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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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Empirical and Molecular Formulas

Empirical and Molecular Formulas

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