Let’s begin with an example:

Suppose I have a glucose molecule. (C6H1206) What percentage of the molecule’s mass is carbon?

The first step is to identify the atomic weights of the various elements in glucose. With the help of a periodic table of elements, we know that carbon‘s atomic weight is roughly 12.011 atomic mass units (amu). Hydrogen‘s atomic weight is roughly 1.008 amu. Oxygen has an atomic weight of 15.999 amu.

The second step is to multiply the element’s atomic weight by the number of atoms in the molecule, and add the new atomic weights together. Let’s take carbon as an example. There are 6 carbon atoms in glucose. Therefore, we multiply 6 by 12.011. We get 72.066 amu. See if you can find the atomic weight of glucose using this strategy.

So you should have gotten about 180.156 amu as the mass for the entire molecule. In order to solve the original question you divide 72.066 (carbon’s mass) by 180.156. The answer is 4/10. Therefore, 40 percent of a glucose molecule’s mass is carbon. We can apply this on a larger scale also. If I had 10 grams of glucose, I could expand 4 grams to be pure carbon.

Using this knowledge, we can predict how matter will change during a chemical reaction. We can also figure out the amount of a substance that will be created during processes such as combustion. This is only the tip of the iceberg. Scientists can deduce more about chemical processes using units such as the mole and employing empirical formulas.

Merely the tip of the iceberg…

Sources: http://www.chem.tamu.edu/class/majors/tutorialnotefiles/percentcomp.htm

http://hyperphysics.phy-astr.gsu.edu/hbase/organic/sugar.html

https://fp.auburn.edu/fire/combustion.htm

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Jael,

How exactly would this apply to combustion? And, you imply that this is not related to the mole or empirical formulas by your statement, “Scientists can deduce more about chemical processes using units such as the mole and employing empirical formulas.” Yet isn’t the very process you state for determining formula weight based on molar relationships quantified by the subscripts in the formula? Isn’t this why 12.01g of carbon is multiplied by 6? Additionally, if you determine that a compound is 40% carbon, isn’t that information one means of determining empirical formulas?

Miss Gardner

Miss Gardner,

All the comments you have made are true. My purpose with the blog was to provide a simple example of percent composition. I felt that stating the link between the mole and empirical formulas to percent composition, though important information, was an unnecessary tangent that would weaken the focus of the blog. Also, since this was a How-To blog, I wanted to emphasis the math and logic used. Thank you for your comment.

Jael