4 Empirical formula

Empirical Formula : 

Empirical Formula Lots of easy math, with lots of rules

Empirical formula : 

Empirical formula The empirical formula is the molecular formula reduce to it’s lowest common denominator Example: Both ethylene (C2H4) and cyclopropane (C4H8) have the same empirical formula: CH2 Ethylene Cyclobutane

% composition by mass : 

% composition by mass We’ll look at H2SO4 H = 1 x 2  2 S = 32 x 1  32 O = 16 x 4  64 TOTAL = 98 g/mol Now find the % H = 2/98  2.0% S = 32/98  32.7% O = 64/98  65.3% The % composition simply is The mass of the individual component The total molecular mass

Empirical formula : 

Empirical formula Quickly find the % composition of both C2H4 and C4H8 What do we notice? Ethylene Cyclobutane

Now find the % composition of these : 

Now find the % composition of these Cyclopropane C3H6 Cyclopentane C5H10 Hint: You can do this in your head

Molecular vs. Empirical formula : 

Molecular vs. Empirical formula Many different molecules can have the same empirical formula! The molecular formula is more specific The molecular formula of benzene is C6H6 What is its empirical formula? Let’s try C6H12O3 This one’s a little trickier

The trick : 

The trick C6H12O3 What is the greatest common factor between 3, 6, and 12 Simply divide all the subscripts by 3 C2H4O is the empirical formula

Going from % composition to empirical formula : 

Going from % composition to empirical formula Determine the empirical formula of a molecule that has is 79.85 % C; 20.15 % H 1) Assume a 100 gram sample 79.85 g carbon 20.15 g hydrogen 2) Convert both into moles 79.85 g C  6.65 mole carbon 20.15 g H  19.95 mole hydrogen

Going from % composition to empirical formula : 

Going from % composition to empirical formula 2) Convert both into moles 79.85 g C  6.65 mole carbon 20.14 g H  19.95 mole hydrogen 3) Divide all mole values by the lowest 6.65 mol C/ 6.65 = 1 mol C 19.95 mol H/ 6.65 = 3 mol H The ratio is 1:3  CH is the empirical formula

Going from % composition to empirical formula with a trick : 

Going from % composition to empirical formula with a trick Determine the empirical formula of a molecule that is 56.34 % 0; 43.66 % P 1) Assume a 100 gram sample 56.34 g O 43.66 g P 2) Convert both into moles 56.34g / 16.0g/mol O  3.52 mole oxygen 43.66g / 31g/mol P  1.41 mole phosphorus

Going from % composition to empirical formula : 

Going from % composition to empirical formula 2) Convert both into moles 56.34g / 16.0g/mol O  3.52 mole oxygen 43.66g / 31g/mol P  1.41 mole phosphorus 3) Divide all mole values by the lowest 3.52 mol O/ 1.41 = 2.5 mol O 1.41 mol P/ 1.41 = 1 mol P Since 2.5 isn’t a whole number you must multiply everything by 2 to make it work

Try these in your group : 

Try these in your group 36.48 % Na; 25.44 % S; 38.08 % O 49.99 % C; 5.61 % H; 44.40 % O 38.76 % Ca; 19.97 % P; 41.27 % O The last two have a trick at the end

The next level : 

The next level A compound analyzes as 79.08 % C; 5.54 % H and 15.38 % N.  What is the molecular formula if the molar mass is 273.36 g/ mol? Determine the empirical formula and then ask me for help.