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What Could It Be? Finding Empirical and Molecular Formulas
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It’s kind of like a fraction reduced to its lowest terms.
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 C6H12O6, but its empirical formula is CH2O 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… 2.199 g copper 0.277 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. Cu2O
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Example 2: A material is found to be composed of 38. 7% carbon, 51
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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Now, divide all the moles by the smallest one, 3.22
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: Now, divide all the moles by the smallest one, 3.22 Oxygen: Hydrogen:
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Example 2: A material is found to be composed of 38. 7% carbon, 51
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 CH3O The molecular weight of the empirical formula is…. C x = g/mol H x 3 = g/mol O x 1 = g/mol 31.04 g CH3O/mol CH3O
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Remember, the empirical formula is not necessarily the molecular formula!
MW of the empirical formula = 31.04 MW of the molecular formula = 62.0
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2 x ( ) CH3O = C2H6O2 Empirical Formula 31.04 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
Example 3: A compound is analyzed and found to contain 2.42 g aluminum and 2.15 g oxygen. Calculate its empirical formula. Aluminum: X 2 = 2 Oxygen: X 2 = 3 Al2O3
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