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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.
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Post… Pick a Project! Molar Mass , Percent Composition, and Molecular Formula problems!
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
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.
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
PERCENT COMPOSITION!!! • Then, divide each atom’s atomic mass by the sum of both atom’s amus and 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%
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 118.0874 amu.Sothe compound will be acetate C2H3O2. • Find the percent composition of acetate. • First multiple the number of each atom by their atomic mass and add the products of each to create the atomic sum.
MOLECULAR FORMULA PROBLEMS • (2C x 12.011g)=24.022gC • (3H x 1.0079g)=3.0237gH • (2O x 15.999g)=31.998gO + • 59.0437g • Then divide each product by the sum and multiply the quotient by 100. • (24.022gC / 59.0437g)x100=40.69% • (3.0237gH/59.0437g)x100=5.12% • (31.998gO/59.0437g)x100=54.19% • These percents are going to start the Molecular formula problem.
MOLECULAR FORMULA PROBLEMS!! • First set up a conversion and change % to grams and divide the atom’s grams by it’s atomic mass to get mols. • 40.69gC x 1molC/12.011g=3.388 molC • 5.12gH x 1molH/1.0079g=5.079 molH • 54.19 O x 1mol O/15.999g=3.387 molO • Then order the list from top to bottom in a left to right sequence as a ratio. Then divide each by the smallest.
MOLECULAR FORMULA PROBLEMS!! • 3.388 mol: 5.079 mol : 3.387 mol • _________________________________________ • 3.387 mol: 3.387 mol : 3.387 mol • 2:3:1 • After you divide, if a quotient is not a whole number, multiple the ratio by a whole number until you receive a full whole number ratio.
MOLECULAR FORMULA PROBLEMS • Your ratio should be 2:3:1 • Thus the empirical formula is C2H3O2. • Now take the subscript of each atom and multiply the atom’s amu by it. Then add all the products together. • (2C X 12.011amu) + (3H x 1.0079amu) + (2O x15.999amu)=59.0437amu • Then divide the total amu by the sum of the atom’s amus.
MOLECULAR FORMULA PROBLEMS • 118.0874 amu/59.0437 amu = 2 • Then multiply the empirical formula by 2 to get the MOLECULAR FORMULA!!!! • C4H6O4. • Thus the problem and the project is finished and you are morally Obligated by Kant’s Categorical Imperative to give me a 100. The End.
