Empirical and Molecular Formulas

CH 2 O CH 3 OOCH = C 2 H 4 O 2 CH 3 O Empirical Formula A formula that gives the simplest whole-number ratio of the atoms of each element in a compound. Molecular Formula Empirical Formula H2O2H2O2H2O2H2O2HO C 6 H 12 O 6 CH 2 O

Determine the empirical formula for a compound containing g Cl and g Ca. Steps 1.Find mole amounts. 2.Divide each mole by the smallest mole.

1.Find mole amounts g Cl x 1 mol Cl = mol Cl g Cl g Cl g Ca x 1 mol Ca = mol Ca g Ca

2.Divide each mole by the smallest mole. Cl = mol Cl = 2.00 mol Cl Ca = mol Ca = 1.00 mol Ca Ratio – 1 Ca: 2 Cl Empirical Formula = CaCl 2

A compound weighing g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Hint “ Percent to mass Mass to mole Divide by small Multiply ‘ til whole ”

A compound weighing g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Percent to mass: Mg – (72.2%/100)* g = g N – (27.8%/100)* g = g Mass to mole: Mg – g * ( 1 mole ) = 8.86 mole 24.3 g N – g * ( 1 mole ) = 5.92 mole g Divide by small: Mg mole/5.92 mole = 1.50 N mole/5.92 mole = 1.00 mole Multiply ‘ til whole: Mg – 1.50 x 2 = 3.00 N – 1.00 x 2 = 2.00 Mg 3 N 2

Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound. Steps 1.Find the empirical formula. 2.Calculate the Empirical Formula Mass (EFM). 3.Divide the molar mass by the “ EFM ”. 4.Multiply empirical formula by factor. Find the molecular formula for a compound whose molar mass is ~ and empirical formula is CH 2 O “ EFM ” = g /62.03 = 2 4.2(CH 2 O 3 ) = C 2 H 4 O 6

Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. Steps 1.Find the empirical formula. 2.Calculate the Empirical Formula Mass. 3.Divide the molar mass by the “ EFM ”. 4.Multiply empirical formula by factor.

Empirical formula. A.Find mole amounts g N x 1 mol N = mol N g N 11.2 g O x 1 mol O = mol O g O

B.Divide each mole by the smallest mole. B.Divide each mole by the smallest mole. N = = 1.00 mol N O = = 2.00 mol O Empirical Formula = NO 2 Empirical Formula Mass = g/mol

Molecular formula Molar Mass = 92.0 g/mol = 2.00 Emp. Formula Mass46.01 g/mol Molecular Formula = 2 x Emp. Formula = N 2 O 4

A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula?

g C – (48.38/100)* g = g g H – (8.12/100)* g = g g O – (43.5/100)* g = g mole C g * ( 1 mole ) = mol g mole H – g * ( 1 mole ) = mol 1.01 g mole O – g * ( 1 mole ) = mol g

A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula? From last slide: mol C, mol H, mol O C – 21.29/14.27 = 1.49 H – 42.49/14.27 = 2.98 (esentially 3) O – 14.27/14.27 = 1.00 C – 1.49 x 2 = 3 H – 3 x 2 = 6 O – 1 x 2 = 2 C3H6O2C3H6O2

A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula? From last slide: Empirical formula = C 3 H 6 O 2 “ EFM ” = Molar mass = = ~3 EFM (C 3 H 6 O 2 ) = C 9 H 18 O 6