WUP#22 A compound composed of 40.68% carbon, 5.08% hydrogen and 54.24% oxygen. Determine the empirical. 40.68 g C 1 mol C = 3.387 mol C = 1.00 x 2 = 2.

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WUP#22 A compound composed of 40.68% carbon, 5.08% hydrogen and 54.24% oxygen. Determine the empirical. 40.68 g C 1 mol C = 3.387 mol C = 1.00 x 2 = 2 mol C 3.387 12.01 g C 5.08 g H 1 mol H = 5.03 mol H = 1.49 x 2 = 3 mol H 1.01 g H 3.387 54.24 g O 1 mol O = 3.390 mol O = 1.00 x 2 = 2 mol O 16.00 g O 3.387 E.F. = C2H3O2

Molecular Formula

HCHO CH2O CH2O C6H12O6 CH2O 30.03 g/mol CH3COOH 30.03 g/mol Compound Empirical Formula Molecular Formula E.F.W M.F.W Formaldehyde HCHO Acetic Acid CH3COOH Glucose C6H12O6 30.03 x 1 = 30.03 g/mol 30.03 g/mol CH2O 30.03 x 2 = 60.06 g/mol 30.03 g/mol CH2O 30.03 x 6 = 180.18 g/mol 30.03 g/mol CH2O

Acetic Acid Formaldehyde Glucose

multiple empirical formula E.F. x 6 = M.F. CH2O x 6 = C6H12O6 Molecular Formula: a formula that tells the actual number of atoms in a compound multiple It is a __________ of its ______________________. empirical formula E.F. x 6 = M.F. CH2O x 6 = C6H12O6

A compound composed of 40. 68% carbon, 5. 08% hydrogen and 54 A compound composed of 40.68% carbon, 5.08% hydrogen and 54.24% oxygen has a molar mass of 118.1 g/mol. Determine the empirical and molecular formula. 40.68 g C 1 mol C = 3.387 mol C = 1.00 x 2 = 2 mol C 3.387 12.01 g C 5.08 g H 1 mol H = 5.03 mol H = 1.49 x 2 = 3 mol H 1.01 g H 3.387 54.24 g O 1 mol O = 3.390 mol O = 1.00 x 2 = 2 mol O 16.00 g O 3.387 E.F. = C2H3O2

E. F. = C2H3O2 = 59.05 g x  = 118.1 g  = 2 So your E.F. x 2 = M.F.  = 2 Calculated molar mass of E.F. given molar mass for the compound in the original problem So your E.F. x 2 = M.F. E. F. x 2 = C2H3O2 X 2 = M.F. = C4H6O4

2) A compound has a molar mass of 462. 8 g/mol and contains 77 2) A compound has a molar mass of 462.8 g/mol and contains 77.87% C, 11.76% H, and 10.37% O. Determine the empirical and molecular formulas. 77.87g C 1 mol C =6.484 mol C 6.484 mol  10 0.6481 mol 12.01g C 11.76g H 1 mol S = 11.64 mol H 11.64 mol  18 0.6481 mol 1.01g S 10.37g O 1 mol O 0.6481 mol = 1 0.6481 mol = 0.6481 mol O 16.00g O E. F. = C10H18O

E. F. = C10H18O = 154.28g x  = 462.8 g  = 3 So your E.F. x 3 = M.F.  = 3 Calculated molar mass of E.F. given molar mass for the compound in the original problem So your E.F. x 3 = M.F. E. F. x 3 = C10H18O X 3 = M.F. = C30H54O3