Biororz in Bombay

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Mole Mountaineering (Avogadro style)

For those of you who teach stoichiometry, you know how hard it can be to get kids to organize a series of conversions that will take them from the information that you have given them to the answer.  This was a huge struggle for my kids last year, so I came up with mole mountaineering.  Here is how it goes.  You start with a balanced equation, a given value, and a desired value.  For example:  ​

4 FeCr2O7 + 8 K2CO3 + O2 ---> 2 Fe2O3 + 8 K2CrO4 + 8 CO2

How many grams of FeCr2O7 are required to produce 44.0g of CO2 ?​

In this problem two molecules have been identified.  The one associated with the given value of 44.0 g, CO2, is mountain X and the ​FeCr2O7 ​will be mountain Y. 

You hate hiking.  You find yourself at the bottom of mountain X where the grams of CO2 are located.​  You have to travel to the grams position on mountain Y to find your answer.  The only way to cross over from mountain X to mountain Y is to take the mole ratio zip line.

Now it is time to plan your hike.  Choose the shortest route which is up CO2 mountain  to moles then across the zip line to the moles of mountain Y, then down FeCr2O7 mountain to the mass box at the bottom.  Three legs of your trip = three conversions. 
Once your trip is planned, it is time to set up your conversions.   The given value that you start with and all destinations go on the top of your conversions 44g CO2 -  Moles of CO2 - moles of FeCr2O7 - grams of FeCr2O7
Every time you go up the mountain, you have to divide your starting quantity by the quantity listed (molar mass of substance or Avogadro's #) if you go down a mountain, you have to multiply the starting number (moles) by the quantity listed ( molar mass or Avogadro's #)  As you can see, you can also calculate # atoms or molecules using mole mountaineering. 

Finally, solve your conversion top/ bottom and Voila! 

If you would like for me to send you a copy of the java file that "Tech Assistant" 3.0 created, just send me an e-mail at newcombr@asbindia.org and I will forward it along...or you could wait for the app! ​