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Chapter 7-4: Determining Chemical Formulas

Chapter 7-4: Determining Chemical Formulas. Coach Kelsoe Chemistry Pages 245-249. Section 7-4 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.

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Chapter 7-4: Determining Chemical Formulas

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  1. Chapter 7-4:Determining Chemical Formulas Coach Kelsoe Chemistry Pages 245-249

  2. Section 7-4 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. Determining Chemical Formulas • When a new substance is synthesized or discovered, it is analyzed quantitatively to reveal its percentage composition. From this data, the empirical formula is then determined. • An empirical formula 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.

  4. Determining Chemical Formulas • For an ionic compound, the formula unit is usually the compound’s empirical formula. • For a molecular compound, however, the empirical formula does not necessarily indicate the actual numbers of atoms present in each molecule. • For example, the empirical formula of diborane gas is BH3, but the molecular formula is B2H6.

  5. Calculation of Empirical Formulas • To determine a compound’s empirical formula from its percentage composition, begin by converting percentage composition to a mass composition. • Assume you have a 100 gram sample of the compound. Then calculate the amount of each element in the sample.

  6. Calculation of Empirical Formulas • For example, the percentage composition of diborane is 78.1% B and 21.9% H. Therefore 100 grams of BH3 contains 78.1 g of B and 21.9 g of H. • Next, the mass composition of each element is converted to a composition in moles by dividing the appropriate molar mass: 78.1 g B x 1 mol B 21.9 g H x 1 mol H = 7.22 mol B = 21.7 mol H 10.81 g B 1.01 g H

  7. Calculation of Empirical Formulas • These values give a mole ratio of 7.22 mol B to 21.7 mol H. However, this is not a ratio of smallest whole numbers. • To find such a ratio, divide each number by the smallest number in the existing ratio. 7.22 mol B 21.7 mol H : = 1 mol B:3.01 mol H 7.22 7.22

  8. Calculation of Empirical Formulas • Because of rounding or experimental error, a compound’s mole ratio sometimes consists of numbers close to whole numbers instead of exact whole numbers. In this case, the differences from whole numbers may be ignored and the nearest whole number taken. Thus, diborane contains atoms in the ratio 1 B:3 H. Its empirical formula is BH3.

  9. Calculation of Empirical Formulas • Sometimes mass composition is known instead of percentage composition. • To determine the empirical formula in this case, convert mass composition to composition in moles. Then calculate the smallest whole-number ratio of atoms.

  10. 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? • Given: percentage compositions of Na, S, and O • Unknown: empirical formula • Step 1: Convert percentages to grams • Step 2: Convert grams to moles • Step 3: Find the ratio of smallest whole numbers • Solution: Na2SO4.

  11. 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? • Given: sample masses of P and O • Unknown: empirical formula • Step 1: Convert sample mass to moles • Step 2: Find the ratio of smallest whole numbers • Solution: P2O5.

  12. Calculation of Molecular Formulas • Remember that the empirical formula contains the smallest possible whole numbers that describe the atomic ratio. • The molecular formula is the actual formula of a molecular compound. • An empirical formula may or may not be a correct molecular formula. • For example, ethene (C2H4) and cyclopropane (C3H6) share an identical atomic ratio, but are very different substances.

  13. Calculation of Molecular Formulas • The relationship between a compound’s empirical formula and its molecular formula can be written as follows: • x(empirical formula) = molecular formula • The number represented by x is a whole-number multiple indicating the factor by which the subscripts in the empirical formula must be multiplied to obtain the molecular formula. The value of x is sometimes 1.

  14. Calculation of Molecular Formulas • To determine the molecular formula of a compound, you must know the compound’s formula mass. For example, experimentation shows the molar mass of diborane to be 27.67 g/mol. • The molar mass for the empirical formula, BH3, is 13.84 g/mol. • Dividing the experimental formula mass by the empirical formula mass gives us x for BH3.

  15. Calculation of Molecular Formulas • Recall that a compound’s molecular formula mass is numerically equal to its molar mass, so a compound’s molecular formula can also be found given the compound’s empirical formula and its molar mass.

  16. Sample Problem 3 • In Sample Problem 2, 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? • Given: empirical formula; molecular formula mass; empirical formula mass mass • Unknown: molecular formula

  17. Sample Problem 3 • x = Molecular FM/Empirical FM • Step 1: Convert molecular molar mass to molecular formula mass • 283.89 g/mol  283.89 amu • Step 2: Find the empirical formula mass by adding the masses of each atom in the formula • 141.94 amu • Step 3: Divide Molecular FM by Empirical FM to find x • x = 283.89 amu/141.94 amu = 2.0001 • Step 4:Plug x into x(EF) = molecular formula • (2)(P2O5) = P4O10

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