Do you feel like this: |
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when you see this: |
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Don't scream. You can do this. Here's how:
The strategy:
1) Get from percentages into units of mass (grams).
2) Use information from the periodic table to convert grams into the number of moles.
3) Do a mole-to-mole comparison so we can get the ratio of carbon to oxygen to hydrogen.
4) Once we have a ratio of each of these, we can determine the formula for the compound.
5) Compare the molar mass of the formula to the given molar mass of the compound. If it's the same, we're done. If it's different, we'll have to adjust things.
Here goes:
1) Get from percentages into mass (grams):
The percentages of carbon, oxygen and hydrogen add up to 100%. (This method will always work. The individual percentages of each component, added together, must equal one-hundred percent!)
So let's assume we have 100 grams of the compound. That way, we'll be able to use the percentages that were given, and it will be easy to determine the number of grams. Here's what I mean:
Using the percentages given: | With 100 grams of the compound: |
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75.5% carbon | 75.5 grams carbon |
15.5% oxygen | 15.5 grams oxygen |
8.8% hydrogen | 8.8 grams hydrogen |
------------------------ | ------------------------ |
100% total | 100 grams total |
2) Use information from the periodic table to convert grams into the number of moles
Using the periodic table, we can get the following information about the molar masses of carbon, oxygen and hydrogen:
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Carbon: 1 mole carbon 75.5grams carbonx ------------------ = 6.29 moles carbon 12.01grams carbonOxygen: 1 mole oxygen 15.5grams oxygenx ------------------ = 0.97 moles oxygen 16.00grams oxygenHydrogen: 1 mole hydrogen 8.8grams hydrogenx ------------------ = 8.71 moles hydrogen 1.01grams hydrogen
3) Do a mole-to-mole comparison so we can get the ratio of carbon to oxygen to hydrogen
Just divide by the smallest number.
In this case, we see that oxygen has the smallest number of moles: 0.97
So divide all the numbers by 0.97:
Carbon: 6.29 moles carbon / 0.97 = 6.5 moles carbon Oxygen: 0.97 moles oxygen / 0.97 = 1.0 moles oxygen Hydrogen: 8.71 moles hydrogen / 0.97 = 9.0 moles hydrogen
4) Once we have a ratio of each of these, we can determine the formula for the compound.
Try putting the number of moles as subscripts for each element:
C6.5H9O1
Hm, if we had whole numbers, everything would be fine. But in this case, that 6.5 moles of oxygen just doesn't look right!
So let's multiply everything by 2, to get rid of that half-mole:
Carbon: 6.5 moles carbon x 2 = 13 moles carbon Oxygen: 1.0 moles oxygen x 2 = 2 moles oxygen Hydrogen: 9.0 moles hydrogen x 2 = 18 moles hydrogenOK, that looks better. Now just use those numbers as your subscripts:
5) Compare the molar mass of the formula to the given molar mass of the compound. If it's the same, we're done. If it's different, we'll have to adjust things
Our chemical formula is:
C13H18O2
Let's determine the molar mass.
[Note: If you're unsure how to calculate molar mass, check out the Moles and Molar Mass Tutorial.]
Carbon: 12.01 grams carbon 13 moles carbon x ------------------ = 156 grams carbon 1 mole carbon Hydrogen 1.01 grams hydrogen 18 moles hydrogen x ------------------- = 18.18 grams hydrogen 1 mole hydrogen Oxygen 16.00 grams oxygen 2 moles oxygen x ------------------ = 32.00 grams oxygen 1 mole oxygenAdd up the masses of carbon, hydrogen and oxygen to get the molar mass of C13H18O2:
Compare this to what was given at the very beginning of the problem:
The molar mass of the compound is approximately 206 grams per mole. |
Guess what? We're done!
Yes, this took awhile. The process is long, but if you follow the steps, you'll eventually get to the correct answer:
1) Get from percentages into units of mass (grams). 2) Use information from the periodic table to convert grams into the number of moles. 3) Do a mole-to-mole comparison so we can get the ratio of carbon to oxygen to hydrogen. 4) Once we have a ratio of each of these, we can determine the formula for the compound. 5) Compare the molar mass of the formula to the given molar mass of the compound. If it's the same, we're done. If it's different, we'll have to adjust things.
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Do some mass percent composition practice problems
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