2A. Percentage Composition EMPIRICAL and MOLECULAR FORMULASA. Percentage Compositionthe percentage by mass of each element in a compound
3A. Percentage Composition Find the % composition of Cu2S.g Cug Cu2S%Cu = 100 =79.852% Cu32.07 g Sg Cu2S%S = 100 =20.15% S
4A. Percentage Composition Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.28 g36 g%Fe = 100 =78% Fe8.0 g36 g%O = 100 =22% O
5A. Percentage Composition How many grams of copper are in a 38.0-gram sample of Cu2S?Cu2S is % Cu(38.0 g Cu2S)( ) = 30.3 g Cu
6A. Percentage Composition Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O?36.04 g H2Og H2O%H2O = 100 =24.51%H2O
7C2H6 CH3 B. Empirical Formula Smallest whole number ratio of atoms in a compoundC2H6reduce subscriptsCH3
8B. Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element.3. Divide moles by the smallest # to find subscripts.4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.
9What ?????????? How can I remember? There is a poem to help Percent to MassMass to MoleDivide by SmallMultiply ‘til WholeSometimes the mole values are still not whole numbers so multiply by the smallest factor that will make them whole numbers
10B. Empirical FormulaFind the empirical formula for a sample of 25.9% N and 74.1% O.25.9 g N1 mol N14.01 g N= 1.85 mol N= 1 N1.85 mol N1.85 mol74.1 g1 mol16.00 g= 4.63 mol O= 2.5 O
11N2O5 N1O2.5 B. Empirical Formula Need to make the subscripts whole numbers multiply by 2N2O5
12Example of Determining Empirical Formula Ex. What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% Oxygen?25.9g N 1 mol N = 1.85 mol N14.0g N74.1g O 1 mol O = 4.63 mol O16.0g ODivide each element by the lowest number ofmoles in the compound:1.85 mol N = 1 mol N1.854.63 mol O = 2.50 mol ONow find the lowest common denominator.1 mol N x 2= N2 2.5 mol O x 2 = O5Therefore, N2O5 is the empirical formula
13Practice ProblemsCalculate the empirical formula for a compound that is 94.1% O and 5.90%HCalculate the empirical formula for a compound that is 79.8% C and 20.2% H
14CH3 C2H6 C. Molecular Formula “True Formula” - the actual number of atoms in a compoundCH3empiricalformula?C2H6molecularformula
15C. Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass.3. Divide the molecular mass by the empirical mass.4. Multiply each subscript by the answer from step 3.
162(CH2) C2H4 C. Molecular Formula The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is g/mol?empirical mass = g/mol28.1 g/mol14.03 g/mol= 2.002(CH2) C2H4
17Example of Molecular Formula Calculate the molecular formula of the compound whose Molar mass is 60.0g and empirical formula is CH4N.Molar Mass of CompoundEmpirical formulaEmpirical formula mass(efm)Molar MassefmMolecular Formula60.0gCH4N= 30.0g CH4N60.0/30.0= 2CH4N x 2= C2H8N2