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.
2. Objectives: Define empirical formula, and explain how the term applies to ionic and molecular compounds Determine an empirical formula from either a percentage or a mass composition Explain the relationship between the empirical formula and the molecular formula of a given compound Determine a molecular formula from an empirical formula
3. Empirical formula Defined 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 compound For ionic compounds – formula unit is usually the compounds empirical formula For molecular compounds – empirical formula does not necessarily indicate the actual numbers of atoms Example: BH3 (diborane) empirical formula B2H6 molecular formula
4. Calculation of empirical formulas Sample 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 number 1. Assume 100 g samples 32.38 g Na x 1 mol Na 22.99 g Na x 1 mol S 32.07 g S x 1 mol O 16.00 g O 2 ÷ o.7063 = 1.993 mol Na = 1.408 mol Na 22.65 g S 1 ÷ o.7063 = 1 mol S = 0.7063 mol S 44.99 g O ÷ o.7063 = 3.981 mol O 4 = 2.812 mol O 4. Round 2. Convert to moles Na2SO4 5. Use as subscripts
5. Sample Problem 2 Sample 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 2 3. Divide by the smallest number 4.433 g P 1 2 = 1 mol P = 0.1431 mol P ÷ 0.1431 x 1 mol P 30.97 g P x 1 mol O 16.00 g O 5.717 g O 2.5 = 2.497 mol O ÷ 0.1431 = 0.3573 mol O 5 1. Convert to moles 4. Round P2O5 5. Use as subscripts
6. Practice Problems A 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: FeS Answer: K2Cr2O7 Answer: CaBr2
7.
8.
9. Practice problems 1. 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 formula 3. Divide by the smallest number 0.44 g H = 0.929mol H 1 = 0.436 mol H ÷ 0.433 6.92 g O 1 = 1 mol O ÷ 0..433 x 1 mol H 1.01 g H x 1 mol O 16.00 g O = 0.433 mol O 1. Convert to moles 4. Round molecular formula mass empirical formula mass 34.00 amu 17.01 amu HO 5. Use as subscripts H 1 x 1.01 amu = 1.01 amu O 1 x 16.00 amu = 16.00 amu = 17.01 amu x = x = = 2 = H2O2 2 x (HO)
10. Practice Problem 2 Determine 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 mass x = 1. Empirical formula mass C 1 x 12.o1 amu = 12.01 amu H 1 x 1.01 amu = 1.01 amu = 13.02 amu x = = 5.999 molecular formula mass empirical formula mass 78.110 amu 13.02 amu So molecular formula would be = C6H6 6 x (CH)
