Post on 06-Mar-2018
Quantitative Composition ofCompounds
Preparation for College ChemistryLuis AvilaColumbia UniversityDepartment of Chemistry
Depending upon Bonding type
Compounds
Ionic (Coulombic forces)
Molecular(Covalent bonds)
MoleculesCations Anions
Careful experimentation lead Proust to demonstrate
H2(g) + Cl2(g)2HCl(g)
H2SO4(l) + 2NaCl(s) 2HCl(g) + Na2SO4(aq)
Proportions by mass of elementsin a compound VARY OVER A CERTAIN RANGE
Proportions by mass of elements in a compound AREFIXED. VARIATIONS ARE DUE TO IMPURITIES.
Claude Berthollet:
Joseph Proust:
THE LAW OF DEFINITE PROPORTIONS (CONSTANT COMPOSITION):
“The proportions by mass of the elements in a compound ARE FIXED,and do not depend on its mode of preparation.”
Wüstite, an iron oxide whose simplest formula is FeO,
with 77.73%Fe.
All gaseous compounds OBEY THE LAW OF DEFINITEPROPORTIONS.
Certain SOLIDS are exceptions of the Law of ConstantComposition: NON STOICHIOMETRIC COMPOUNDS(BERTHOLLIDES)
Its composition truly ranges from Fe0.95O (76.8% Fe) to
Fe0.85O (74.8% Fe) depending of the method of preparation.
The composition of a compound is shown by its CHEMICALFORMULA.
C + O2 A
C + O2 B
CHEMICAL ANALYSIS:
If A is CO then B = CO2
For a FIXED mass of C the ratio of O in A and B is:
If A is CO2 then B is C2O4
Let’s take the elements C and O:
(1.000 g C and 1.333 g O)
(1.000 g C and 2.667 g O)
1.333 : 2.667 or 1: 2
We are unable to say which one is the right formula, but weknow the ratio C : O
is the QUOTIENT OF INTEGERS.
Molecules
Types of Formulas
Composition
Composition
Usually made up of nonmetal atoms
Held together by covalent bonds
Types of Formulas
Empirical
Molecular
Structural
CH3
C2H6
C
H
H
CH
H
H
H
Atomic and Formula Masses
Meaning of Atomic Masses
Masses of Individual Atoms
Formula Mass
Masses of Individual Atoms
The atomic masses of H, Cl, and Ni are
H = 1.008 amu
Cl = 35.45 amu
Ni = 58.69 amu
Therefore 1.008g H, 35.45g Cl, and 58.69g Ni all have thesame number of atoms: NA
NA = Avogadro’s number = 6.022 x 1023
Meaning of Atomic Masses
• Give relative masses of atoms based on C–12 scale• The Most common isotope of carbon is assigned an atomic mass of 12 amu.• The amu is defined as 1/12 of the mass of one neutral carbon atom
1amu = 1dalton =1
1212g 6
12Cmol 6
12C¥
1mol 612C
6.0221¥1023atoms 612C
Ê
Ë Á ˆ
¯ ˜ = 1.66054 ¥10-24 g/ atom 6
12C
Mass of H atom:
1 H atom x = 1.674 x 10–24g
Number of atoms in one gram of nickel:
1.00g Ni x = 1.026 x 1022 atoms
Masses of Individual Atoms
1.008g H6.022 x 1023 atoms
6.022 x 1023 atoms Ni58.69g Ni
Formula Mass
The formula for water is H2O. What is its molar mass?
