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Chemical Quantities

Chemical Quantities. Empirical Formulas. Class Information. If you are planning on retesting the last exam you must sign up for a time slot to retest by end of day tomorrow. I have a sheet I will pass around class. (See district policy)

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Chemical Quantities

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  1. Chemical Quantities Empirical Formulas

  2. Class Information • If you are planning on retesting the last exam you must sign up for a time slot to retest by end of day tomorrow. I have a sheet I will pass around class. (See district policy) • Exam retakes must be done during Prime Time on Wednesday or Thursday of next week. • ALL ASSIGNMENTS FOR THIS CURRENT UNIT ARE DUE THE DAY OF THE EXAM. WE WILL NOT BE ACCEPTING ANY WORK AFTER THIS DATE. PLAN ACCORDINGLY.

  3. Empirical Formulas The empirical formula gives the lowest whole number ratio of the elements in a compound. Reduces the subscripts. The empirical is sometimes and sometimes not the same as the molecular or “actual” formula. For ionic compounds, the formula unit is the empirical formula. Example when NOT: The empirical formula for H2O2 is HO.

  4. Empirical Formulas The molecular formula for vitamin C is C6H8O6. What is the empirical formula? A. CHO B. C3H4O3 C. CH2O D. C2H4O2

  5. Empirical Formulas Which of the following is an empirical formula? A. C3H6 B. NaCl C. CH4 D. C4H10

  6. Calculating Empirical Formulas • (Review): empirical formula tells relative number of atoms of each element • One can calculate the empirical formula from the percent composition. • The ratio of the number of moles of each element in a compound gives the subscripts in a compound’s empirical formula. “Percent to mass, mass to mol, divide by small, multiply ‘til whole!”

  7. 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.

  8. Calculating Empirical Formulas Assume 100.00 g of para-aminobenzoicacid, find mass of each element: C: 61.31 % = 61.31 g C H: 5.14 % = 5.14 g H N: 10.21 % = 10.21 g N O: 23.33 % = 23.33 g O

  9. From mass, calculate the number of moles of each element: 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 12.01 g 1 mol 14.01 g 1 mol 16.00 g 1 mol 1.01 g Calculating Empirical Formulas

  10. 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 5.105 mol 0.7288 mol 1.458 mol 0.7288 mol Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: Empirical formula: C7H7NO2

  11. Molecular Formula from Empirical Formula • If we are given molar mass of the compound, we can determine the molecular formula • The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in the empirical formula: whole number multiple = molar mass of actual compound molar mass of empirical formula • Example: • The empirical formula of a compound is found to be NaS2O3. The molar mass of the compound is 270.4 g/mol. Determine the molecular formula. • molar mass of empirical formula = 135.1 g/mol • 270.4 g/mol • 135.1 g/mol • The molecular formula is Na2S4O6. Multiply each subscript in empirical formula by 2 to get subscripts of molecular formula = 2.001

  12. Example: Empirical and Molecular Formulas Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass. Its molar mass is 206 g/mol. Determine its empirical and molecular formulas.

  13. Example: Empirical and Molecular Formulas Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass. Its molar mass is 206 g/mol. Determine its empirical and molecular formulas. Empirical formula = C13H18O2 Molecular formula = C13H18O2

  14. Chemical Quantities Empirical Formulas

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