Presentation on theme: "Chapter 3 Percent Compositions and Empirical Formulas"— Presentation transcript:
1 Chapter 3 Percent Compositions and Empirical 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 100Molar 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 grams/mole = g/molN 2 x (14.0) grams/mole = 28.0g/molO 6 x (16.0) grams/mole = g/mol164.1 g/mole
4 Step 2: Divide the mass of each element by the molar mass and multiply by 100 to get the percent. % Ca = g x 100% = % Calcium164.1g% N = g x 100% = % Nitrogen164.1 g% O = g x 100% = % Oxygen
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 AgX 100 = 87.1 % Ag33.3 g totalTotal = 100 %4.30 g SX 100 = 12.9 % S33.3 g total
6 FormulasEmpirical formula: the lowest whole number ratio of atoms in a compound.Molecular formula: the true number of atoms of each element in the formula of 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.
7 Formulas (continued)Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = cannot be reduced).Examples:NaClMgCl2Al2(SO4)3K2CO3
8 Formulas (continued)Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio).Molecular:H2OC6H12O6C12H22O11(Correct formula)Empirical:H2OCH2OC12H22O11(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 ExampleCalculate the empirical formula of a compound composed of % C, % H, and %N.Step 1 : Assume 100gram sampleStep 2 : Convert grams to moles for each element38.67 g C x 1mol C = 3.22 mole C g C16.22 g H x 1mol H = mole H g H45.11 g N x 1mol N = 3.22 mole N g NNow 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 mol H = 5 mol H mol N 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 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/moleSet up ratio g = 297gMultiply each subscript in the empirical formula by 2.C8H10N4O2 is the molecular formula.
14 Empirical formula from combustion reactions Example: grams of an organic compound containing carbon, hydrogen, and oxygen was analyzed by combustion. The amount of carbon dioxide produced was grams and the amount of water produced was grams. Determine the empirical formula of the compound.Combustion reaction looks like the following:CHO + O2 CO H2O
15 Assume all of the Carbon in the compound turned into carbon dioxide, and determine the mass percent of carbon in the compoundg CO2 1mole mol C g C44g CO2 1 mol CO mol= grams Carbong x = % Carbon7.321 g
16 Assume all of the hydrogen in the compound went to water, and determine the mass percent of hydrogen in the compound7.316 g H2O 1mol H2O 2 mols H 1.0 g H 18g H2O 1mol H2O 1 mol H = grams Hydrogen 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.(66.58% C % H) =22.32 % OxygenNow use these percents to find the empirical formula!C4H8O