Empirical and Molecular Formulas Empirical Formula What are

Empirical and Molecular Formulas

Empirical Formula • What are we talking about? ? ? • Empirical Formula represents the smallest ratio of atoms in a formula. • In other words it represents the simplest chemical formula for a particular compound.

Steps to Solve! • You need to follow these basic steps to determine your answer. • 1) Make sure you are starting in grams • 2) Convert all grams to moles (Molar Mass) • 3) Divide each by the smallest number of moles. • 4) Create whole number ratio of subscripts • 5) Multiply if necessary

Example 1 • In an unknown molecule, there are 4. 15 grams of Carbon and 1. 38 grams of Hydrogen. Determine the empirical formula for the substance! • Step 1) Make sure you are starting in grams • Step 2) Convert each mass in grams to moles.

Now What? • Step 3) Divide each by the smallest number of moles. Set up the Ratio of elements

Solution! • The 1 and the 4 represent the whole number ratio of the elements in the chemical formula. • The 1 represents that there is only one Carbon atom • The 4 represents that there are four Hydrogen atoms • The Solution: The empirical formula is CH 4

• Phenol, a general disinfectant, is 76. 57 % Carbon, 6. 43 % Hydrogen, and 17. 00 % Oxygen. Determine it’s empirical formula. • Step 1) Make sure you are starting in grams……. • How can we go from percent to grams? • Easy! Assume we have a 100 gram sample so there would be how much of each element?

• Step 2) Convert each mass in grams to moles.

Try this! • Element “A” is 78. 1% abundant and has a molar mass of 10. 81 g/mol. Element “B” is 21. 9% abundant and has a molar mass of 1. 01 g/mol. Determine the empirical formula. • • 78. 1 g A X 1 mol/10. 81 g = 7. 22 mol B • 21. 9 g B X 1 mol/1. 01 g = 21. 7 mol H • 7. 22 mol ÷ 7. 22 mol =1 • 21. 7 mol ÷ 7. 22 mol = 3. 01 • Ratio of 1: 3 • AB 3

Molecular Formula • The molecular formula is related, but different to the empirical formula. • Remember that Empirical Formula represents the lowest ratio of the atoms in a compound. • The Molecular Formula is the actual, or true ratio of the elements in the compound.

Needs and Steps! • 1. In order to solve you need the molar mass of the molecular formula for a compound in g/mol. • 2. Determine the empirical formula. • 3. Calculate the molar mass of the empirical formula • 4. Divide the molar mass of the molecular formula by the molar mass of the empirical formula. Apply ratio to all subscripts.

Example • The empirical formula for hydroquinone, a chemical used in photography, is C 3 H 3 O. The molecular weight of the compound Is 110 g/mol. Determine the molecular formula. • Step 1) Determine the molecular weight of the compound: Given at 110 g/mol.

• Step 2) Determine the empirical formula: Given C 3 H 3 O. • Step 3) Determine the molar mass of the empirical formula.

Think…I Know it’s hard… • Step 4) Think about the molar mass of the molecular formula compared to the molar mass of the empirical formula. • Mol. Formula Emp. Formula • 110 g/mol 55. 06 g/mol • Divide and get the ratio • 110 g/mol / 55. 06 g/mol = 2

So…. • That 2 represents what you will multiply all the subscripts by to determine the correct molecular formula. • C 3 H 3 O x 2 = C 6 H 6 O 2 Molecular Formula

Try this • The empirical formula for a compound is C 3 H 7. If the molecular weight is 86 g/mol, then what is the molecular formula? • 1. The molecular weight is given: – 86 g/mol • 2. The empirical formula is given: - C 3 H 7

• 3. Calculate the molar mass for the empirical formula.

Compare the two…. • The Molecular weight was 86 g/mol • The Empirical weight was 43. 1 g/mol • Divide Molecular by Empirical to determine the ratio! • 86 g/mol / 43. 1 g/mol = 2

• So 2 is the ratio, multiply all the subscripts by 2 • C 3 H 7 x 2 = C 6 H 14 is the Mol. Formula

Practice! A) Empirical Formula is S Molecular weight is 256 grams per mol B) Empirical Formula is NO 2 Molecular Weight is 46 g/mol