Quantitative+Chemistry

Looks at the mole (not the underground furry kind)- what it is, how it is used and the relevance and importance of the numerical aspects of chemistry When you have been through this and understand the theory for AS 2.3 check out the Titration page for AS 2.2 - Acid Base titrations. Remember, these are 2 separate internal achievement standards. Any Q's see me.

__Introduction:__ A mole is a unit



This somewhat cheesy but amusing youtube clip gives us an idea of just how big a mole really is! media type="youtube" key="1R7NiIum2TI" height="385" width="480"

-Go to Demo mode if you don't yet have your log in, click on Stoichiometry in the left hand column. Under stoichiometry you will find a list of subtopics based on this unit. Start with reviewing balancing equations. Then move onto Mole Basics and so on :)
 * Also recall:** [|**Bestchoice**]

//Just how relevant really is Avogadro's number?// //How easy is it to work out the cost of one atom of aluminium in a roll of aluminium foil? Is this a valid assignment? Some students argued not - read the pdf attached. Then see if you can work out the cost of one atom of Al in the same roll of aluminium foil.//



__//Moles, moles and more moles...//__
//__S'more Fun!__//

//__An introduction to Empirical and Molecular Formulae, and Percentage Composition:__// Powerpoint covered in class.

__Example of an Empirical Formula Problem__ Find the empirical formula for a compound consisting of 63% Mn and 37% O Assuming 100 g of the compound, there would be 63 g Mn and 37 g O Look up the number of grams per mole for each element using the Periodic Table. There are 54.94 grams in each mole of manganese and 16.00 grams in a mole of oxygen. 63 g Mn × (1 mol Mn)/(54.94 g Mn) = 1.1 mol Mn 37 g O × (1 mol O)/(16.00 g O) = 2.3 mol O Find the smallest whole number ratio by dividing the number of moles of each element by the number of moles for the element present in the smallest molar amount. In this case there is less Mn than O, so divide by the number of moles of Mn: 1.1 mol Mn/1.1 = 1 mol Mn 2.3 mol O/1.1 = 2.1 mol O The best ratio is Mn:O of 1:2 and the formula is MnO2 The empirical formula is MnO2
 * Solution for Finding the Empirical Formula**

// __Home learning -20th July 2010.__// Empirical Formulae, Molecular Formulae and Percentage Composition.

Extra Practise on Empirical & Molecular Formula

Practical looked at in class, using nuts and bolts to give a visual and kinesthetic aid to learning empirical formulae. Have a go at the problems at the end of the practical if you didn't get chance in class, or use as a revision tool to refresh your memory if you've already had a go once.



From here on you want to check out the Titration page - this looks at standard solutions, how to set up a titration and volumetric analysis ready for your practical assessment in Term 3, Week 6.