# Empirical Formulas n Empirical formulas smallest whole number

Empirical Formulas n Empirical formulas: smallest whole number ratio of atoms present in a substance n Molecular formula: actual number of each type of atom present in a molecule n Many molecular compounds have different empirical and molecular formulas n N 2 O 4 (molecular) NO 2 (empirical)

Empirical Formulas n The empirical formula for a molecular compound can be found by dividing all subscripts in the molecular formula by the greatest common factor. n Examples: n C 6 H 12 O 6 n C 4 H 10 n Ca 2 SO 4

Percent Composition n Empirical formulas are generally obtained by determining the percent composition of a compound: n Percent composition: n the percentage of the mass contributed by each element in a substance n % Element X = (# atoms of X)(AW) x 100% FW of compound

Percent Compositon Example: Calculate the % composition of C 2 H 4 O 2 (i. e find %C, %H, and %O). % C = (# atoms of C)(AW) x 100% FW of compound First, find the FW: FW = 2(12. 0 amu) + 4(1. 0 amu) + 2(16. 0 amu) FW = 60. 0 amu

Percent Composition n Calculate the % composition for each element. % C = 2(12. 0 amu) x 100% = 40. 0% C 60. 0 amu % H = 4(1. 0 amu) x 100% = 6. 7 % H 60. 0 amu % O = 2(16. 0 amu) x 100% = 53. 3 % O 60. 0 amu

Percent Composition Example: Calculate the % nitrogen in dinitrogen tetroxide. N 2 O 4 FW = 2(14. 0 amu) + 4(16. 0 amu) = 92. 0 amu % N = 2(14. 0 amu) x 100% = 30. 4% 92. 0 amu

Calculating Empirical Formulas n Percent composition data is commonly used to determine the empirical formula of a compound: n Four steps: n percent to mass n mass to mole n divide by smallest n multiple ‘til whole

Calculating Empirical Formulas Example: Calculate the empirical formula for a substance that contains 34. 63% C, 3. 875% H and 61. 50% O. Step 1: % to mass If you assume 100. 00 g of substance, then 34. 63 % C 34. 63 g C 3. 875 % H 3. 875 g H 61. 50 % O 61. 50 g O

Calculating Empirical Formulas Step 2: Mass to moles # moles C = 34. 63 g C x 1 mole C = 2. 883 mol C 12. 01 g C # moles H = 3. 875 g H x 1 mole H = 3. 844 mol H 1. 008 g H # moles O = 61. 50 g O x 1 mole O = 3. 844 mol O 16. 00 g O

Calculating Empirical Formulas Step 3: Divide by smallest (this gives the molar ratio of the elements) C: 2. 883 moles C = 1. 000 2. 883 moles C H: 3. 844 mol H = 1. 334 2. 883 mol C O: 3. 884 mol O = 1. 334 2. 883 mol C

Calculating Empirical Formulas Step 4: Multiply ‘til Whole (this step is necessary only when step 3 does not give whole number molar ratios) C: 2. 883 moles C = 1. 000 x 3 = 3. 000 mol C 2. 883 moles C H: 3. 844 mol H = 1. 334 x 3 = 4. 000 mol H 2. 883 mol C O: 3. 884 mol O = 1. 334 x 3 = 4. 000 mol O 2. 883 mol C

Calculating Empirical Formulas n Use these numbers to write the empirical formula: 3. 000 mol C 4. 000 mol H 4. 000 mol O C 3 H 4 O 4

Calculating Empirical Formulas Example: An iron compound contains 69. 943 % Fe and 30. 057 % O. Calculate its empirical formula. n. Step 1: % to mass n 69. 943 % Fe 69. 943 g Fe n 30. 057 % O 30. 057 g O

Calculating Empirical Formulas n Step 2: Mass to Moles mol Fe = 69. 943 g Fe x mol Fe 55. 845 g Fe mol O = 30. 057 g O x O 15. 9994 g O = 1. 2524 mol Fe mol O = 1. 8786 mol

Calculating Empirical Formulas n Step 3: Divide by smallest O: 1. 8786 mol O = 1. 5000 1. 2524 mol Fe Fe: 1. 2524 mol Fe = 1. 0000 1. 2524 mol Fe

Calculating Empirical Formulas n Step 4: Multiply ‘til whole Fe. O 1. 5 doesn’t make sense. 1. 5000 mol O x 2 = 3. 000 mol O 1. 0000 mol Fe x 2 = 2. 000 mol Fe Fe 2 O 3

Using Empirical Formulas to Find Molecular Formulas n Empirical formula for a molecule is the smallest whole number ratio of atoms in the molecule. n The subscripts in the molecular formula must be some whole number multiple of the subscripts in the empirical formula: CH 2 O C 2 H 4 O 2, C 3 H 6 O 3, C 4 H 8 O 4 X 2 X 3 X 4

Using Empirical Formulas to Find Molecular Formulas n Steps: n Find the empirical formula n Calculate the formula weight for the empirical formula. n MW number FW (empirical formula) MW and FW = the whole ratio between n Multiply the subscripts in the empirical

Using Empirical Formulas to Find Molecular Formulas Example: A certain substances has an empirical formula of CH 2 O. If its molecular weight is 180. 0 amu, what is its molecular formula? Step 1: Empirical Formula = given = CH 2 O

Using Empirical Formulas to Find Molecular Formulas Step 2: FW = 1 (12. 0 amu) + 2 (1. 0 amu) + 1 (16. 0 amu) = 30. 0 amu Step 3: Whole number ratio = 180. 0 amu = 6 30. 0 amu Step 3: Molecular formula: C(1 x 6)H(2 x 6)O(1 x 6) = C 6 H 12 O 6

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