Empirical Formula Is the formula with the smallest whole-number mole ratio of the elements in a compound.

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Chapter 11 Empirical and Molecular Formulas

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Empirical Formula Is the formula with the smallest whole-number mole ratio of the elements in a compound.

Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula

Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula CH 2 O.

Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula CH 2 O. 2)Succinic acid– Molecular formula C 4 H 6 O 4. Empirical formula

Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula CH 2 O. 2)Succinic acid– Molecular formula C 4 H 6 O 4. Empirical formula C 2 H 3 O 2.

Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula CH 2 O. 2)Succinic acid– Molecular formula C 4 H 6 O 4. Empirical formula C 2 H 3 O 2. 3) Sulfuric acid -- Molecular formula H 2 SO 4. Empirical formula

Examples: 1) Glucose – Molecular formula C 6 H 12 O 6. Empirical formula CH 2 O. 2)Succinic acid– Molecular formula C 4 H 6 O 4. Empirical formula C 2 H 3 O 2. 3) Sulfuric acid -- Molecular formula H 2 SO 4. Empirical formula H 2 SO 4.

8 Practice: Determine the Empirical Formula of Benzopyrene, C 20 H 12. Empirical formula (C 5 H 3 ) 4 = Find the greatest common factor (GCF) of the subscripts: 20 factors = 12 factors = 201,2,4,5,10, 1,24,3,6,12 GCF =4 Divide each subscript by the GCF to find the empirical formula: C 20/4 H 12/4 = C 5 H 3 C 20 H 12

1) What is the empirical formula for a compound which contains g of iron, g of sulfur and g of oxygen? Fe = g x S = O = g x g x = mol (O) = mol (Fe) = mol (S) 1 st – Determine the # of moles of each element:

FeSO 3 Fe = S = O = = 3 mol (O) = 1 mol (Fe) = 1 mol (S) Empirical Formula 2 nd – Calculate the simplest ratio (dividing each value by the smallest one):

2) Determine the empirical formula for a compound that contains % magnesium, % chlorine, and 57,34 % oxygen?

1 st – Determine the # of moles of each element: (Mg = % ; Cl = % & O = %) Mg = g x Cl = O = g x g x = 3.59 mol (O) = mol (Mg) = mol (Cl)

2 nd – Calculate the simplest ratio (dividing each value by the smallest one): (Mg = ; Cl = & O = 3.59) = 8 mol (O) = mol (Mg) = mol (Cl) = mol (O) = 1 mol (Mg) = 2 mol (Cl) Empirical Formula MgCl 2 O 8

3) A laboratory analysis of aspirin determined the following mass percent composition carbon (60.00 %), hydrogen (4.48 %) and oxygen (35.53 %). Calculate its empirical formula. = mol (C) = mol (O) = mol (H) O = g x C = g x H = 4.48 g x

3) A laboratory analysis of aspirin determined the following mass percent composition carbon (60.00 %), hydrogen (4.48 %) and oxygen (35.53 %). Calculate its empirical formula. C = mol H = 4.44 mol O = mol Dividing by the smallest one:

3) A laboratory analysis of aspirin determined the following mass percent composition carbon (60.00 %), hydrogen (4.48 %) and oxygen (35.53 %). Calculate its empirical formula. C = mol H = 4.44 mol O = mol C = ÷ = H = 4.44 ÷ = O = ÷ = Dividing by the smallest one:

3) A laboratory analysis of aspirin determined the following mass percent composition carbon (60.00 %), hydrogen (4.48 %) and oxygen (35.53 %). Calculate its empirical formula. C = mol H = 4.44 mol O = mol C = ÷ = H = 4.44 ÷ = 2.00 multiply by 4 O = ÷ = Dividing by the smallest one:

3) A laboratory analysis of aspirin determined the following mass percent composition carbon (60.00 %), hydrogen (4.48 %) and oxygen (35.53 %). Calculate its empirical formula. C = ÷ = x 4 = 9 H = 4.44 ÷ = 2.00 x 4 = 8 O = ÷ = x 4 = 4 C9H8O4C9H8O4 (Empirical Formula)