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Empirical and Molecular Formulas. Determining Empirical Formulas Definition: the lowest whole number ratio of elements in a compound. Steps to determine.

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Presentation on theme: "Empirical and Molecular Formulas. Determining Empirical Formulas Definition: the lowest whole number ratio of elements in a compound. Steps to determine."— Presentation transcript:

1 Empirical and Molecular Formulas

2 Determining Empirical Formulas 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 2 0 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.

3 Examples: Calculate EF of a compound with 25.9% N and 74.1% O. 1) %=assume mass mass  mols 25.9g N x 1 mol N = 1.85 mols N 14.007g N 74.1g 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

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

5 … examples continued Given: 36.5g Na, 25.4g S, and 38.1g O 36.5g Na x 1 mol Na = 1.59 mols Na 22.989g Na 25.4g S x 1 mol S =.79 mols S 32g S 38.1g O x 1 mol O = 2.4 mols O 16g O Na: 1.59 = 2 S:.79 = 1 O: 2.4 = 3.79.79.79 = Na 2 SO 3

6 An oxide of aluminum is formed by the reaction of 4.151g of aluminum with 3.692g of oxygen. Find EF. 4.151g Al x 1 mol Al = 0.1539 mol Al atoms 26.98 g Al 3.692g O x 1 mol O = 0.2308 mol O atoms 16 g O 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

7 Determining Molecular Formulas 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.

8 Examples: Determine the molecular formula of a compound whose EF is CH 2 O and molecular mass is 120 g/mol 120 g/mol = 4 30 g/mol -Distribute that 4 throughout the empirical formula = C4H8O4C4H8O4

9 …continued A compound is 64.9% carbon, 13.5% hydrogen, and 21.6% oxygen. Its molc mass is 74 g/mol. What is its MF? 64.9g C x 1 mol C = 5.40 mols C = 4 C 12.01 g C 13.5g H x 1 mol H = 13.37 mols H = 10 H 1.01 g H 1.35 21.6g 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

10 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 = 4H 2.28 mols 2.28 mols O = 1O 2.28 mols = C 4 H 8 O 2

11 Ex) A compound has an empirical formula of ClCH 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


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