Empirical & Molecular Formula Notes Chemistry 4/8/15

Drill Calculate the percent composition of Na2SO4. HW: Finish pg. 3 (Empirical Formulas) MM=142.05 g/mol 32.37% Na, 22.58% S, 45.05% O

Objectives IWBAT Calculate the empirical formula of a compound, given percent composition or mass composition. Calculate the molecular formula of a compound, given: Empirical formula and molar mass OR Percent composition and molar mass

HW Review I’ll go over the answers—see if we need to put any on the board!

Empirical and Molecular Formulas

Empirical Formula Empirical vs. Molecular Formulas Molecular Formula – a formula that specifies the actual number of atoms of each element in one molecule of a compound. Empirical Formula – a formula with the smallest whole-number mole ratio of the elements that make up a compound.

Empirical Formula Empirical Formula May or may not be the same as the molecular formula Molecular formula is always a simple multiple of the empirical formula ex. H2O2 Empirical formula is HO Molecular formula is TWO times the empirical formula

How to calculate an empirical formula STEP 1: You will be given either masses or percent composition. STEP 2: If you are given % composition, turn it into grams by assuming a 100.0 g sample. NOTE: If you are given mass, you do not need to do this step. STEP 3: Convert the masses to the number of moles of each element.

STEP 4: Figure out the proportion of moles of each element in the compound by dividing each by the smallest number of moles. STEP 5: If step 4 resulted in whole numbers, you are done! However, if there were decimals, you will need to multiply by small, whole numbers until you have whole numbers.

An example: STEP 1: Compound is 40.05% S and 59.95% O STEP 2: I assume 100 g of the compound, so I have: 40.05 g S and 59.95 g O STEP 3: 40.05 g S•(1 mol S/32.07 g S) = 1.249 mol S 59.95 g O•(1 mol O/16.00 g O) = 3.747 mol O

Continued… STEP 4: 1.249 mol S : 3.747 mol O Divide each by 1.249 (smallest number in ratio) 1 mol S : 3 mol O STEP 5: SO3 You are done! The compound is sulfur trioxide.

A way to remember those steps: A Poem by Joel Thompson: Percent to mass Mass to mole Divide by small Multiply ‘til whole

Molecular Formula Molecular Formula – this tells us how many atoms of each type there really are in the compound. Can two substances have the same empirical formula but be different? YES! Benzene vs. acetylene: C6H6 vs. C2H2 What is their empirical formula? How is this different from ionic compounds?

Calculating Molecular Formula STEP 1: You will be given the molar mass of the compound and the empirical formula. STEP 2: Calculate the empirical mass (mass of the empirical formula). STEP 3: Divide the given molar mass by the empirical mass. You should get a small whole number. STEP 4: Multiply the subscripts of the empirical formula with the number obtained.

Molecular Formula Example STEP 1: The empirical formula is CH2O and the molar mass is 180.18 g. STEP 2: The empirical mass is 12.01g + 2(1.01g) + 16.00g = 30.03 g STEP 3: 180.18 g/ 30.03 g = 6 STEP 4: CH2O becomes C6H12O6

Empirical Formula Practice Calculate the empirical formula for a compound with a percent composition of 53.73% Fe and 46.27% S. Fe2S3

Practice The formula mass of a compound is 92 u. Analysis shows that there are 0.608 g N and 1.388 g O. What is the molecular formula? N2O4

Closure Acetylene (C2H2) is a gas that is used as a fuel for welding. Benzene (C6H6) is a liquid solvent. How are they similar? How are they different? Why are they different? Why is one a gas at room temperature and one a liquid?