemCAL on-line module on stoichiometry is also useful (username: 1405 password: helium).

Another useful resource of stoichiometry questions and explanations is the workshop W1. Here are the full worked answers for the workshop and the post-work for the workshop.

Periodic table with links to atomic emission spectra is here.

Week 2 homework problem sheet: a number of people were asking me about the complex equation to balance in Q9. This is how I did it:

What we are told is:
Na3(PO4)  +  CaCl2    →  Ca5(PO4)3(OH)

There is something missing that will generate the OH, present as the anion OH.  We are precipitating the hydroxyapatite from water solution and so water seems the most likely source of the hydroxide, a H+ being left over.

So adding in water we get:
Na3(PO4)  +  CaCl2  + H2O   →  Ca5(PO4)3(OH)  +  H+

Now if we balance for Ca and add the Cl ions to the product side we get:
Na3(PO4)  +  5CaCl2  + H2O   →  Ca5(PO4)3(OH)  +  H+    +  10Cl

Now if we balance for phosphate and add the Na+ ions to the product side we get:
3Na3(PO4)  +  5CaCl2  + H2O   →  Ca5(PO4)3(OH)  +  H+    +  10Cl    +  9Na+

We have the correct number of each type of atom and could do a bit of tidying up by writing:
3Na3(PO4)  +  5CaCl2  + H2O   →  Ca5(PO4)3(OH)  +  HCl  +  9NaCl

My preference would be to leave this as a ‘formula equation’ with no state symbols as I believe it is incorrect to write NaCl(aq) as you see in many books.

A true ionic equation would show just the ions involved in the reaction, ie:
5Ca2+ +   3PO43– + H2O  →  Ca5(PO4)3(OH)  +  H+

And here we could put in the state symbols:
5Ca2+(aq) +   3PO43–(aq)   + H2O(l)   →  Ca5(PO4)3(OH) (s)   +  H+(aq)

For part b) take the following steps:

• work out the number of moles of each reagent present
• work out which reagent is 'limiting'
• calculate moles of product based on limiting reagent

For part c) take the following steps:

• note chloride ions are not used in the reaction so the initial amount you start with is the amount in the final solution
• determine the number of moles of chloride ions present from initial moles of calcium chloride noting that 1 mole of CaCl2 give 2 moles of Cl- ions
• work out the final total volume
• work out the concentration in moles / L

Topic 2 - Stoichiometry and bonding

Topic 3 - Molecular geometry, intermolecular forces and colloids

Note the colloids section will not be examined.

Topic 4 - Thermodynamics and equilibrium

Topic 5 - Acids & bases and solution chemistry

Topic 6 - Electrochemistry and kinetics