Empirical and Molecular Formulas

Definitions Empirical formula – the lowest whole-number ratio of the atoms of the elements in a compound. Molecular formula – chemical formula that shows the actual number of atoms in a compound.

Which is it: empirical, molecular, or possibly both?? NO N2O4 C6H12O6 Ba(OH)2 C4H8

Empirical Formulas H2O2 H2O H2O CH2O CaCl2 CaCl2 HO C6H12O6 The simplest whole-number ratio of atoms in a compound. H2O2 HO Both are divisible by 2. H2O H2O Already simplified CH2O C6H12O6 All are divisible by 6. CaCl2 Correct ionic formulas are always empirical. CaCl2

Finding an Empirical Formula Grams  mole Divide by smallest Round ‘til whole An unknown compound contains: 0.0806 grams of C 0.01353 grams of H 0.1074 grams of O What is the empirical formula of the compound?

Step 1: Change grams to moles 0.0806 g C 0.01353 g H 0.1074 g O 1 mol O 0.0806 g C 1 mol C 0.01353 g H 1 mol H 0.1074 g O 12.01 g C 1.01 g H 16.00g O 0.0068 mol C 0.01340 mol H 0.0067 mol O

Step 2: Divide by smallest mole value Step 3: Round to the whole 0.0068 mol C 0.01340 mol H 0.0067 mol O 0.0067 0.0067 0.0067 1.0149 mol C 2 mol H 1 mol O 1 mol C 2 mol H 1 mol O

1 mol C 2 mol H 1 mol O CH2O This is the empirical formula.

Let’s try another – This time, using percent composition!!

An unknown compound is known to have the following composition: 11.11% hydrogen 88.89% oxygen What is the empirical formula of the compound? In these problems we will choose to use a 100 gram sample. So, in this sample we have 11.11 grams H and 88.89 grams O.

Step 1: Change grams to moles. 11.11 g H 1 mol H 1.01 g H = 11.00 mol H 88.89 g O 1 mol O 16.00 g O = 5.56 mol O

Step 2: Divide by smallest mole value. Step 3: Round to a whole number. 5.56 mol O 11.00 mol H 5.56 5.56 2 1

The ratio of moles of hydrogen to moles of oxygen is 2:1 H2O

Going one step further… Finding molecular formulas from empirical formulas. Remember: The empirical formula is the simplest whole number ratio; whereas, the molecular formula shows the actual number of each atom. Step 1: Find molar mass of empirical formula. Step 2: Divide the molar mass of the compound by the molar mass of the empirical formula. Step 3: Multiply the empirical formula by that number.

Practice Problem CH2O is the empirical formula. The molar mass of the actual compound is 180 g/mol. What is the molecular formula?

Step 1: Find molar mass of empirical formula. (1 x 12.01) + (2 x 1.01) + (1 x 16.00) = 30. 03 g/mol Step 2: Divide the molar mass of the compound by the molar mass of the empirical formula. 180 g / 30 g = 6 Step 3: Multiply the empirical formula by that number. 6 (CH2O) = C6H12O6 The Molecular Formula

You try one… Empirical formula – CH4 The actual molecular mass is 64g. Find the molar mass of empirical formula  C + 4(H) = 16g Divide molecular mass by empirical mass  64g / 16g = 4 Multiply the empirical formula by that number  4 (CH4) C4H16  molecular formula

One more thing to mention… Hydrates vs. Anhydrates Hydrates – salts with water molecules attached BaCl2 • 2 H2O Name the salt, then use a prefix + word “hydrate”—Barium Chloride Dihydrate Anhydrates – the salt without its water BaCl2

Finding the Formulas of Hydrates --Similar to Finding an Empirical Formula 1. Find the mass of the water by subtraction. The mass of the hydrate and the anhydrate will be given. 2. Change g to mol Mass of water to mol Mass of salt to mol 3. Divide by the smallest mole value 4. Round to a whole # if necessary.

Finding the Formulas of Hydrates BaCl2 • X H2O Hydrate mass (mass with water) – 24.40 g Anhydrate mass (mass w/o water) - 20.80 g Step 1: Find the mass of the water. Mass H2O = 24.40 g – 20.80 g = 3.60 g H2O For the formula we want the ratio of moles of water to moles of anhydrate.

Step 2: Change grams to moles Grams of Salt 20.80 g BaCl2 208 g BaCl2 1 mol = 0.1 mol BaCl2 Grams of Water 3.60 g H2O 18 g H2O 1 mol = 0.2 mol H2O

BaCl2 • 2 H2O 1 2 Barium chloride dihydrate Step 3: Divide by the smallest mole value. 0.1 mol BaCl2 0.2 mol H2O 0.1 0.1 1 2 Step 4: Round to a whole # if necessary. BaCl2 • 2 H2O Barium chloride dihydrate

Calculating Percent Water in a Hydrate Calculating the percent water in a hydrate is similar to calculating percent composition: % H2O = mass of the water x 100 mass of the hydrate

Example Problem In the lab, a five gram sample of hydrous copper (II) nitrate is heated. If 3.9 grams of the anhydrous salt remains, what is the percent water in the hydrate? Mass of water: 5 g – 3.9 g = 1.1 g H2O Mass of hydrate: 5 g (1.1 g H2O / 5 g total) x 100 = 22 % H2O