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**Empirical and Molecular Formulas**

Let’s use those mole conversions!

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**Empirical and Molecular Formulas**

Empirical formula tells you the lowest ratio of atoms within a molecule Molecular formula tells you the actual ratio of atoms within a molecule The empirical formula CAN equal the molecular formula, but it doesn’t have to.

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**Empirical and Molecular Formulas**

Example: A compound contains 39.99% carbon, 6.73% hydrogen, and 53.28% oxygen. The molar mass is g/mol. What is the empirical and molecular formulas for this compound?

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**Empirical and Molecular Formulas**

Step 1: Convert grams to moles (If the problem gives you percents, assume that you have 100g of sample and use the percents as grams) C: 𝑔 1 × 1 𝑚𝑜𝑙 12.01𝑔 =3.33 𝑚𝑜𝑙 H: 6.73𝑔 1 × 1 𝑚𝑜𝑙 1.01𝑔 =6.66 𝑚𝑜𝑙 O: 𝑔 1 × 1 𝑚𝑜𝑙 16.00𝑔 =3.33 𝑚𝑜𝑙

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**Empirical and Molecular Formulas**

Step 2: Divide all of the moles from Step 1 by the lowest number of moles from Step 1. C: 𝑔 1 × 1 𝑚𝑜𝑙 12.01𝑔 =3.33 𝑚𝑜𝑙 H: 6.73𝑔 1 × 1 𝑚𝑜𝑙 1.01𝑔 =6.66 𝑚𝑜𝑙 Lowest # O: 𝑔 1 × 1 𝑚𝑜𝑙 16.00𝑔 =3.33 𝑚𝑜𝑙

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**Empirical and Molecular Formulas**

Step 2: Divide all of the moles from Step 1 by the lowest number of moles from Step 1. C: 𝑔 1 × 1 𝑚𝑜𝑙 12.01𝑔 = 3.33 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 = 1 These numbers tell you how many of each atom you have in the molecule. H: 6.73𝑔 1 × 1 𝑚𝑜𝑙 1.01𝑔 = 6.66 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 =2 O: 𝑔 1 × 1 𝑚𝑜𝑙 16.00𝑔 = 3.33 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 =1

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**Empirical and Molecular Formulas**

Sometimes you may end up with non-whole numbers in Step 2. If this happens: multiply all of the moles by 2 if any end in .5 multiply all of the moles by 3 if any end in .3

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**Empirical and Molecular Formulas**

Step 3: Write the formula using the numbers found in Step 2. This is the empirical formula: C: 𝑔 1 × 1 𝑚𝑜𝑙 12.01𝑔 = 3.33 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 = 1 CH2O H: 6.73𝑔 1 × 1 𝑚𝑜𝑙 1.01𝑔 = 6.66 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 =2 O: 𝑔 1 × 1 𝑚𝑜𝑙 16.00𝑔 = 3.33 𝑚𝑜𝑙 3.33 𝑚𝑜𝑙 =1

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**Empirical and Molecular Formulas**

If the problems asks for molecular formula, move on to Step 4: Find the molar mass of the empirical formula and divide the GIVEN molar mass by this number. Molar mass of empirical formula: CH2O 12.01g + 2*1.01g g = g/mol GIVEN molar mass = g/mol 𝑔/𝑚𝑜𝑙 30.03𝑔/𝑚𝑜𝑙 =6 multiplier

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**Empirical and Molecular Formulas**

Step 5: Multiply the number of atoms in the empirical formula by the multiplier. (CH2O) * 6 = C6H12O6 𝑔/𝑚𝑜𝑙 30.03𝑔/𝑚𝑜𝑙 =6 multiplier This is the molecular formula.

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