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Chapter 12: Chemical Quantities

Chapter 12: Chemical Quantities. Section 12.2: Using Moles (part 3). Mass Percent. Steps: 1) Calculate mass of each element 2) Calculate total mass 3) Divide mass of element/ mass of compound. Calcualte the mass % of C and H in C 2 H 6 2 mol C x 12 g C = 24 g C

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Chapter 12: Chemical Quantities

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  1. Chapter 12: Chemical Quantities Section 12.2: Using Moles (part 3)

  2. Mass Percent Steps: 1) Calculate mass of each element 2) Calculate total mass 3) Divide mass of element/ mass of compound

  3. Calcualte the mass % of C and H in C2H6 • 2 mol C x 12 g C = 24 g C • 6 mol H x 1 g H = 6 g H Total mass = C2H6 = 30 g % C = 24 g C x 100 = 80% 30 g C2H6 % H = 6 g H x 100 = 20% 30 g C2H6 * Should add up to 100

  4. Example 2 Calculate the mass % of C, H, Br in C6H5Br • 6 mol C x 12.011 = 72.066 • 5 mol H x 1.0079 = 5.0395 • 1 mol Br x 79.904 = 79.904 = 157.0095 g % C = 72.066 g C x 100 = 45.9% 157.0095 g C2H5 Br % H = 5.0395 g H x 100 = 3.2% 157.0095 g C2H5 Br % Br = 79.904 g Br x 100 = 50.9% 157.0095 g C2H5 B

  5. Chemical Formulas • Empirical Formulas • Molecular Formulas Empirical: NaCl Molecular: K2C2O4 Empirical: KCO2 Molecular: C20H20O4 Empirical: C5H5O

  6. Empirical formula Formula that shows the smallest whole-number ratio of atoms for the compound Steps: 1) Switch % into grams 2) Convert grams to moles for each element 3) Divide by the smallest mole

  7. EMPIRICAL FORMULA A compound is ~26 % N and 74 % O. What is the empirical formula? 26 g N x 1 mol N = 1.86 mol N / 1.86 = 1 14.007 g N 74 g O x 1 mol O = 4.63 mol O / 1.86 = 2.5 15.999 g O Can’t have decimal #s (N1O2.5) 2 = N2O5

  8. What is the empirical formula of a compound that is 80% C and 20% H of 100 g compound? 80 g C x 1 mol C = 6.67 mol C 12 g C 20 g H x 1 mol H = 20 mol H 1 g H Divide both by smallest: 6.67 mol C = 1.0 mol C 6.67 20 mol H = 3.0 mol H 6.67 The mole ratio for this compound is 1 mol C: 3 mol H Formula: CH3(Empirical formula)

  9. Molecular formula To determine the molecular formula you need the mass of the compound Steps: 1) First find the empirical formula 2) Divide the total mass by the empirical formula mass 3) Multiply the empirical formula by this #

  10. Example continued: Molar mass is 30 g/mol →molecular mass is 30u Molecular mass of CH3 = 15 u By dividing the molecular mass you can find the multiple: 30 u = 2 15 u Molecular formula is C2H6

  11. An oxide of Cr is 68.4% Cr by mass. The molar mass is 152 g/mol. What is the formula? 100-68.4 = 31.6 g O 68.4 g Cr x 1 mol Cr = 1.32/1.32 = 1 51.996 g Cr 31.6 g O x 1 mol O = 1.98/1.32 = 1.5 15.999 g O Cr1O1.5 Cr= 51.996 O= 23.9985 = 75.9945 152/ 75.99945 = 2 (Cr1O1.5) 2 = Cr2O3

  12. What is the molecular formula for C4H4O with a mass = 136 g/mol? C: 4 x 12.011 = 48.044 H: 4 x 1.0079 = 4.0316 O: 1 x 15.999 = 15.999 = 68.07 136/68.07 = 2 (C4H4O)2 = C8H8O2

  13. What % of O is NO? N:14 O:16 = 30 16/30 = 53.3 % O

  14. A nitride of Cr is 84.8 %. It has a mass of ~ 184 g. What is the formula? 100 - 84.8 = 15.2 % N 84.8 g Cr x 1 mol Cr = 1.63/1.09 = 1.5 51.996 g Cr 15.2 g N x 1 mol N = 1.09/1.09 = 1 14.007 g N Cr1.5N 77.994 + 14.007 = 92.001 184/92.001 = 1.99 (2) (Cr1.5N) 2 = Cr3N2

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