1 / 10

Post… Pick a Project!

Post… Pick a Project!. Molar Mass , Percent Composition, and Molecular Formula problems!. Molar Mass Conversions!!!. Find the Molar mass of Calcium Oxide. (Ca 2 O 2 ) First take the subscript of Calcium and use that the amount of mols for Calcium.

mandell
Download Presentation

Post… Pick a Project!

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. Post… Pick a Project! Molar Mass , Percent Composition, and Molecular Formula problems!

  2. Molar Mass Conversions!!! • Find the Molar mass of Calcium Oxide. (Ca2O2) • First take the subscript of Calcium and use that the amount of mols for Calcium. • Then set up a conversion with Calciums atomic mass in grams on the top and mols on the bottom. When you multiply the mols cancel out and you’re left with grams. • 2 mols Ca x 40.078gCa/1mol Ca=80.156gCa

  3. Molar Mass Conversions!!! • Now repeat the same process, except find the molar mass of oxygen. • 2 mols O x 15.999g O/1 mol O=31.998g O • Now add both the molar mass of oxygen (31.998g) and Calcium (80.156g) to get the molar mass of Calcium oxide. Remember the units are g/mol. • (31.998g O) + (80.156g Ca)=112.154 g/mol.

  4. PERCENT COMPOSITION! Find the percent composition of Lithium fluoride. (LiF) First, label and write down the Atomic mass formula for Lithium and Fluoride, and then multiply the amount of each Lithium and fluoride atom by their atomic mass. Then add the Amu’s. 1 Li x 6.941g Li= 6.941g Li 1 F x 18.998g F=+18.998g F 25.939g

  5. PERCENT COMPOSITION!!! • Then, divide each atom’s atomic mass by the sum of both atom’s amusand then multiply the quotient by 100 to get the percent composition. • (6.941gLi/25.939g) x 100 = 26.76% • (18.998gF/25.939g) x 100 =73.24% • Thus the answer is Li=26.76% and F=73.24%

  6. MOLECULAR FORMULA PROBLEMS! • These problems are very long and very hard. However, I have mastered them, so all is well. The compound has a formula mass of 820.482amu. • Find the molecular formula of 23.50 % O, 53% C, 23.50 % N. • First set up a conversion and change % to grams and divide the atom’s grams by it’s atomic mass to get mols.

  7. MOLECULAR FORMULA PROBLEMS!! • 23.50g O x 1mol/15.999g=1.469 mols O • 53 C x 1mol/12.011g = 4.413 mols C • 23.50g N x 1mol/14.007g=1.678 mols N • Then order the list from top to bottom in a left to right sequence as a ratio. Then divide each by the smallest. • 1.469 mols: 4.413 mols : 1.678 mols • _________________________________________ • 1.469 mols : 1.469 mols : 1.469 mols

  8. MOLECULAR FORMULA PROBLEMS • Your ratio should be • 6: 18 : 7. However this ratio results from multiplying the original ratio by 6 to receive whole numbers. • Thus the empirical formula is O6C18N7. • Now take the subscript of each atom and multiply the atom’s amu by it. Then add all the products together.

  9. MOLECULAR FORMULA PROBLEMS • O 6x(15.999amu)+C 18x(12.011amu)+ N 7x(14.007amu)=410.241amu. • Then divide 820.482 amu’s by 410.241amu. • 820.482amu/410.241 amu= 2 • Then multiply the empirical formula by 2. • O12C36N14 Thus is the molecular formula!!

  10. By Jonathan Horvat 1/10/11. Period 7. Folder 8

More Related