Transcript Notes - Empirical and Molecular Formula

CHEMICAL FORMULAE
A way of expressing the number of atoms of a compound
Write the chemical formula of the following:
1.
2.
3.
4.
Sodium chloride
Lithium oxide
Iron (II) bromide
Ammonium sulphide
____________
____________
____________
____________
CHEMICAL FORMULAE
A way of expressing the number of atoms of a compound
Write the chemical formula of the following:
1.
2.
3.
4.
Sodium chloride
Lithium oxide
Iron (II) bromide
Ammonium sulphide
NaCl
Li2O
FeBr2
(NH4)2S
TYPES OF CHEMICAL FORMULAE
• Molecular Formula
• Identifies the number of each type of atom
• Example: Hexane: C6H14
• Empirical Formula
• Simplest whole number ratio of atoms of each element
• Example: Hexane: C3H7
• Structural Formula
• Shows the structure of the molecule
FIND THE EMPIRICAL FORMULA
Substance
Molecular Formula
Water
H2O
Methane
CH4
Benzene
C6H6
Sulfur
S8
Glucose
C6H12O6
Empirical Formula
FIND THE EMPIRICAL FORMULA
Substance
Molecular Formula
Empirical Formula
Water
H2O
H2O
Methane
CH4
CH4
Benzene
C6H6
CH
Sulfur
S8
S
Glucose
C6H12O6
CH2O
FIND THE MOLECULAR AND EMPIRICAL FORMULA’S
EXAMPLE 1:
A sample of compound contains 5.723 g Ag, 0.852 g S, and 1.695
g O. Determine its empirical formula.
EXAMPLE 1:
A sample of compound contains 5.723 g Ag, 0.852 g S, and 1.695
g O. Determine its empirical formula.
mol Ag = 5.723 g x
1 mol
= 0.05305 mol
107.87g
EXAMPLE 1:
A sample of compound contains 5.723 g Ag, 0.852 g S, and 1.695
g O. Determine its empirical formula.
1 mol
= 0.05305 mol
107.87g
1 mol
mol S = 0.852 g x
= 0.02657 mol
32.07g
1 mol
mol O = 1.695 g x
= 0.10594 mol
16.00g
mol Ag = 5.723 g x
EXAMPLE 1:
A sample of compound contains 5.723 g Ag, 0.852 g S, and 1.695
g O. Determine its empirical formula.
1 mol
0.05305 mol
mol Ag = 5.723 g x
= 0.05305 mol ;
»2
107.87g
0.02657 mol
1 mol
0.02657 mol
mol S = 0.852 g x
= 0.02657 mol ;
»1
32.07g
0.02657 mol
mol O = 1.695 g x
1 mol
0.10594 mol
= 0.10594 mol ;
»4
16.00g
0.02657 mol
EXAMPLE 1:
A sample of compound contains 5.723 g Ag, 0.852 g S, and 1.695
g O. Determine its empirical formula.
1 mol
0.05305 mol
mol Ag = 5.723 g x
= 0.05305 mol ;
»2
107.87g
0.02657 mol
1 mol
0.02657 mol
mol S = 0.852 g x
= 0.02657 mol ;
»1
32.07g
0.02657 mol
mol O = 1.695 g x
1 mol
0.10594 mol
= 0.10594 mol ;
»4
16.00g
0.02657 mol
Ans: Ag2SO4
EXAMPLE 2:
A sample of compound contains 18.7% Li, 16.3% C and 65.0% O
by mass. Determine its empirical formula.
EXAMPLE 2:
A sample of compound contains 18.7% Li, 16.3% C and 65.0% O
by mass. Determine its empirical formula.
1 mol
2.6945 mol
mol Li = 18.7g x
= 2.6945 mol ;
»2
6.94g
1.3572 mol
1 mol
1.3572 mol
mol C = 16.3 g x
= 1.3572 mol ;
»1
12.01g
1.3572 mol
1 mol
4.0938 mol
mol O = 65.5 g x
= 4.0938 mol ;
»3
16.00g
1.3572 mol
Ans: Li2CO3
MOLECULAR FORMULA
To find the Molecular Formula we use the following relationship
MolarMass
N=
EmpircalMass
Where Molar Mass is the mass of the Molecular Formula
Where N = the whole number multiple by which the empirical
mass is increased
EXAMPLE 3:
A compound has an empirical formula of NH2 and a molar mass of
32.1 g/mol.
List 3 possible molecular formulae.
What is the compound’s molecular formula?
EXAMPLE 3:
A compound has an empirical formula of NH2 and a molar mass of
32.1 g/mol.
List 3 possible molecular formulae.
NH2, N2H4, N3H6
What is the compound’s molecular formula?
N 2 H4
EXAMPLE 4:
A compound containing 40.0% carbon, 6.7% hydrogen and 53.3%
oxygen has an molar mass of 180.2 g/mol. What is the molecular
formula?
EXAMPLE 4:
A compound containing 40.0% carbon, 6.7% hydrogen and 53.3%
oxygen has an molar mass of 180.2 g/mol. What is the molecular
formula?
1 mol
6.633 mol
mol H = 6.7 g x
= 6.633 mol ;
»2
1.01g
3.3306 mol
1 mol
3.3306 mol
mol C = 40.0g x
= 3.3306 mol ;
»1
12.01g
3.3306 mol
1 mol
3.33313
mol O = 53.3g x
= 3.33313 mol ;
»1
16.00g
3.3306 mol
EXAMPLE 4:
A compound containing 40.0% carbon, 6.7% hydrogen and 53.3%
oxygen has an molar mass of 180.2 g/mol. What is the molecular
formula?
MolarMass
180.2g / mol
N=
=
= 6!
EmpircalMass 30.02g / mol
EXAMPLE 4:
A compound containing 40.0% carbon, 6.7% hydrogen and 53.3%
oxygen has an molar mass of 180.2 g/mol. What is the molecular
formula?
Molecular Formula: C6H12O6
IN-CLASS WORK
Download and complete the Empirical and Molecular
worksheet from my Website
AND…
Hebden Questions: p. 91-95 Q’s 46 a,b,c,m,n 47-52 and 54