Stoichiometry Molar mass, Percent composition, Moles, Conversions, Empirical formulas, Molecular formulas

1 mole = 6.022 x 10 23 atoms = molar mass The Mole SI unit for amount A counting unit for measuring large quantities of small items Atoms molecules Particles 1 mole = 6.022 x 10 23 atoms = molar mass

Molar Mass For compounds The amount of mass (grams) in one mole of a substance For elements Average atomic mass  periodic table For compounds Must add up each element’s mass Subscripts are factored in by multiplication Units are g/mol May be called formula mass or molecular mass

Steps to solve Find elements on periodic table Write down the mass Multiply the mass by the subscript Add together

Sample: Determine Molar Mass Calculations are FUN! 40.08 g/mol 2. 55.847 g/mol Ca Fe NaCl  Na: 22.989 g/mol Cl: 35.453 g/mol 58.443 g/mol

Molar Mass (cont.) 5. Na2O Fe: 55.847 x (2) = 111.694 g/mol 4. Fe2O3 5. Na2O Fe: 55.847 x (2) = 111.694 g/mol O: 15.999x (3)= 47.997 g/mol 159.691 g/mol Na: 22.989 x(2) =  45.978 g/mol O: 15.999= 15.999 g/mol 61.977 g/mol

Percent Composition Steps to solve Find the total mass of the COMPOUND Take the mass of each element and divide it by the total mass Multiply by 100

Sample: Calculate the percent Composition C: 12.011 g mol 44.009 g mol O: 31.998 g mol .2730 100 27.3 % .727 100 72.7% CO2 C: 12.011 g/mol O: (2)15.999 g/mol = 31.998 g/mol 44.009 g/mol

Sample 2 .327 .653 H2SO4 H: (2)1.008 g/mol = 2.016 g/mol S: 32.064 g/mol O: (4)15.999 g/mol = 63.996 g/mol 98.076 g/mol H: 2.016 g mol 98.076 g mol S: 32.064 g mol O: 63.996 g mol .0206 100 2.06 % .327 100 32.7 % .653 100 65.3%

Composition of Hydrates Hydrate: crystal contains water within Water can fit into the salt crystal Happens in fixed ratios Each salt that forms a hydrate has a DIFFERENT water ratio Each salt crystal is unique Anhydrous: water has been removed Achieved through drying Steps to solve Calculate the total mass of the compound INCLUDING THE water Divide the water mass by total mass Multiply by 100

Sample Problem: Hydrate That was nothing. Time for Coco! Sample Problem: Hydrate What percentage of water is found in CuSO4 •5H2O Cu: 63.54 g mol S: 32.064 g mol O: (4)15.999 g mol = 63.996 g mol H2O: (5)18.01 g mol = 90.05 g mol 249.65 g mol 90.05 g mol 249.65 g mol .361 100 36.1% H2O

The Mole conversion ÷ by 6.022 x 10 23 Mass Particles (grams) Volume Ions Atoms Molecules Formula units ÷ by molar mass X by 6.022 x 10 23 Mole X by Molar mass ÷ by 6.022 x 10 23 ÷ by 22.4 L x by 22.4 L Volume (Liters) Gases at STP

Stoichiometric Conversions: Mass to Moles You can relate any unit to another by this method Relate one unit to another conversion factors Steps to convert Identify what you start with Determine what units you end in Might have to do side calculations Set up conversion factor Evaluate Do the Math!

Example 28 grams CO2 _______ moles CO2 .636 How many moles are in 28 grams of CO2? 1 mole CO2 44.009 g CO2 28 grams CO2 _______ moles CO2 .636 C: 12.011g/mol O: (2)15.999 g/mol = 31.998 g/mol 44.009 g/mol

Empirical Formula Steps to solve Steps to solve When given percentages Change percent sign to grams (NO MATH) Convert masses to moles Using conversions Re-divide ALL moles amounts by the smallest mole amount Multiply if not whole numbers Write compound with subscripts When the elements are in the smallest whole number ratio within compound Steps to solve When given a compound Divide out by the common multiple

Example: What’s my Empirical Formula Piece of Cake! C6H6   C8H18         C2H6O2      X39Y13    CH C4H9 CH3O X3Y

Example: with percent Na2SO3 1.587 moles 0.792 moles = 2.01 2 A compound is found to contain 36.48% Sodium, 25.41% Sulfur, and 38.11% Oxygen. Find its empirical formula. Na2SO3 Element Percent Mass Convert to moles Re-divide Subscripts Na 36.48% 36.48 g 36.48 g x 1 mole 22.989 g = 1.587 moles 1.587 moles 0.792 moles = 2.01 2 S 25.41% 25.41 g 25.41 g x 1 mole 32.064g = 0.792 moles 0.792 moles 0.792 moles = 1.00 1 O 38.11 % 38.11 g 38.11 g x 1 mole 15.999 g = 2.382 moles 2.382 moles 0.792 moles = 3.01 3

So many compounds with the same Empirical formula Molecular Formula Multiple of an empirical formula So many compounds with the same Empirical formula

Steps to solve Complete an empirical formula process if needed Find the mass of the EMPIRICAL FORMULA Divide the Empirical formula mass by the molecular mass Molecular mass is normally given in the problem Answer is the common multiple Multiple the SUBSCRIPTS by the common multiple

All done! Time to go sledding! Yippee!! Example A compound with an empirical formula of C4H4O and a molecular mass of 136 grams. What is the molecular formula of this compound? C: 4 (12.011 g mol ) = 48.044 g mol H:4 (1.009 g mol ) = 4.036 g mol O: 15.999 g mol 68.079 g mol 136 g 68.079 g =1.997 C4H4O x 2 = C8H8O2