This again uses the crucible/clay triangle/clamp/ring stand setup:
Here's one lab that you might find useful:
Now I'm a fan of the "less is more" approach, but it's difficult to find lab procedures that are brief for this experiment.
One concept that can be applied here is that of determining the empirical formula for the magnesium oxide formed. An empirical formula is the lowest whole number ratio of the elements in a compound. All ionic compounds (metal/nonmetal compounds) are empirical formulas (for example, CaCl2 ) but some molecular formulas can be reduced to an empirical formula -- e.g C2H4 can be reduced to CH2.
Mass data can be collected, converted to moles, the lowest ratio calculated, then reduce to a whole number ratio. Phew, that's a lot of steps! Let's break that down:
1) collect mass data -- 80 g C and 20 g G
2) convert to moles
80 g C x 1 mole C / 12 g C = 6.7 mol C
20 g H x 1 mole H / 1 g H = 20 mol H
3) calculate the lowest mole ratio-- by dividing both values by the smallest number
6.7 mol C / 6.7 = 1 mol C
20 mol H / 6.7 = 2.99 mol H
4) Get a whole number ratio by (a) rounding off the numbers or (b) multiplying the ratios by 2 or 3, then r.
Rounding off works well in this case:
1 mol C 2.99 mol H rounds to 3 mol H
So the empirical formula in this case is CH3 !
I hope this lab works out well for you. Please ask if you have any questions or ideas for other topics
Check out my lab book "Chemistry on a Budget" !