This document discusses determining chemical formulas, including: 1) Defining empirical and molecular formulas, and how to calculate empirical formulas from percentage or mass composition data. 2) Explaining how to determine a molecular formula from an empirical formula using the relationship between molecular formula mass and empirical formula mass. 3) Providing examples of calculating empirical formulas from composition data and determining molecular formulas from empirical formulas and molar masses.

1. Determining chemical formulasChapter 7.4

2. Objectives:Define empirical formula, and explain how the term applies to ionic and molecular compoundsDetermine an empirical formula from either a percentage or a mass compositionExplain the relationship between the empirical formula and the molecular formula of a given compoundDetermine a molecular formula from an empirical formula

3. Empirical formulaDefined as: consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compoundFor ionic compounds – formula unit is usually the compounds empirical formulaFor molecular compounds – empirical formula does not necessarily indicate the actual numbers of atomsExample:BH3 (diborane) empirical formulaB2H6 molecular formula

4. Calculation of empirical formulasSample problem 1:Quantitative analysis shows that a compound contains 32.38 % sodium, 22.65 % sulfur, and 44.99% oxygen. Find the empirical formula of this compound.3. Divide by the smallest number1. Assume 100 g samples32.38 g Nax 1 mol Na 22.99 g Nax 1 mol S 32.07 g Sx 1 mol O 16.00 g O2÷ o.7063= 1.993 mol Na= 1.408 mol Na22.65 g S1÷ o.7063= 1 mol S= 0.7063 mol S44.99 g O÷ o.7063= 3.981 mol O4= 2.812 mol O4. Round2. Convert to molesNa2SO45. Use as subscripts

5. Sample Problem 2Sample problem 2:Analysis of a 10.150 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 g. What is the empirical formula of this compound.5. Multiply by 23. Divide by the smallest number4.433 g P12= 1 mol P= 0.1431 mol P÷ 0.1431x 1 mol P 30.97 g Px 1 mol O 16.00 g O5.717 g O2.5= 2.497 mol O÷ 0.1431= 0.3573 mol O51. Convert to moles4. RoundP2O55. Use as subscripts

6. Practice ProblemsA compound is found to contain 63.52 % iron and 36.48 % sulfur. Find its empirical formula.Find the empirical formula of a compound found to contain 26.56 % potassium, 35.41 % chromium, and the remainder oxygen.Analysis of 20.0 g of a compound containing only calcium and bromine indicates that 4.00 g of calcium are present. What is the empirical formula of the compound formed?Answer: FeSAnswer: K2Cr2O7Answer: CaBr2

7. Calculation of Molecular formulasRemember: empirical formula is smallest possible whole number ratio of atoms in a compoundMolecular formula is the ACTUAL formulaRelationship can be written as:

8. x(empirical formula mass) = molecular formula massx = whole number multipleFactor which the subscripts are multiplied to give empirical formula.Sample Problem 1: Molecular formulaIn the last sample problem, the empirical formula of a compound of phosphorus and oxygen was found to be P2O5. Experimentation shows that the molar mass of this compound is 283. 89 g/mol. What is the compound’s molecular formula?x(empirical formula mass) = molecular formula massx =molecular formula massempirical formula mass283.89 amu141.94 amu1. Empirical formula massP 2 x 30.97 amu = 61.94 amu O 5 x 16.00 amu = 80.00 amu = 141.94 amux == 2.0001So molecular formula would be= P4O102 x (P2O5)

9. Practice problems1. A sample of a compound with a formula mass of 34.00 amu is found to consist of 0.44 g H and 6.92 g O. Find its molecular formula.First: find the empirical formula3. Divide by the smallest number0.44 g H= 0.929mol H1= 0.436 mol H÷ 0.4336.92 g O1= 1 mol O÷ 0..433x 1 mol H 1.01 g Hx 1 mol O 16.00 g O= 0.433 mol O1. Convert to moles4. Roundmolecular formula massempirical formula mass34.00 amu17.01 amuHO5. Use as subscriptsH 1 x 1.01 amu = 1.01 amu O 1 x 16.00 amu = 16.00 amu = 17.01 amux =x == 2= H2O22 x (HO)

10. Practice Problem 2Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.110 amu.x(empirical formula mass) = molecular formula massx =1. Empirical formula massC 1 x 12.o1 amu = 12.01 amu H 1 x 1.01 amu = 1.01 amu = 13.02 amux == 5.999molecular formula massempirical formula mass78.110 amu 13.02 amuSo molecular formula would be= C6H66 x (CH)