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You were asked to calculate the percentage composition of these two substances:
If you did, you may have noticed something interesting.
They both have the same percentage composition and it's also the same as that of glucose, example #2 in part one.
Here is CH3COOH.
First, figure out the molar mass from the formula. It is 60.05 g/mol.
Second, figure out the grams each atom contribues by multiplying the atomic weight by the subscript.
Carbon = 2 x 12.011 g = 24.022 g
Hydrogen = 4 x 1.008 = 4.032 g
Oxygen = 2 x 16.00 = 32.00 g
Third, divide the answer for each atom by the molar mass and multiply by 100 to get a percentage.
Carbon's percentage: (24.022 g / 60.05 g) x 100 = 40.00 %
Hydrogen's percentage: (4.032 g / 60.05 g) x 100 = 6.71 %
Oxygen's percentage: (32.00 g / 60.05 g) = 53.29 %
You could have done the last of the three by subtraction, if you wished. I just put up all three calculations.
All three substances (C6H12O6, CH2O, CH3COOH) have the same percentage composition because they all have the same empirical formula. "Empirical formula" is a REAL IMPORTANT concept and we will get to it in the next tutorial.
Percent Composition Practice Problems
1) Benzene has the formula C6H6. Calculate its percentage composition.
2) Acetylene has the formula C2H2. What is its percentage composition?
3) There are nine different compounds all of which have the formula C7H16. What do we know about the percentage composition for each of the nine compounds? Calculate the percent composition for the formula C7H16.
Instructions for #4 and 5: find an Internet site other than the ChemTeam to answer these two questions. You are welcome to study the ChemTeam (or textbook) definitions before you begin searching. Include the full URL of the site where you got the definition.
4) Look up the definitions of "empirical formula" and "molecular formula" and write them down.
5) Find a second site with the same two definitions worded differently.
Go to the Answers
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Go to Empirical Formula discussion.