2H = 2(1.008 g/mol) = 2.016 g/mol1O = 1(16.00 g/mol) = 16.00 g/mol
18.02 g/mol = molar mass of water
The Mole
Meaning
Molar Mass
Mole - Mass Conversions
Meaning 1 mol = 6.022 x 1023 items
1 molar mass Cl1 molar mass
H
1 molar mass
HCl
1 molar mass Cl2
1 at-gr Cl1 at-gr H1 mol HCl1 mol Cl2
35.45g Cl1.008g H36.46g HCl70.90 g Cl2
6.022 x 1023
atoms
6.022 x 1023
atoms
6.022 x 1023
molecules
6.022 x 1023
molecules
ClHHClCl2
Molar Mass
Generalizing from the previous examples, the molar mass, M,is numerically equal to the formula mass
formula mass molar massCaCl2 110.98 amu 110.98 g/mol
C6H12O6 180.18 amu 180.18 g/mol
Mole-Mass Conversions
110.98g CaCl2
1 mol CaCl2
mass = 13.2 mol CaCl2 x = 1.47 x 103g
Calculate mass in grams of 13.2 mol CaCl2
Calculate number of moles in 16.4g C6H12O6
1 mol C6H12O6
180.18g C6H12O6moles = 16.4g C6H12O6 x = 9.10 x 10-2mol
Calculating Composition
% Composition from Formula
Empirical Formula from % Composition
Molecular Formula from Empirical Formula
% Composition from Experimental Data
Mass % from Formula
Percent composition of potassium dichromate, K2Cr2O7?
molar mass K2Cr2O7 = (78.20 + 104.00 + 112.00)g/mol = 294.20g/mol
78.20294.20
%K = x 100 = 26.58%
112.00294.20
%O = x 100 = 38.07%
Note that percents must add to 100
104.00294.20
%Cr = x 100 = 35.35%
% Composition from Experimental Data
q Calculate mass of compound formed
q Divide mass of each element by total mass of compound and multiply by 100.
Aluminum chloride is formed by reacting 13.43 g aluminum with53.18 g chlorine. What is the % composition of the compound?
13.43 gAl + 53.18 gCl = 66.61 gAlCl3
13.43 gAl66.61 g AlCl3
Ê
Ë Á
ˆ
¯ ˜ ¥100 = 20.16%Al
53.18 gCl66.61 g AlCl3
Ê
Ë Á
ˆ
¯ ˜ ¥100 = 79.84%Cl
Empirical Formula from % Composition
Empirical formula of compound containing26.6% K, 35.4% Cr, 38.0% O
moles K = 26.6g x = 0.680 mol K1 mol39.10g
moles Cr = 35.4g x = 0.681 mol Cr1 mol52.00g
work with 100g sample:26.6 g K, 35.4 g Cr, 38.0 g O
Empirical Formula from % Composition
moles O = 38.0g x = 2.38 mol O1 mol16.00g
Note that 2.38 / 0.680 = 3.50 = 7 / 2
Empirical formula: K2Cr2O7
Potassium Dichromate?
Empirical Formula from Analytical Data
A sample of acetic acid (C, H, O atoms) weighing 1.000 g burnsto give 1.446 g CO2 and 0.6001 g H2O. Empirical formula?
Solution:
find mass of C in sample (from CO2)
find mass of H in sample (from H2O)
find mass of O by difference
Empirical Formula from Analytical Data
2.02g H18.02g H2O
mass H = 0.6001g H2O x = 0.0673g H
mass O = 1.00g – 0.394g – 0.067g = 0.539g O
mass C : 1.446 gCO2 ¥1mol CO2
44.01 g CO2¥
1molC1mol CO2
¥12.01gC1molC
= 0.394gC
Simplest Formula from Analytical Data
1 mol C12.01g C
moles C = 0.394g C x = 0.0328 mol C
1 mol H1.008g H
moles H = 0.0673g H x = 0.0668 mol H
1 mol O16.00g O
moles O = 0.533g O x = 0.0333 mol O
Empirical formula is CH2O
Molecular Formula from Empirical Formula
Must know molar mass
n =molar mass
mass of empirical formula= number of empirical formula units
Calculate empirical and molecular formulas of a compound that contains80%C, 20%H, and has a molar mass of 30.00 g/mol.
C : 80.0 gC ¥1 mol C atoms
12.01 gÊ
Ë Á
ˆ
¯ ˜ = 6.661 mol C
Molecular Formula from Empirical Formula
H : 20.0 gH ¥1 mol H atoms
1.008 gÊ
Ë Á
ˆ
¯ ˜ = 19.84 mol H
Divide each value by smaller number of moles
H : 19.846.661
= 2.97 ~ 3.00 C : 6.6616.661
= 1.00
Empirical Formula: CH3
n =molar mass
mass of empirical formula= 30.0g
15.01g~ 2
Molecular Formula: (CH3)2 = C2 H 6