C H What is the Empirical mass if it has 65.5% C, 5.5% H AND 29.0% O STEP 1: assume a 100 gram sample. STEP 2: express the % of each element as grams. STEP 3: follow the chart below. ELEMENT MASS MOLES (raw ratio) REDUCE (divide by smallest) ROUNDED RATIO C 65.5 g H 5.5 g O 29.0 g

C H What is the Empirical mass if it has 65.5% C, 5.5% H AND 29.0% O Mol = MASS/MOLAR MASS Mol C = 65.5 g = 5.45 mol C 12.01g/mol STEP 1: assume a 100 gram sample. STEP 2: express the % of each element as grams. STEP 3: follow the chart below. ELEMENT MASS MOLES (raw ratio) REDUCE (divide by smallest) ROUNDED RATIO C 65.5 g 5.45 mol H 5.5 g 5.44 mol O 29.0 g 1.81 mol

C H What is the Empirical mass if it has 65.5% C, 5.5% H AND 29.0% O STEP 1: assume a 100 gram sample. STEP 2: express the % of each element as grams. STEP 3: follow the chart below. ELEMENT MASS MOLES (raw ratio) REDUCE (divide by smallest) ROUNDED RATIO C 65.5 g 5.45 mol 5.45/1.81=3.011 H 5.5 g 5.44 mol 5.44/1.81=3.001 O 29.0 g 1.81 mol 1.81/1.81=1.00

What is the Empirical mass if it has 65.5% C, 5.5% H AND 29.0% O The final mole ratio is C3H3O STEP 1: assume a 100 gram sample. STEP 2: express the % of each element as grams. STEP 3: follow the chart below. ELEMENT MASS MOLES (raw ratio) REDUCE (divide by smallest) ROUNDED RATIO (moles) C 65.5 g 5.45 mol 5.45/1.81=3.011 3 H 5.5 g 5.44 mol 5.44/1.81=3.001 O 29.0 g 1.81 mol 1.81/1.81=1.00 1