Transcript Empirical Formulas

Empirical Formulas
Simplest Chemical formula for a
compound
Writing Empirical Formulas
1. Drop the % sign and add g (for grams) to the
element
75% Hg and 25% Cl
75 g Hg and 25 g Cl
Why can we do this?
Writing Empirical Formulas
2. Change all gram amounts to moles using the
substances molar mass (look on periodic
table)
75 g Hg ( 1 mole ) = 0.375 mol Hg
200 g Hg
25 g Cl ( 1 mole) = 0.714 mol Cl
35 g Cl
Writing Empirical Formulas
3. Divide all mole amounts by the smallest mole
amount
75 g Hg ( 1 mole ) = 0.375 mol Hg = 1
200 g Hg
0.375
25 g Cl ( 1 mole) = 0.714 mol Cl = 1.9  2
35 g Cl
0.375
Writing Empirical Formulas
4. The answer from step 3 now will be the
subscripts for each element
75 g Hg ( 1 mole ) = 0.375 mol Hg = 1
200 g Hg
0.375
25 g Cl ( 1 mole) = 0.714 mol Cl = 1.9  2
35 g Cl
0.375
HgCl2
Mole Rounding Rules
If the mole amount ends in this decimal…
• .25 x 4
• .33 x 3
• .5 x 2
• If you multiply you must multiply to all
numbers
Example with Rounding
36.84% N and 63.16% O
36.84 g N (1 mole) =
63.16 g O (1 mole) =
Example with Rounding
36.84% N and 63.16% O
36.84 g N (1 mole) = 2.63 mol N
14 g N
63.16 g O (1 mole) = 3.95 mol O
16 g O
Example with Rounding
36.84% N and 63.16% O
36.84 g N (1 mole) = 2.63 mol N = 1
14 g N
2.63
63.16 g O (1 mole) = 3.95 mol O = 1.5
16 g O
2.63
Example with Rounding
36.84% N and 63.16% O
36.84 g N (1 mole) = 2.63 mol N = 1 x2 =2
14 g N
2.63
63.16 g O (1 mole) = 3.95 mol O = 1.5 x 2= 3
16 g O
2.63
N2O3
Molecular Formula
Based on the actual number of atoms of each
type in the compound
Empirical formula C3H8
Molecular Formula C6H16
C12H48
Calculating Molecular Formula
Molecular Formula mass = #
Empirical Formula mass
The number now needs to be multiplied to the
subscripts in the empirical formula
Molecular Formula example
A substance has a molecule formula mass of 42
amu and the empirical formula for a
compound is CH2. What is the molecule
formula for this compound
Molecular Formula Mass = 42 amu = 3
Empirical Formula Mass 14 amu
CH2  C3H6