Empirical Formula When scientists discover a new compound they need to experiment to determine the chemical formula. They can do experiments where the percent composition of the new compound is found or they can measure how many grams of each element is present directly. Based on this data they can determines the lowest whole number ratio between particles using the chemistry counting unit=the mole. It is important to remember that empirical formulas are experimental formulas and will always be the lowest common ratio between the elements. CHO is an empirical formula. C6H12O6 is NOT an empirical formula; all of the elements can be divided by 6! Empirical Formula: The simplest ratio of the atoms present in a molecule. Circle the 6 empirical formulas in the list below: N2O2 CH4 H2O2 C6H12O6 MgSO4 NH3 H2O NO CaCO3 C12H22O11 The goal is for you to actually calculate an empirical formula when given either the % composition or the mass of each element present. Directions: Step 1: If you are given percent composition pretend the percents are grams and go to the next step. (Really- just change the % to a g for grams!!!) Step 2: Convert the mass of each element to moles of each element using the atomic masses from the periodic table. Step 3: Find the ratio of the moles of each element by dividing the number of moles of each by the smallest number of moles. (Your answers should be whole numbers and are used as the subscripts in the chemical formula) Step 4: Use the mole ratio to write the empirical formula Example: Find the empirical formula for the compound that contains 42.05 g of nitrogen and 95.95 g of oxygen. Step 1: Skip –you were given grams! (If you were given % you would just change the % sign to a g for grams=EASY!!) Step 2: N = 42.05g N ( 1 mol N ) = 3.001 mol N ( 14.01 g N) O = 95.95g O ( 1 mol O ) = 5.997 mol O ( 16.00 g O) Step 3: 3.001 is smaller then 5.997 so divide BOTH moles by 3.001 N = 42.05g N ( 1 mol N ) = 3.001 mol N / 3.001 mol = 1 ( 14.01 g N) O = 95.95g O ( 1 mol O ) = 5.997 mol O / 3.001 mol = 2 ( 16.00 g O) Step 4: Answer : NO2 Use this example to complete the worksheet and do your homework. You can also read p 174- 182 in your book for more explanation and examples. Molecular Formula An empirical formula is the lowest common ratio between the elements in a compound. The molecular formula is an ACTUAL, real formula. In order to calculate a molecular formula you need an empirical formula (which can be given or calculated by YOU) and the actual molecular mass of a compound. Directions: Step 1: Calculate the molar mass of the empirical formula (empirical mass) Step 2: Divide the actual formula mass by the empirical mass Step 3: Multiply the subscripts in the empirical formula by the answer in Step 2 Example: The empirical formula for vitamin C is C3H4O3. Experimental data indicates that the molecular mass of vitamin C is about 180g/mol. What is the molecular formula of vitamin C? Step 1: C= 3 x 12.01 = 36.03 g/mol H= 4 x 1.01 = 4.04 g/mol O= 3 x 16.00 = 48.00 g/mol Total= 88.07 g/mol Step 2 : Actual / Empirical 180 g/mol / 88.07 g/mol = 2.0 ~2 Step 3: Answer is : C6H8O6 Use this example to complete the worksheet and do your homework. You can also read p183- 185 in your text for more explanation and examples. .