Empirical Formula PDF

Empirical Formula Empirical Molecular Formula Formula A formula that shows the A formula that shows elements symbols and the simplest exact number of each whole-number ratio of...

Empirical Formula Empirical Molecular Formula Formula A formula that shows the A formula that shows elements symbols and the simplest exact number of each whole-number ratio of type of atom in a elements in a molecular compound compound Empirical Molecular Formula Formula A formula that shows the A formula that shows elements symbols and the simplest exact number of each whole-number ratio of type of atom in a elements in a molecular compound compound analogy..  It is possible for different compounds to have the same empirical formulas Molecular Formula C6H6 C2H2 Benzene Acetylene Empirical Formula: ?  It is possible for different compounds to have the same empirical formulas Molecular Formula C6H6 C2H2 Benzene Acetylene Empirical Formula: CH  Many compounds have molecular formulas that are the same as their empirical formulas Example:  Many compounds have molecular formulas that are the same as their empirical formulas Example: Ammonia Molecular Formula: NH3 Empirical Formula: NH3 Empirical Formula Recall: Percentage composition gives the proportion of masses of the elements in a compound The empirical formula gives the simplest ratio of the atoms or ions of each element in the compound If we know the percentage composition of a compound, we can determine the empirical formula Determining the Empirical Formula from Percent Composition: 1.Calculate the empirical formula of a compound that is 85.6% carbon and 14.4% hydrogen. Calculate the empirical formula of a compound that is 85.6% carbon and 14.4% hydrogen. Solution: Since you have the percent composition, assume 100 g of compound Mass of carbon = 85.6 g Mass of hydrogen = 14.4 g # of moles of C in 100 g sample = 85.6 g = 7.13 mol 12.01 g/mol # of moles of H in 100 g sample = 14.4 g = 14.3 mol 1.01 g/mol # of moles can be converted to lowest terms ratio of elements to obtain empirical formula C H ◊C H ◊ CH2 7.13/7.13 14.3/7.13 1.00 2.01 Check Your Solution Work backwards. Calculate the percentage composition of CH2 Mass % of C = 12.01 g/mol x 100% 14.03 g/mol = 85.6% Mass % of H = 2 x 1.01 g/mol x 100% 14.03 g/mol = 14.0 % Tips for Solving Empirical Formula Problems In the previous problem, numbers were rounded to simplify calculations because the terms were very close to whole numbers In order to get an accurate answers, try to keep the maximum number of significant digits Rounding too soon may result in an incorrect answer Tips for Solving Empirical Formula Problems 0.95 – 0.99 - may be rounded up to the nearest whole number 0.01 – 0.05 - may be rounded down to the nearest whole number Tips for Solving Empirical Formula Problems What if you have the empirical formula C HO? 1.5 3 1 Tips for Solving Empirical Formula Problems What if you have the empirical formula C HO? 1.5 3 1 C H O -CHO 1.5x2 3x2 1x2 3 6 2 A ratio that involves a decimal ending in 0.5 must be doubled If a decimal ends in 0.45 – 0.55, round the decimal so that it ends in 0.5 - then double the ratio Multiplying the subscripts by the fraction denominator gives whole numbers e.g., CH 1 1.33 C H 1x 3 1.33 x 3 C3H4 Determining the Empirical Formula from Percent Composition with Fractions: 2.The percentage composition of a fuel is 81.7% carbon and 18.3% hydrogen. Find the empirical formula of the fuel. Solution CH -C H →CH 1 2.66 1x3 2.66x3 3 8 More Practice Problems Determine the empirical formulas for the compounds with the following percentage compositions or mass. 3.52.2 % carbon, 6.15% hydrogen, and 41.7% oxygen 4.58.2 g of magnesium and 41.8 g of chlorine Solutions 3. C5H7O3 4. Mg2Cl