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Aim: What is the difference between empirical and molecular formulas?

Aim: What is the difference between empirical and molecular formulas?. Do Now: Complete Worksheet Homework: Castle Learning #3, test Friday, Quiz Wednesday. VIII. Percent Composition, Molecular and Empirical Formulas.

walter-hyde
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Aim: What is the difference between empirical and molecular formulas?

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  1. Aim: What is the difference between empirical and molecular formulas? Do Now: Complete Worksheet Homework: Castle Learning #3, test Friday, Quiz Wednesday

  2. VIII. Percent Composition, Molecular and Empirical Formulas a) Percent Composition: the composition of a compound in terms of the percentage of each component present with respect to the whole. Ex #1: What is the percent composition by mass of the elements in sodium sulfate? • Find the formula of sodium sulfate. • Find the molar mass of sodium sulfate. • Using the formula for % composition by mass (reference table T), calculate the % composition of each element in the compound.

  3. Sodium Sulfate - Na2SO4 Na2SO4 Na = 2 X 23 = 46 S = 1 X 32 = 32 O = 4 X 16 = 64 • Molar Mass = 142 g/mol • % composition = part/whole X 100 Na = 46/142 X 100 = 32.4% S = 32/142 X 100 = 22.5% O = 64/142 X 100 = 45.1% 100%

  4. Ex #2: What is the percent composition by mass of the elements in Potassium permanganate? Ex #3: What is the percent composition by mass of the water in Copper (II) sulfate pentahydrate?

  5. b) Types of Formulas • Molecular Formulas • Represents the type and number of elements in a covalent compound. • The subscripts do not need to be reduced in a molecular formula. Ex:C2H4Ethene C6H12O6 Glucose

  6. 2. Empirical Formulas • Represents the lowest whole number ratio of atoms in a compound. • For ionic compounds, the formula is always empirical. This is due to the fact that one must reduce the subscripts in an ionic compound to the lowest whole number ratio. • For covalent compounds (molecules), reducing its molecular formula to an empirical formula only shows the ratio of elements within the compound. Ex: C6H12O6Molecular Formula C1H2O1Empirical Formula Ratio: C:H:O = 1:2:1

  7. c) Determining empirical formulas from percent composition • One can determine the chemical formula for a substance if given the percent composition of its elements. Ex: What is the empirical formula of a compound that consists of 58.80% barium, 13.75% sulfur, and 27.45% oxygen by mass? 1. Convert the percentages to grams. Ba 58.80% 58.80g S 13.75% 13.75g O 27.45% 27.45g

  8. 2. Convert from Mass to Moles Ba 137g = 58.80g = 0.43 mol 1 mol x mol S 32g = 13.75g = 0.43 mol 1 mol x mol O 16g = 27.45g = 1.71 mol 1 mol x mol

  9. 3. Simplify the mole ratios to the lowest whole number Ba = 0.43/0.43 = 1 S = 0.43/0.43 = 1 O = 1.71/0.43 = 4 4. The simplified ratio of elements is equal to the subscripts of the atoms within the compounds. Ba1S1O4 or BaSO4

  10. Practice: Given the following % compositions, determine the formulas for the following compounds. • 27.3 % carbon; 72.7 % oxygen • 30.43 % nitrogen; 69.57% oxygen

  11. c) 37.6% carbon; 12.5% hydrogen; 49.9% oxygen d) 32.4 % sodium; 22.5% sulfur; 45.1% oxygen

  12. c) Determining molecular formulas from percent composition KEY IDEA: The procedures for determining the molecular formula of a compound (given its elements % composition) is the same as finding its empirical formula with one exception. • YOU MUST DETERMINE THE EMPIRICAL MASS OF THE COMPOUND. • YOU MUST DIVIDE THE MOLECULAR MASS BY THE EMPIRICAL MASS TO DETERMINE THE ACTUAL NUMBER OF MOLES OF ATOMS WITHIN THE COMPOUND.

  13. Ex #1: In analyzing 30g of a compound, it was determined to consist of 80% carbon and 20% hydrogen by mass. • Determine the compound’s empirical and molecular formulas. 1. Convert percentages to grams 80% C = 80g C 20% H = 20g H 2. Convert Mass to Moles 80g C/12g per mole = 6.66 moles 20g H/1g per mole = 20 moles 3. Simplify mole ratios C = 6.66/6.66 = 1 H = 20/6.66 = 3

  14. 4. The mole ratios represent the subscripts for the empirical formula C1H3 5. Calculate the empirical mass C = 1 X 12 = 12 H = 3 X 1 = 3 Empirical Mass 15g/mole 6. Divide the molecular mass (given) by the empirical mass to find the formula unit Molecular Mass = 30 = 2 Empirical Mass 15

  15. 7. Distribute the formula unit 2 C1H3 = C2H6 C2H6 MOLECULAR FORMULA =

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