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Published byTomas Pay Modified over 2 years ago

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What Could It Be?

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Empirical Formulas The empirical formula is the simplest whole number ratio of the atoms of each element in a compound. Note: it is not necessarily the true formula of the compound. Example: The molecular formula for glucose is C 6 H 12 O 6, but its empirical formula is CH 2 O It’s kind of like a fraction reduced to its lowest terms.

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Molecular Formulas The molecular formula gives the actual numbers of each element, and thereby represents the true formula of the compound.

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Calculating the Empirical Formula Example 1: A compound is found to contain the following… g copper g oxygen Calculate its empirical formula.

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Calculating the Empirical Formula Step 1: Convert the masses to moles. Copper: Oxygen:

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Calculating the Empirical Formula Step 2: Divide all the moles by the smallest value. This gives the “mole ratio”

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Calculating the Empirical Formula Step 3: Round off these numbers, they become the subscripts for the elements. Cu 2 O

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Example 2: A material is found to be composed of 38.7% carbon, 51.6% oxygen, and 9.7% hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol Calculate the empirical and molecular formulas. If you assume a sample weight of 100 grams, then the percents are really grams.

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Example 2: A material is found to be composed of 38.7% carbon, 51.6% oxygen, and 9.7% hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol Calculate the empirical and molecular formulas. Carbon: Oxygen: Hydrogen: Now, divide all the moles by the smallest one, 3.22

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Example 2: A material is found to be composed of 38.7% carbon, 51.6% oxygen, and 9.7% hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol Calculate the empirical and molecular formulas. So, the empirical formula must be CH 3 O The molecular weight of the empirical formula is…. C x 1 = g/mol H 1.01 x 3 = 3.03 g/mol O x 1 = g/mol g CH 3 O/mol CH 3 O

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Remember, the empirical formula is not necessarily the molecular formula! MW of the empirical formula = MW of the molecular formula = 62.0

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CH 3 O 2 x ( ) = C 2 H 6 O 2 Empirical Formula g/mol Molecular Formula 62.0 g/mol

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Remember, the molecular formula represents the actual formula.

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What if the mole ratios don’t come out even?

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Example 3: A compound is analyzed and found to contain 2.42 g aluminum and 2.15 g oxygen. Calculate its empirical formula. Aluminum: Oxygen: X 2 = 2 X 2 = 3 Al 2 O 3

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