Presentation on theme: "Percentage Composition and Empirical Formula"— Presentation transcript:
1 Percentage Composition and Empirical Formula OBJECTIVES:Describe how to calculate the percent by mass of an element in a compound.Why do we care?allows us to convert from one to the other,,,but most helpful in the lab…I can mass grams and use the PT to help me determine moles and I can use moles to compare, cannot compare masses, only moles…thus we can calculate formula relationships from experimental dataUse example of a crucible.
2 Percentage Composition and Empirical Formula OBJECTIVES:Interpret an empirical formula.
3 Percentage Composition and Empirical Formula OBJECTIVES:Distinguish between empirical and molecular formulas.
4 Percentage Composition If given the masses of each constituent element, divide the mass of each by the total massMultiply by 100 to get a percentage
5 ExampleCalculate the percentage 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 Getting it from the formula If we know the formula, assume you have 1 moleYou know the mass of the elements and you can calculate the mass of the whole compound (these values come from the periodic table).
7 Examples Calculate the percent composition of C2H4 Find the molar mass of compoundDivide the molar mass of each element by the molar mass of the compound (x 100)85.6% C, 14.4 % H
8 Try One!How about aluminum carbonate?Al2 (CO3)3Remember to consider that you have 3 molecules of CO3 ( so 3 atoms of carbon and 9 of oxygen)Molar mass of the compound:g mol-1 (could be if used different PT table)23.02% Al, 15.41% C, and 61.57% O
9 Empirical Formula AKA simplest formula Indicates the elements present in the compoundIndicates the simplest whole number ratio of these elementsC6H12O6 is the molecular formula for glucose ( the “true” formula)The empirical formula is CH2O (all divided by the highest common factor of 6)
10 DistinctionAll ionic compound formulas are already in empirical formula “mode”Covalent compound formulas may or may not be in empirical formula “mode”Examples:H2O is the empirical and molecular formula for waterC6H6 is the molecular formula for benzene, but CH is the empirical formula
11 Calculating Empirical We can get a ratio from the percent composition.Assume you have a 100 g sample and the percentages 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.
12 Calculate the Empirical formula If given grams: convert all grams to molesDivide each by the smallest number of molesYou will now have the ratio and therefore the coefficients of the elementsIf this doesn't give you whole numbers, then multiply by the smallest value that will give you whole numbers.
13 ExampleCalculate the empirical formula of a compound composed of % C, % H, and %N.Assume 100 g sample38.67 g C x mol C = mole C g C16.22 g H x mol H = mole H g H45.11 g N x mol N = mole N g NNow divide each value by the smallest value
14 Example ContinuedThe ratio is mol C = 1 mol C mol N mol NThe ratio is mol H = mol H mol N mol NThe coefficients are C = 1, H = 5, N = 1 soC1H5N1 which is = CH5N
15 PracticeA compound is 43.64% P and 56.36% O. What is the empirical formula?P2O5Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?C4H5N2O
16 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 formulaCaffeine has a molar mass of 194 g. what is its molecular formula?= C8H10N4O2
17 ExampleA compound made of C and H contains 92.24% by mass of C. Its Mr = 78.1 Determine its molecular formula92.24% = g C = mol C100% % = 7.76 % = 7.76 g H = mol HEmpirical formula is CHRelative empirical mass is 13.02
18 Continued Mr = 78.1 and the relative empirical mass = 13.02 Divide 78.1 by = 6Multiply each subscript by 6Molecular formula is C6H6
19 Experimental Methods All the carbon is converted to carbon dioxide One way to find the percentage composition of organic compounds is to burn a known mass of the compound with excess O2All the carbon is converted to carbon dioxideAll the hydrogen is converted to waterMass of oxygen can be found by subtracting the sum of these two from the initial mass
20 Example2.80 g of an organic compound containing only C and H forms 8.80 g of CO2 and 3.60 g of H2O when it undergoes complete combustion. Determine the empirical formula.Amount of CO2 = amount of CAmount of H2O = ½ amount of HConvert each original amount to moles
21 Example Continued 8.80 g of CO2 = 0.200 mol CO2 = mol C 3.60 g of H2O = mol H2O, so the amount of H is 2 x = mol HDivide by the smallest value and you get 1 for C and 2 for HThe empirical formula is CH2
22 Another ExampleVitamin C contains C, H, and O only. On combustion of 1.00g of Vitamin C, 1.50 g CO2 and g H2O are produced. What is the empirical formula for vitamin C?1.50 g CO2 = mol CO2 = mol C0.408 g H2O = mol H2O2 x = mol H
23 ContinuedNeed to find how many grams of C and H are present to find how many grams of O are presentmol C = g Cmol H = g H1.00 g – ( ) = 0.54 g ONow change g of O to mol of O =Divide by the smallest value ( )
24 Continued You will get coefficients of 1, 1.3, and 1 Multiply each by 3 to get 3, 4, 3Empirical formula for vitamin C is C3H4O3