1 / 17

Chapter 3 Percent Compositions and Empirical Formulas

Chapter 3 Percent Compositions and Empirical Formulas. AP Chemistry. Calculating Percent Composition of a Compound. The following general formula is used to determine the percent composition of each element in a compound: Mass of element

shanae
Download Presentation

Chapter 3 Percent Compositions and Empirical Formulas

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 3Percent CompositionsandEmpirical Formulas AP Chemistry

  2. Calculating Percent Composition of a Compound • The following general formula is used to determine the percent composition of each element in a compound: • Mass of element • (in 1 mole of compound) X 100 • Molar Mass of Compound

  3. What is the percent composition of each element in Ca(NO3)2 ? • Step 1: Find the molar mass of Ca(NO3). • Ca 40.1 grams/mole = 40.1 g/mol • N 2 x (14.0) grams/mole = 28.0g/mol • O 6 x (16.0) grams/mole = 96.0 g/mol • 164.1 g/mole

  4. Step 2: Divide the mass of each element by the molar mass and multiply by 100 to get the percent. % Ca = 40.1g x 100% = 24.4 % Calcium 164.1g % N = 28.0g x 100% = 17.1% Nitrogen 164.1 g % O = 96.0g x 100% = 58.5 % Oxygen 164.1g

  5. *Note percent composition can also be done given masses of a compound and each element in the compound. • Calculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S. 29.0 g Ag X 100 = 87.1 % Ag 33.3 g total Total = 100 % 4.30 g S X 100 = 12.9 % S 33.3 g total

  6. Formulas Empirical formula: the lowest whole number ratio of atoms in a compound. • The empirical formula for C3H15N3 is CH5N. • you are basically dividing by the greatest common factor , which in this case is 3, for each element subscript. Molecular formula: the true number of atoms of each element in the formula of a compound.

  7. Formulas(continued) Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = cannot be reduced). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

  8. Formulas(continued) Formulas for molecular compoundsMIGHT be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 (Correct formula) Empirical: H2O CH2O C12H22O11 (Lowest whole number ratio)

  9. Calculating Empirical • We can get a ratio from the percent composition. • Assume you have a 100 g sample - the percentage become grams (75.1% = 75.1 grams) • Convert grams to moles. • Find lowest whole number ratio by dividing each number of moles by the smallest value.

  10. Example • Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. • Step 1 : Assume 100gram sample • Step 2 : Convert grams to moles for each element • 38.67 g C x 1mol C = 3.22 mole C 12.0 g C • 16.22 g H x 1mol H = 16.22 mole H 1.0 g H • 45.11 g N x 1mol N = 3.22 mole N 14.0 g N Now divide each value by the smallest value

  11. Example • The ratio is 3.22 mol C = 1 mol C 3.22 mol N 1 mol N • The ratio is 16.22 mol H = 5 mol H 3.22 mol N 1 mol N =C1H5N1 which is = CH5N

  12. Empirical to molecular • Since the empirical formula is the lowest ratio, the actual molecule would weigh more. • By a whole number multiple. • Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula

  13. Caffeine has a molar mass of 194 g. and its empirical formula is C4H5N2O, what is its molecular formula? Find the empirical formula molar mass. 97 g/mole Set up ratio 194g = 2 97g Multiply each subscript in the empirical formula by 2. C8H10N4O2 is the molecular formula.

  14. Empirical formula from combustion reactions Example: 7.321 grams of an organic compound containing carbon, hydrogen, and oxygen was analyzed by combustion. The amount of carbon dioxide produced was 17.873 grams and the amount of water produced was 7.316 grams. Determine the empirical formula of the compound. Combustion reaction looks like the following: CHO + O2 CO2 + H2O

  15. Assume all of the Carbon in the compound turned into carbon dioxide, and determine the mass percent of carbon in the compound 17.873 g CO2 1mole 1 mol C 12g C 44g CO2 1 mol CO2 1 mol = 4.8744 grams Carbon 4.8744 g x 100 = 66.58% Carbon 7.321 g

  16. Assume all of the hydrogen in the compound went to water, and determine the mass percent of hydrogen in the compound 7.316 g H2O 1mol H2O 2 mols H 1.0 g H 18g H2O 1mol H2O 1 mol H = 0.8129 grams Hydrogen 0.8129 x 100 = 11.10% Hydrogen 7.321

  17. With the mass percent of Carbon and Hydrogen, you can figure out the mass percent of oxygen in the compound. 100 - (66.58% C + 11.10% H) = 22.32 % Oxygen Now use these percents to find the empirical formula! C4H8O

More Related