Percent composition is the percentage by mass of each element in a compound. first step to finding percent composition is calculating the formula/molecular weight or the molar mass of the compound you want to know the percent composition of. Formula weight is the sum of atomic masses of the atoms in the Ionic compound. Molecular weight is the sum of atomic masses of the atoms in a molecule. Both weights are Identical to the molar mass of a substance in amu’s or grams as ratios are similar. For example MgCl2:

Magnesium: (1)24.305

Chlorine : + (2) 35.45= 70.9

Magnesium Chloride= 95.205 g/mol or amu

Now that the mass is found you divide the part by the whole multiplied by a hundred to find the percentage. In this case you want to find the percentage of Magnesium in Magnesium Chloride.

Magnesium/Magnesium Chloride = 24.305g Mg/95.205 MgCl2 x 100 = 25.53%

Another example:

Finding the mass of an element in a substance is very similar to the finding percent composition. You just put the substance’s mass over one and multiply it by the molar mass of the element divided by the molar mass of the whole formula. For example 24 grams of MgCl2 and you want to find how much magnesium and chlorine make up the mass of the compound.

24.00g/1 x 70.9g of 1 mole of Cl2/95.205 molar mass of MgCl2 = 17.873grams of Chlorine

24.00/1 x 24.305 of 1 mole of Mg/95.205 molar mass of MgCl2 = 6.126 grams of Magnesium

Adding up the two masses together you get 23.999g which is only off due to rounding, but that is how you calculate how much mass of each element is in a substance.

Just as one can calculate percents from formulas you can also calculate formulas from percents. These are called empirical formulas and Molecular formulas. Molecular formulas are all about finding molar ratios between atoms. In Molecular formula equations if percentages are only given for each element assume the sample is a hundred grams. For example 50%C and 50%O would be 50 grams oxygen and 50 grams Carbon. First step of finding Molecular formulas is to divide the mass of the element given by one. Secondly multiply the mass of the single element by 1 mole divided by that elements molar number. Repeat this process for all elements, then divide all products by the smallest product to get molar numbers. If molar numbers such as 0.5, 0.4, 0.3, 0.7 or 0.6 are given multiply all numbers by two in till numbers are close to a whole number. The empirical formula is the simplified version of the molecular formula. So if the molecular formula is C6H12O6 then the empirical formula would be CH2O. For example 10 grams carbon, 15 grams nitrogen, and 20 grams of oxygen are given.

10g of C/1 x 1mole/12.011g= 0.8325 C/0.8325 = 1 x 4 = 4 mole

15g of N/1 x 1mole/14.007g= 1.0708 N/0.8325 = 1.2862 x 4 = 5.1448 mole

20g of O/1 x 1mole/15,9994g= 1.2500 O/0.8325 = 1.5015 x 4 = 6.006 mole

The Molecular formula and Empirical formula would be C4N5O4.

Another example:

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

That is how percent composition, Empirical formulas, and Molecular formulas are calculated.

Sources:

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

http://virtualgardner2.weebly.com/h-chemistry-unit-3.html

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

Chris,

One question: how can I have grams and moles in the same step of a calculation?

“24.00g/1 x 70.9g of 1 mole of Cl2/95.205 molar mass of MgCl2 = 17.873grams of Chlorine”

Doesn’t that mess up my units and therefore my calculation?

And, is there a difference between empirical and molecular formula? You used it interchangeably in your opening paragraph on the topic, but are they really the same thing? Lastly, your example for the formula calculation has a ‘2.5’ for the O. Are you supposed to round that? Can there be a 2.5 in a formula?

Miss Gardner

Sorry for the confusion. I said there was one mole of Cl2 which in turn is 70.98g. This can be done because the two are the same thing and therefore the interchangeable. So no it does not mess up the units in the calculation.

I actually did state in my blog that Empirical formulas are the simplified version of Molecular formulas. The example I used was if the Molecular formula is C6H12O6 then the empirical formula is CH2O. The way I stated it was pretty poor on my part so sorry for that.

And to address your last question about the 2.5 of Oxygen I assume you mean the picture used. The picture did all the math right except the last step of doubling the moles till you get to a whole number or very close to it. So all the mole numbers should be 14 for C 18 for H 2 for N and 5 for O.

thanks for your questions,

Chris