School example: M&M’s example: Tylenol: Given the ratio of students to teachers to administratives to maintenance people, can you say how many people are in the school? M&M’s example: Given the ratio of colours only, can you say how many candies make up the bag? Tylenol: Given the ratio of each type atom in the active ingredient, can you say how many of each type of atom there are in a molecule of it?

MOLECULAR FORMULA C?H?O?

Molecular Formula Definition: A formula that tells you the exact number of atoms in one molecule of a compound.

Molecular Formula vs. Empirical Formula Empirical formula – a formula that tells you the lowest ratio of atoms in a compound Molecular formula – a formula that tells you the exact number of atoms in a molecule of a compound

Steps to determining the molecular formula Determine the empirical formula Determine the molar mass of the empirical formula Determine the molar mass of the compound Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula Multiply the value of the ratio by the subscripts in the empirical formula

EXAMPLE 1 What is the molecular formula of a compound with empirical formula CH3O and molar mass 93.12 g/mol ? Step 1 - Determine empirical formula Given - CH3O

Step 2 - Determine the molar mass of the empirical formula M CH3O = 1(12.01g/mol) + 3(1.01 g/mol) + 1(16.00 g/mol) M CH3O = 31.04 g/mol ___________________________________ Step 3 - Determine the molar mass of the compound Given – Mcompound = 93.12 g/mol

Step 4 - Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula Mcompound = 93.12 g/mol = 3 Mempirical formula 31.04 g/mol

Therefore, the molecular formula of the compound is C3H9O3 Step 5 - Multiply the value of the ratio by the subscripts in the empirical formula Molecular formula = 3(empirical formula) = 3(CH3O) = C3H9O3 Therefore, the molecular formula of the compound is C3H9O3

EXAMPLE 2 The percentage composition of a compound, as determined by a combustion analyzer, is 40.03% carbon, 6.67% hydrogen, and 53.30% oxygen. Using a mass spectrometer, the molar mass of the compound is found to be 180.18 g/mol. What is the molecular formula of the compound?

Remember this? Percent to mass, Mass to mole, Divide by the smallest, Multiply ‘til whole.

Step 1 – Empirical formula? C = 40.03% mC = 40.03 x 100g = 40.03 g 100 H = 6.67% mH = 6.67 x 100g = 6.67 g O = 53.30% mO = 53.30 x 100g = 53.30 g

nC = 40.03g C x 1mol C = 3.33 mol C 12.01 g C nH = 6.67g H x 1 mol H = 6.60 mol H 1.01 g H nO = 53.30g O x 1 mol O = 3.33 mol O 16.00 g O

Therefore, the empirical formula of the compound is CH2O nC : nH : nO = 3.33 : 6.60 : 3.33 3.33 3.33 3.33 nC : nH : nO = 1 : 1.98 : 1 nC : nH : nO = 1 : 2 : 1 Therefore, the empirical formula of the compound is CH2O

Step 2 - Determine the molar mass of the empirical formula M CH2O = 1(12.01g/mol) + 2(1.01 g/mol) + 1(16.00 g/mol) M CH2O = 30.03 g/mol ___________________________________ Step 3 - Determine the molar mass of the compound Given – Mcompound = 180.18 g/mol

Step 4 - Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula Mcompound = 180.18 g/mol = 6 Mempirical formula 30.03 g/mol

Therefore, the molecular formula of the compound is C6H12O6 Step 5 - Multiply the value of the ratio by the subscripts in the empirical formula Molecular formula = 6(empirical formula) = 6(CH2O) = C6H12O6 Therefore, the molecular formula of the compound is C6H12O6