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Percent Composition

Empirical Formula

Determination of percent competition is nothing hard. You simply determine the total molar mass of the compound and the amount each element contributes to the total mass (you probably wrote that down while determining the molar mass. Then, you divide the mass of each individual element by the total and express as percent. The trick is going back to emperical formula from percent composition.
Emperical formula, you will recall, is simply the ratios of all the atoms to each other in a compound. In an ionic compound, that's all you need. However, in molecular compounds, there can be many different molecules with the same ratio of atoms. You can determine the ratio of atoms simply by determining the ratio of the moles of each element in the compound. You start by assuming you have 100g of the compound. Each percentage then tells you the number of grams of each element. You convert that to moles of each element. At that point you have the mol ratio, which is the same as the ratios in the empirical formula. However, they are unlikely to be integers. To get to integers, you divide through by the lowest value. Usually, that will get integers. But, you may have to multiply each by some integer to get whole numbers. The steps as outlined on the slide are below.

  1. Base calculation on 100 grams of compound.
  2. Determine moles of each element in 100 grams of compound.
  3. Divide each value of moles by the smallest of the values.
  4. Multiply each number by an integer to obtain all whole numbers.

Here is an easy practice: You have a compound with only Potassium, Oxygen and Nitrogen (the very astute among you may already have an idea what it is likely to be). The percentages are: O, 47.47%; N, 13.85%; and K, 38.67%. What is the formula? You can get the
solution here.

Here is a harder one. I'll post the solution later: C, 42.11%; H, 6.48%; O, 51.42% And the mass of the molecule is 342.3AMU. What is the empirical and molecular formula?

Now, here's a really hard one we will be looking at in class: Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers.
It contains only C, H, and O. Combustion of 19.81 mg terephthalic acid produces 41.98 mg CO2 and 6.45 mg H2O. If 0.250 mol of terephthalic acid has a mass of 41.5 g, determine the molecular formula of terephthalic acid. (Type your answer using the format CO2 for CO2.) Have fun


The answers will be below. Consider the following equation for burning pentane in oxygen.
5H12 + O2 --> CO2 + H2O

First you have to balance that equation. Here are the instructions:
  1. Choose an element that is only in one compound on each side (ie not oxygen in this case). I'll choose carbon. Balance that (put a 5 before CO2 to match the five carbons in pentane). DON'T TOUCH EITHER OF THEM AGAIN until the last step, if necessary. We have 1 pentane for 5 CO2 and that ratio will stay the same, even if it ends up 2:10 or 3:15.
  2. Do the next thing that is only in one compound on each side…Hydrogen. 12 H in pentane, 2 in each water…I need 6 before the H2O. DON'T TOUCH IT IN THE NEXT STEP.
  3. Now look at the oxygens. I've got 5CO2 (10 oxygens) and 6 waters (6 oxygens) for a total of 16. So, I put an 8 in front of O2 on the left to provide those.

5H12 + 8O2 --> 5CO2 + 6H2O

Now, imagine I mix 5.45g of C5H12 and 11.34g of O2 How many grams of water will I make? And which ingredient is left over AND how much of that in grams is left?

Answers C
5H12 is 72.15 g/mol water is at 18.02g/mol O2 is 32g/mol For C5H12, 5.45g x 1mol/72.15=0.0755mol 11.42g of O2 is 0.3569mol. You clearly have more O2 than C5H12. However, you don't have 8x as much. In other words, for 0.0755 mol of C5H12 you would need 0.604mol. You don't have that, so Oxygen is limiting. So, for 0.3569 mol of oxygen, you would use 1/8 that in C5H12 or 0.0446 mol. To answer the last part first, that means you would have left 0.0755-0.0446 mol of C5H12 which is 0.309 mol or 2.23g left over. For water, you have 0.3569 mol O2 x 6 mol H2O/8molO2 x 18.02g H2O/mol for 4.82g H2O.
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