2 Definition: the lowest whole number ratio of elements in a compound. Steps to determine empirical formulas:1) find moles of each element [find the moles of H20 if compound is hydrated].2) divide by smallest number of moles to get a whole number.3) multiply by a common whole number, if necessary.
3 Calculate EF of a compound with 25.9% N and 74.1% O. %=assume mass mass mols25.9g N x 1 mol N = mols N14.007g N74.1g O x 1 mol O = 4.63 mols O16 g O2) N: = 11.85O: = 2.53) Multiply by common whole numberN: 1 x 2 =2O: 2.5 x 2 = = N2O5
4 Just remember this!!Percents to grams Grams to moles Divide by smallest Multiply ‘til whole
5 Given: 36.5g Na, 25.4g S, and 38.1g O 25.4g S x 1 mol S = .79 mols S 36.5g Na x 1 mol Na = 1.59 mols Na22.989g Na25.4g S x 1 mol S = .79 mols S32g S38.1g O x 1 mol O = 2.4 mols O16g ONa: = S: .79 = O: 2.4 = 3= Na2SO3
6 An oxide of aluminum is formed by the reaction of 4 An oxide of aluminum is formed by the reaction of 4.151g of aluminum with 3.692g of oxygen. Find EF.4.151g Al x 1 mol Al = mol Al atoms26.98 g Al3.692g O x 1 mol O = mol O atoms16 g Omol Al = mol Al atoms0.1539mol O = mol O atoms1.500 O x 2 = = 3 O atoms1.000 Al x 2 = 2.000= 2 Al atoms= Al2O3
7 Multiple = molecular mass x empirical formula (whole number) Molecular formulas show the number of atoms in a compound. In order to determine the molecular, you must have the empirical formula first.The molecular mass of molecule will always be given.Multiple = molecular mass xempirical formula (whole number)The molecular formula is always an integer multiple of the empirical formula.
8 Determine the molecular formula of a compound whose EF is CH2O and molecular mass is 120 g/mol -Distribute that 4 throughout the empirical formula =C4H8O4
9 A compound is 64. 9% carbon, 13. 5% hydrogen, and 21. 6% oxygen A compound is 64.9% carbon, 13.5% hydrogen, and 21.6% oxygen. Its molc mass is 74 g/mol. What is its MF?64.9g C x 1 mol C = 5.40 mols C = 4 C12.01 g C13.5g H x 1 mol H = mols H = 10 H1.01 g H21.6g O x 1 mol O = 1.35 mols O = 1 O16.0 g O74 g/mol = 174 g/mol = C4H10O
10 Ex) a compound is 54. 5% carbon, 9. 1% hydrogen, and 36. 4% oxygen Ex) a compound is 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen. It’s molc mass is 88 g/mol. What is its molecular formula?54.5 C x 1 mol C = 4.54 mols C12.01 g C9.1 g H x 1 mol H = 9 mols H1.01 g H36.4 g O x 1 mol O = 2.28 mols O16 g O4.54 mols C = 2 C2.28 mols9 mols H = 4H2.28 mols O = 1O= C4H8O2
11 Ex) A compound has an empirical formula of ClCH2 and a molecular weight of g/mol. What is it’s molecular formula?Mass Cl + mass C + 2(mass H)Mass of empirical unit= (2.008) = g/mol98.96 g/mol = 2.00049.47 g/mol= Cl2C2H4