It ain’t that bad. Just the darn numbers. If you read the last part of my previous post:
1 mole = 6.022 exp 23 = X grams
What is a mole, anyhow? No, it’s not a burrowing furry animal, the bane of gardeners. It is a count of 6.022 exp 23 of……whatever. The number is named for Amedeo Avogadro, a 19 century Italian chemist. That number, being that HUUUGE, must count something very small. What could be smaller than atoms or molecules? (Well, a lot of things, actually – how about a mole of virus, or is it virae????).
So, if we could, given an imaginery set of molecular tweezers and a near infinite amount of time, count that number of, let’s say, carbon atoms, we would accumulate a mole of carbon atoms. Since this is, to say the least, impractical, we need a better way. Turns out that a mole of carbon can be easily weighed out – 12.011 grams is the weight of a mole, 6.022 exp 23 carbon atoms. All you need is a Periodic Table and a balance (scale)! As a practical matter, you don’t even need to bother with Avogadro’s Constant to solve the vast majority of problems in stoichimetry (there’s that word again…).
I am old enough to remember the days before calculators were invented. We used slide rules and logarithms (remember those)? It wasn’t until I got my first professional job as an analytical chemist with FDA in 1960 that I got to use a rudimentary calculator. Many of the stoichiometry problems we worked on in chemistry courses involved setting up the equations, but not solving them.
Back to the present. Grams to moles calculations are necessary because (along with molecular tweezers) there is no such thing as a mole weighing device (balance). Since 12.011 grams (say) of carbon equals one mole, we need to be able to convert from one to the other. (Incidentally, one mole of sulfur weighs 32.06 grams; one mole of uranium weighs 238.03 grams). How much does 0.45 moles of sulfur weigh? You guessed it! 14.43 grams! So, if you want to know how much does 3.45 moles of sulfur weighs, simply multiply 3.45 moles x 32.06 grams per mole = 110.61 grams. Ain’t calculators grand!
We suffered through balancing equations last post. The coefficients for each chemical in a balanced equation represent the moles. We convert to grams by multiplying moles by molecular weight (the sum of weights from the Periodic Table for each of the compounds). What could be simpler? In a similar fashion, we can convert grams to moles by dividing by molecular weight.
OK, consider a reaction often used to produce oxygen (that is, molecular oxygen, O2). Potassium chlorate is heated to produce potassium chloride and oxygen:
KClO3 —–> KCl + O2
Balancing is simple: 2KClO3 ——> 2KCl + 3O2
Supposing I need to produce 2,000 grams of oxygen. How much KClO3 do I need? From the balanced equation, for every 3 moles of oxygen I need to react 2 moles of KClO3. How many moles are there in 2,000 grams of oxygen? If we set up an expression (whatever that means) such that 2.000 grams x 32.0g/mole = 62.5 moles O2. Since 3 moles of O2 are produced from 2 moles of KClO3, we need 41.7 moles of KClO3. Since the molecular weight of KClO3 is 122.6 (39.1+35.5+48.0), the answer is………………………5,112 grams.
A necessary skill one needs to master is the art of recognition of a correct answer to a problem. When I was teaching, I had to spend an inordinate amount of time trying to teach students how to use their calculator. Many of the problems involved the numerous operating systems in use at that time. I doubt that that has much improved today. For example, say the correct answer was 2.84 exp -3. It is conceivable that another student might get 0.00284. What is the right answer??
A colleague of mine at O’Connell H.S. coined a phrase, which should be a mantra of sorts, Chem is Try. It’s not rocket science, in the current idiom. Just need to work at it.
Many of you might ask, as my high school kids used to, “Mr Canaff, when do we blow stuff up?” I’ll deal with that in a future post.