Molecular Formula vs Empirical Formula Different compounds can

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Molecular Formula vs Empirical Formula

Molecular Formula vs Empirical Formula

 • Different compounds can have the same empirical formula but different molecular formulas.

• Different compounds can have the same empirical formula but different molecular formulas. • Empirical Formula is a reduced form of Molecular formula • Molecular Formula may be the same as the empirical formula but is probably different.

An empirical formula is: simplest whole-number ratio of atoms in a compound- Reduced form

An empirical formula is: simplest whole-number ratio of atoms in a compound- Reduced form A molecular formula is: The actual formula of the molecule

Therefore…. CH 2 O is an empirical formula C 6 H 12 O 6

Therefore…. CH 2 O is an empirical formula C 6 H 12 O 6 is a molecular formula

Empirical Formulas Step 1: Change % to grams (g) - if you are given

Empirical Formulas Step 1: Change % to grams (g) - if you are given grams, skip this step Step 2: Convert masses to moles using molar mass Step 3: Divide all # of moles by the one that is the smallest Step 4: If dividing gave you. 5, then multiply by 2 Step 5: If dividing gave you. 3 or. 7, then multiply by 3 Once you know the ratios, place them as subscripts in the formula

Empirical Practice #1: Find the empirical formula of a compound that is 33. 38%

Empirical Practice #1: Find the empirical formula of a compound that is 33. 38% Na, 22. 65% S, and 44. 9% O. What is the important information from this problem? ( the given)

Empirical Practice #1: Find the empirical formula of a compound that is 33. 38%

Empirical Practice #1: Find the empirical formula of a compound that is 33. 38% Na, 22. 65% S, and 44. 9% O. 33. 38 g. Na 22. 65 g S 44. 9 g O

Empirical Pracitice #2: A compound contains 3. 26 g of arsenic and 1. 04

Empirical Pracitice #2: A compound contains 3. 26 g of arsenic and 1. 04 g of oxygen. What is the empirical formula? What is the important information from this problem? ( the given)

Empirical Pracitice #2: A compound contains 3. 26 g of arsenic and 1. 04

Empirical Pracitice #2: A compound contains 3. 26 g of arsenic and 1. 04 g of oxygen. What is the empirical formula? 3. 26 g As 1. 04 g O

Empirical Pracitice #2: A compound contains 3. 26 g of arsenic and 1. 04

Empirical Pracitice #2: A compound contains 3. 26 g of arsenic and 1. 04 g of oxygen. What is the empirical formula? 3. 26 g As 1 mole 74. 92 g 1. 04 g O 1 mole 16. 00 = = 0. 0435 mole 0. 0435 0. 065 mole 0. 0435 =1 =1. 49 Step 4: If dividing gave you. 5, then multiply by 2 Step 5: If dividing gave you. 3 or. 7, then multiply by 3

Empirical Pracitice #2: A compound contains 3. 26 g of arsenic and 1. 04

Empirical Pracitice #2: A compound contains 3. 26 g of arsenic and 1. 04 g of oxygen. What is the empirical formula? 3. 26 g As 1 mole 74. 92 g 1. 04 g O 1 mole 16. 00 = = 0. 0435 mole 0. 0435 0. 065 mole 0. 0435 =1*2=2 =1. 49*2=3 As 2 O 3

What is the empirical formula of a compound that is 62. 10% C, 13.

What is the empirical formula of a compound that is 62. 10% C, 13. 80% H & 24. 10% N?

Molecular Formulas 1. To find the molecular formula you must: 2. Find the empirical

Molecular Formulas 1. To find the molecular formula you must: 2. Find the empirical formula if not given 3. Determine the molar mass of the empirical formula 4. MM molecular formula =X MM empirical formula 5. multiply each subscript in the empirical formula by “X”

Molecular Practice 4: The empirical formula of a compound is CH; the molecular molar

Molecular Practice 4: The empirical formula of a compound is CH; the molecular molar mass is 78. 11 g/mol. What is its molecular formula? What is the important information from this problem? ( the given)

Molecular Practice 4: The empirical formula of a compound is CH; the molecular molar

Molecular Practice 4: The empirical formula of a compound is CH; the molecular molar mass is 78. 11 g/mol. What is its molecular formula? Empirical: CH MMmolecular formula = 78. 11 g/mole

Molecular Practice #5: A compound has an empirical formula of CH 3 O and

Molecular Practice #5: A compound has an empirical formula of CH 3 O and a molecular mass of 62 g/mol. What is its molecular formula? What is the important information from this problem? ( the given)

Molecular Practice #5: A compound has an empirical formula of CH 3 O and

Molecular Practice #5: A compound has an empirical formula of CH 3 O and a molecular mass of 62. 00 g/mol. What is its molecular formula? Empirical: CH 30 MMmolecular formula = 62. 00 g/mole

Molecular Practice #5: A compound has an empirical formula of CH 3 O and

Molecular Practice #5: A compound has an empirical formula of CH 3 O and a molecular mass of 62. 00 g/mol. What is its molecular formula? Empirical: CH 30 MMmolecular formula = 62. 00 g/mole MMCH 3 O = 31. 04 g/mole 62. 00 31. 04 =2 C 2 H 6 O 2

A compound is 26. 70% C, 2. 2. 0%O & 71. 1% O. If

A compound is 26. 70% C, 2. 2. 0%O & 71. 1% O. If its molecular molar mass is 90. 00 g/mole. What is its molecular formula?

Summary Determine the molecular and empirical formula for each: a. C 6 H 8

Summary Determine the molecular and empirical formula for each: a. C 6 H 8 b. N 2 O 6 c. C 6 H 12 O 6 d. BCl 3

Summary Determine the molecular and empirical formula for each: a. C 6 H 8

Summary Determine the molecular and empirical formula for each: a. C 6 H 8 b. N 2 O 6 c. C 6 H 12 O 6 d. BCl 3 molecular empirical

Summary Determine the molecular and empirical formula for each: a. C 6 H 8

Summary Determine the molecular and empirical formula for each: a. C 6 H 8 b. N 2 O 6 c. C 6 H 12 O 6 d. BCl 3 molecular -C 3 H 4 (empirical) molecular –NO 3 (empirical) molecular- CH 2 O (empirical) empirical

Explain why some compounds do not have an empirical formula: ?

Explain why some compounds do not have an empirical formula: ?