1. CALCULATING EMPIRICAL FORMULA FROM PERCENT GIVEN COMPOSITION 1.OBJECTIVE: TO MASTER THE PROCESS OF CALCULATING THE EMPIRICAL FORMULA OF A COMPOUND IF YOU ARE GIVEN THE % BY MASS COMPOSITION. 2.PERCENTAGE COMPOSITION IS AN INTENSIVE PROPERTY, IT IS CONSTANT EVEN AS THE SIZE (MASS) OF YOUR SAMPLES VARY IN SIZE. 3.TO MAKE THE MATH EASY, YOU CAN ASSUME YOU HAVE A 100 GRAM SAMPLE. 4.IF YOU ASSUME A 100 GRAM SAMPLE, YOUR PERCENTAGE MASS COMPOSITION FOR EACH ELEMENT BECOMES GRAMS. FOR EXAMPLE 30.0% BECOMES 30.0 GRAMS.

2. CALCULATING EMPIRICAL FORMULA FROM PERCENT GIVEN COMPOSITION THE STEPS OF THE PROCEDURE ARE: ASSUME A 100 GRAM SAMPLE, WHEN YOU ARE GIVEN % COMPOSITION. EXPRESS THE % COMPOSITION OF EACH ELEMENT AS MASS IN GRAMS. CONVERT THE MASS OF EACH ELEMENT TO MOLES. DIVIDE EACH ELEMENT BY THE SMALLEST NUMBER OF MOLES, THIS SHOULD GIVE A RATIO OF INTEGERS IN MOST REDUCED FORM. WRITE THE INTEGER RATIO AS SUBSCRIPTS FOR THE APPROPRIATE ELEMENT.

3. EXAMPLE: A SUBSTANCE IS FOUND TO HAVE 40.0% C, 6.71% H AND 53.29 % O STEP 1: assume a 100 gram sample. STEP 2: express the % of each element as grams. STEP 3: follow the chart below. ELEMENTMASSMOLES (raw ratio) REDUCE (divide by smallest) ROUNDED RATIO C 40.0 g H 6.71 g O 53.29 g

4. STEP 1: assume a 100 gram sample. STEP 2: express the % of each element as grams. STEP 3: follow the chart below. Mol = MASS/MOLAR MASS Mol C = 40.0 g = 3.330 mol C 12.01g/mol ELEMENTMASSMOLES (raw ratio) REDUCE (divide by smallest) ROUNDED RATIO C 40.0 g 3.33055 H 6.71 g 6.71 O 53.29 g3.332708

5. STEP 1: assume a 100 gram sample. STEP 2: express the % of each element as grams. STEP 3: follow the chart below. ELEMENTMASSMOLES (raw ratio) REDUCE (divide by smallest) ROUNDED RATIO C 40.0 g 3.33055 1 1 H 6.71 g 6.71 6.71/3.33055 = 2.01 2 O 53.29 g3.332708 3.332708/3.33055 =1.00064 1 The final mole ratio is CH 2 O

6. 1)THE MOLAR MASS OF THE COMPOUND IS GIVEN IN THE QUESTION AS 60.05 G/MOL, THIS WILL BE AN INTEGER MULTIPLE OF THE EMPIRICAL MASS (THE MASS OF THE EMPIRICAL FORMULA). 2) THE MASS OF THE EMPIRICAL FORMULA (CH 2 O) IS 30.0 G/MOL. 3) TO FIND WHAT MULTIPLE TO MULTIPLY THE SUBSCRITS OF THE EMPIRICAL FORMULA, DIVIDE THE MOLECULAR MASS OVER THE EMPIRICAL MASS. MOLECULAR MASS 60.05 = 2 THEREFORE THE MOLECULAR EMPIRICAL MASS 30.00 FORMULA IS C 2 H 4 O 2