Empirical and Molecular Formulas: Calculation and Comparison, Slides of Chemistry

Download Empirical and Molecular Formulas: Calculation and Comparison and more Slides Chemistry in PDF only on Docsity! 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. CgH120¢ 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: N2Oz2 CH, H20z2 CegHi206 MgSO, NH3 H20 NO CaCO3 Ci2H22011 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 ag 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)