Empirical and Molecular Formulas Determining Empirical Formulas l

Determining Empirical Formulas l l Definition: the lowest whole number ratio of elements in

Examples: Calculate EF of a compound with 25. 9% N and 74. 1% O.

Just remember this!! Percents to grams Grams to moles Divide by smallest Multiply ‘til

… examples continued • Given: 36. 5 g Na, 25. 4 g S, and

An oxide of aluminum is formed by the reaction of 4. 151 g of

Determining Molecular Formulas l l Molecular formulas show the number of atoms in a

Examples: l Determine the molecular formula of a compound whose EF is CH 2

…continued l A compound is 64. 9% carbon, 13. 5% hydrogen, and 21. 6%

l Ex) a compound is 54. 5% carbon, 9. 1% hydrogen, and 36. 4%

l Ex) A compound has an empirical formula of Cl. CH 2 and a

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Empirical and Molecular Formulas

Determining Empirical Formulas l l Definition: the lowest whole number ratio of elements in a compound. Steps to determine empirical formulas: 1) find moles of each element [find the moles of H 20 if compound is hydrated]. 2) divide by smallest number of moles to get a whole number. 3) multiply by a common whole number, if necessary.

Examples: Calculate EF of a compound with 25. 9% N and 74. 1% O. 1) %=assume mass mols 25. 9 g N x 1 mol N = 1. 85 mols N 14. 007 g N 74. 1 g O x 1 mol O = 4. 63 mols O 16 g O 2) N: 1. 85 = 1 1. 85 O: 4. 63 = 2. 5 1. 85 3) Multiply by common whole number N: 1 x 2 =2 O: 2. 5 x 2 = 5 = N 2 O 5

Just remember this!! Percents to grams Grams to moles Divide by smallest Multiply ‘til whole

… examples continued • Given: 36. 5 g Na, 25. 4 g S, and 38. 1 g O 36. 5 g Na x 1 mol Na = 1. 59 mols Na 22. 989 g Na 25. 4 g S x 1 mol S =. 79 mols S 32 g S 38. 1 g O x 1 mol O = 2. 4 mols O 16 g O Na: 1. 59 = 2. 79 S: . 79 = 1. 79 O: 2. 4 = 3. 79 = Na 2 SO 3

An oxide of aluminum is formed by the reaction of 4. 151 g of aluminum with 3. 692 g of oxygen. Find EF. 4. 151 g Al x 1 mol Al = 0. 1539 mol Al atoms 26. 98 g Al 3. 692 g O x 1 mol O = 0. 2308 mol O atoms 16 g O l 0. 1539 mol Al = 1. 000 mol Al atoms 0. 1539 0. 2308 mol O = 1. 500 mol O atoms 0. 1539 1. 500 O x 2 = 3. 000 = 3 O atoms 1. 000 Al x 2 = 2. 000= 2 Al atoms = Al 2 O 3

Determining Molecular Formulas l l Molecular formulas show the number of atoms in a compound. In order to determine the molecular, you must have the empirical formula first. The molecular mass of molecule will always be given. Multiple = molecular mass x empirical formula (whole number) • The molecular formula is always an integer multiple of the empirical formula.

Examples: l Determine the molecular formula of a compound whose EF is CH 2 O and molecular mass is 120 g/mol = 4 30 g/mol -Distribute that 4 throughout the empirical formula = C 4 H 8 O 4

…continued l A compound is 64. 9% carbon, 13. 5% hydrogen, and 21. 6% oxygen. Its molc mass is 74 g/mol. What is its MF? 64. 9 g C x 1 mol C = 5. 40 mols C = 4 C 12. 01 g C 13. 5 g H x 1 mol H = 13. 37 mols H = 10 H 1. 01 g H 1. 35 21. 6 g O x 1 mol O = 1. 35 mols O = 1 O 16. 0 g O 1. 35 74 g/mol = 1 74 g/mol = C 4 H 10 O

l Ex) a compound is 54. 5% carbon, 9. 1% hydrogen, and 36. 4% oxygen. It’s molc mass is 88 g/mol. What is its molecular formula? 54. 5 C x 1 mol C = 4. 54 mols C 12. 01 g C 9. 1 g H x 1 mol H = 9 mols H 1. 01 g H 36. 4 g O x 1 mol O = 2. 28 mols O 16 g O 4. 54 mols C = 2 C 2. 28 mols 9 mols H = 4 H 2. 28 mols O = 1 O 2. 28 mols = C 4 H 8 O 2

l Ex) A compound has an empirical formula of Cl. CH 2 and a molecular weight of 98. 96 g/mol. What is it’s molecular formula? Mass Cl + mass C + 2(mass H) Mass of empirical unit= 35. 45 +12. 00 + 2(2. 008) = 49. 47 g/mol 98. 96 g/mol = 2. 000 49. 47 g/mol = Cl 2 C 2 H 4