This lab is a classic and hopefully a roll of magnesium ribbon is available in your department. I've worked in departments where the grams per centimeter had been measured, so a small length of magnesium ribbon could be used. However, if the ribbon sample must be massed, a much longer sample of magnesium ribbon may need to be used (depending on the balance available) -- and you may have crumple it up to fit into the crucible.
This again uses the crucible/clay triangle/clamp/ring stand setup:
This again uses the crucible/clay triangle/clamp/ring stand setup:
Safety: Laboratory goggles and an apron are to be worn by students. Notice that the crucible is glowing red from the Bunsen burner flame. This is a very hot setup, and it takes time for the crucible to cool down. I've told this tale here before, but I've had students touch this set-up while it was still hot, flinch and knock it over. Bring the back of your hand close to the setup to see if it's still warm.
Here's one lab that you might find useful:
http://stahlchem.wikispaces.com/file/view/MgO+aca+09.PDF
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" !
http://www.amazon.com/Chemistry-Budget-Marjorie-R-Heesemann/dp/0578129159/ref=sr_1_1?s=books&ie=UTF8&qid=1389410170&sr=1-1&keywords=chemistry+on+a+budget
Here's one lab that you might find useful:
http://stahlchem.wikispaces.com/file/view/MgO+aca+09.PDF
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" !
http://www.amazon.com/Chemistry-Budget-Marjorie-R-Heesemann/dp/0578129159/ref=sr_1_1?s=books&ie=UTF8&qid=1389410170&sr=1-1&keywords=chemistry+on+a+budget
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