Transcript Empirical_and_Molecular_Formulas

UNIT 6: CHEMICAL
QUANTITIES
Chapter 10: Empirical and Molecular Formulas
Empirical Formulas

Empirical Formulas
 The lowest whole-number ratio of atoms of
the elements in a compound.
 Empirical
Formulas are much like creating a
recipe of elements.
Juice  1 can of concentrate to 3 cans
of water = (OJ)1(H2O)3 = 1:3 ratio
 Double Batch
 (OJ)2(H2O)6 = 2:6 = 1:3 ratio
 Orange
Empirical Formulas

Steps in determining Empirical Formulas :
1.

2.
3.
4.
5.
Assume that the percent given is equal to the number
of grams in a 100 gram sample.
Meaning…remove the % and add g for the units.
Convert the grams to moles using the molar mass.
Divide each mole value by the smallest mole calculated
to determine a ratio.
If all the values are not near a whole number (_.8-_.2
range), multiply ALL the values by 2 or 3 until the
values are all whole numbers.
Give the Empirical Formula
Empirical Formulas

Practice with Steps
 A compound is analyzed and found to contain
23.9 % nitrogen and 74.1 % oxygen. Find the
empirical formula of the compound.
Step 1: Assume percents are grams
 Nitrogen = 23.9% = 23.9g
 Oxygen = 74.1% = 74.1g
Empirical Formulas

Practice with Steps
 A compound is analyzed and found to contain
23.9 % nitrogen and 74.1 % oxygen. Find the
empirical formula of the compound.
Step 2: Convert to moles using molar mass

Nitrogen = 23.9 g 1 mol =1.85 mol Nitrogen
14 g

Oxygen = 74.1 g 1 mol = 4.63 mol Oxygen
16 g
Empirical Formulas

Practice with Steps
 A compound is analyzed and found to contain
23.9 % nitrogen and 74.1 % oxygen. Find the
empirical formula of the compound.
Step 3: Divide each mole by the smallest
mole value
 Nitrogen
= 1.85 mol ÷ 1.85 mol = 1mol of N
 Oxygen = 4.63 mol ÷ 1.85 mol = 2.5 mol of O
Empirical Formulas

Practice with Steps
 A compound is analyzed and found to contain
23.9 % nitrogen and 74.1 % oxygen. Find the
empirical formula of the compound.
Step 4: Multiply the moles by a coefficient to
have whole number ratios.
 Nitrogen
= 1 mol x 2 = 2 mol of N
 Oxygen = 2.5 mol x 2 = 5 mol of O
Step
5: Give Empirical Formula
 N2O5
Molecular Formulas

Molecular Formulas
 Based
on Empirical Formulas
 Most molecular formulas are empirical formulas, but
some molecular formulas are not the lowest ratios.
 Example: Hydrogen Peroxide
 Empirical Formula  HO
 Molecular Formula  H2O2

Requires two things
 Mass
of the Compound
 Empirical Formula of the Compound
Molecular Formulas

Practice
 Calculate the molecular formula of a compound
whose mass is 60.0 g and empirical formula is
CH4N.
Find
the molar mass of the empirical formula
CH4 N = 30.0 g/mol
Molecular Formulas

Practice
 Calculate the molecular formula of a compound
whose mass is 60.0 g and empirical formula is
CH4N.
Divide
the molecular mass by the empirical mass
Molecular mass/ empirical mass
(60.0 / 30.0 ) = 2
Molecular Formulas

Practice
 Calculate the molecular formula of a compound
whose mass is 60.0 g and empirical formula is
CH4N.
Multiply
the empirical formula subscripts by the
coefficient found in step 2.
C (1x2) H (4x2) N (1x2) = C2H8N2