1 / 14

Chapter 12: Chemical Quantities

Learn how to calculate mass percent of elements in compounds, identify empirical and molecular formulas, and solve empirical formula problems step-by-step. Examples provided for better understanding.

vergies
Download Presentation

Chapter 12: Chemical Quantities

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related