I have always tried to impress upon students that reactions are between molecules (particles) an when 6.02 x 10^23 particles are put together we have a mole. Reactions are between moles and not grams. Grams, of course, get into the equations because moles = g/molar mass but the reaction is between moles and not grams. We don't have balances that weigh in mols; our balances weigh in grams so ultimately we express input and output in grams but the stoichiometry is by mols. That's why we convert from grams to moles before we start the stoichiometric calculations. We CAN express the equations in grams, and by a little slight of hand, make it appear that grams are doing all the work but that is trickery. For example,
2KClO3 ==> 2KCl + 3O2
Suppose we have 12.25 grams KClO3. How much O2 will we collect? 12.25/122.5 = 0.1 mole. 0.1 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.1 x 3/2 = 0.15 moles oxygen an that is 0.15 mol x (32 g/mol) = 4.8 g oxygen. But watch what we can do.
2KClO3 ==> 2KCl + 3O2
molar mass KClO3 = 122.5 so
2 x 122.5 grams KClO3 will produce 3 x 32 grams oxygen OR 245 g KClO3 will produce 96 g oxygen. So how much oxygen will we get? We will obtain 96 grams x (12.25 g/245) = 4.8 grams. In fact, that is the way I learned to do it when I was a freshman in college MANY years ago and I used that method throughout graduate school. At first glance it appears we never make the conversion from moles to grams and that grams are going to grams. BUT, that first step of 2 x 122.5 = 245 and 3 x 32 = 96 put the mole ratios into gram ratios. From there it was just ratio and proportion. This may be more than you ever wanted to know about how molecule react but it's free information and that's the best kind.
Why do we need to use amount in moles to solve stoichiometry problems? Why can't I just convert from mass to mass?
I'm not sure how to answer. I think I know the answer and just can't formulate it into words.
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The element nickel has ccp packing with a face-centered cubic unit cell. The volume of the unit cell is 4.38 x 10-23 cm3. Calculate the density (kg/m3) of the element.
1 answer