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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