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Finding Empirical Formulas

Unit 1-9 Chapter 3.5. Finding Empirical Formulas. Formulas. Empirical formula : the lowest whole number ratio of atoms in a compound. molecular formula = (empirical formula) n [ n = integer] molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH.

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Finding Empirical Formulas

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  1. Unit 1-9 Chapter 3.5 Finding Empirical Formulas

  2. Formulas Empirical formula: the lowest whole number ratio of atoms in a compound. • molecular formula = (empirical formula)n [n = integer] • molecular formula = C6H6 = (CH)6 • empirical formula = CH Molecular formula: the true number of atoms of each element in the formula of a compound.

  3. Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

  4. Formulas (continued) Formulas for molecular compoundsMIGHTbe empirical (lowest whole number ratio). Molecular: C6H12O6 H2O C12H22O11 Empirical: H2O CH2O C12H22O11

  5. Calculating Empirical Formulas by mass % One can calculate the empirical formula from the percent composition: • Base calculation on 100 grams of compound. • Determine moles of each element in 100 grams of compound. • Divide each value of moles by the smallest of the values. • Multiply each number by an integer to obtain all whole numbers.

  6. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

  7. Assuming 100.00 g of para-aminobenzoic acid, C: 61.31 g x = 5.105 mol C H: 5.14 g x = 5.09 mol H N: 10.21 g x = 0.7288 mol N O: 23.33 g x = 1.456 mol O 1 mol 16.00 g 1 mol 14.01 g 1 mol 1.01 g 1 mol 12.01 g Calculating Empirical Formulas

  8. Calculate the mole ratio by dividing by the smallest number of moles: C: = 7.005  7 H: = 6.984  7 N: = 1.000 O: = 2.001  2 5.09 mol 0.7288 mol 0.7288 mol 0.7288 mol 1.458 mol 0.7288 mol 5.105 mol 0.7288 mol Calculating Empirical Formulas

  9. Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2

  10. Combustion Analysis (finding mass %) • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this • C is determined from the mass of CO2 produced • H is determined from the mass of H2O produced • O is determined by difference after the C and H have been determined (law of conservation of mass)

  11. Elemental Analyses • Compounds containing other elements are analyzed using methods analogous to those used for C, H and O • Example: Ag2O

  12. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  13. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 2. Divide the molecular mass by the mass given by the emipirical formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  14. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C3H5O2) x 2 = C6H10O4

  15. Homework • Page 98: 3.45, 47, 49, 51, 53, 55

  16. Chapter 3.6 Stoichiometric Calculations

  17. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products

  18. Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

  19. Stoichiometric Calculations C6H12O6 + 6 O2 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams

  20. Homework • Page 99: 57, 59, 61, 63

  21. Chapter 3.7 Limiting Reactants and yield

  22. How Many Cookies Can I Make? • You can make cookies until you run out of one of the ingredients • Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

  23. How Many Cookies Can I Make? • In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

  24. Limiting Reactants/reagents • The limiting reactant is the reactant present in the smallest stoichiometric amount • In other words, it’s the reactant you’ll run out of first (in this case, the H2)

  25. Limiting Reactants In the example below, the O2 would be the excess reactant/reagent

  26. Theoretical Yield • The theoretical yield is the amount of product that can be made • In other words it’s the amount of product possible as calculated through the stoichiometry problem • This is different from the actual yield, the amount one actually produces and measures

  27. Actual Yield Theoretical Yield Percent Yield = x 100 Percent Yield A comparison of the amount actually obtained to the amount it was possible to make

  28. Homework • Page 100: 67, 75, 77, 79

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