In order to do percent Composition it is imperative to know what it is. According to Anne Marie Helmenstine, Ph.D. “percent composition is the percentage by mass of each element in a compound. ” That means the end result will tell what percentage of the weight is a certain element such as carbon, hydrogen, or water. There are two equations that will find this number. They are exactly the same thing it just depends on the definition that is easier for you to remember. They look like this:

The best way to understand this is by example, so we will start with a practice problem to solve. The Problem is:

**What is the Percent Composition of Carbon in Ethane(C2H6)?**

Step number 1.Multiply the subscribes of Carbon (AKA number of atoms) times the atomic weight from the periodic table (it is written in amu) and that is 2 x 12.011=24. 022

Step number 2. Find the Formula weight. The only difference this time is it won’t just be the part we are looking for, it is the entire equation and that looks like: C:2(12.011)+ H:6(1.0079) = 30.0694

Step number 3. Use your answer from step one or the Carbon weight (in this case Carbon) divided by Step two or the Formula weight. It looks like 24.022/30.0694= .798885

Step number 4. Lastly find percent by multiplying the quotient by 100. So that looks like .798885 x 100= 79.8885%

Most likely I would round that to 80% or 79.89% depending on the situation and how accurate the information needs to be.

Now its time to think about how this is useful in chemistry. Remember “percent composition is the percentage by mass of each element in a compound. ” Or in other words the answer is the number of grams of Carbon in a one hundred gram sample of Ethane. So using that ratio it is possible to determine how many grams of Carbon is in any sample of Ethane. So if you needed to know how much Carbon is in 36.17g of Ethane you would use the percent composition to determine this. it looks like :

Therefore for this particular problem to know how much Carbon is in 36.17g of Ethane the calculation would be

36.17g C (79.8885g C)/(100g)= 28.8957 g of Carbon.

From these calculation and more it is possible to calculate the Empirical Formula of a substance. So in order to demonstrate this we will again use an example using the Percent calculation to go to Empirical Formula.

If 40.0% is calcium, 12.0% is carbon, and 48.0% is oxygen, what is the empirical formula? To figure this out use the steps below and I will show the calculations below to help clarify what to do.

Steps | Ways to Complete the Step |

1. Mass % of Elements | Assume 100 gram sample |

2. Grams of Each element | Use Molar Mass |

3. Moles of each element | Calculate mole ratio using the smallest mole |

4. Empirical Formula |

Ca 40%= 40g, 40g (1 mole/ 40.08g)= .998004moles, .998004/.998004= 1

C 12%= 12g, 12g(1 mole/12.011g)= .999084moles, .999084/.998004= 1.00108 or 1

O 48%= 48g, 48g(1 mole/15.999g)= 3.00019moles, 3.00019/.998004= 3.00619 or 3

So the empirical formula is CaCO3.

The reason this is important is because the last number could be and in this case is the identity of the substance. The only way it wouldn’t be is if the actual formula or molecular formula was something like glucose that is C6H12O6 which is not the smallest ratio like the empirical formula which is CH2O. But this formula tells us what the element is. And it is Calcium Carbonate.

http://chemistry.about.com/od/chemistryglossary/g/Percent-Composition-Definition.htm

http://upload.wikimedia.org/wikipedia/commons/d/d3/Calcium_carbonate.png

https://sweetpea1096.files.wordpress.com/2013/10/b1596-caco3.gif

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