preparation for college chemistry luis avila columbia university department of chemistry...
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Preparation for College ChemistryLuis AvilaColumbia UniversityDepartment of Chemistry
Quantitative Composition of Compounds
Meaning of Atomic Masses
Atomic Mass from Isotopic Composition
Masses of Individual Atoms
Formula Mass
Atomic and Formula Masses
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=112
12g612C
mol612C
×1mol6
12C6.0221×1023atoms6
12C
⎛
⎝ ⎜ ⎞
⎠ ⎟=1.66054 ×10−24 g/ atom6
12C
Isotope Atomic Mass PercentNe-20 20.00 amu 90.92Ne-21 21.00 amu 0.26Ne-22 22.00 amu 8.82
Isotope Atomic Mass PercentNe-20 20.00 amu 90.92Ne-21 21.00 amu 0.26Ne-22 22.00 amu 8.82
A.M.=(A.M.isotope1 ×%
100+A.M.isotope2 ×
%100
+...A.M.=(A.M.isotope1 ×%
100+A.M.isotope2 ×
%100
+...
Atomic Mass from Isotopic Composition
20.00 (0.9092) +21.00 (0.0026)22.00 (0.0882)
20.18 amu
A.M. Ne = 20.18g/mol
Atomic Mass from Isotopic Composition
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 the same number of atoms: NA
NA = Avogadro’s number = 6.022 x 1023
1.008g H6.022 x 1023 atoms
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 atoms6.022 x 1023 atoms Ni
58.69g Ni
Masses of Individual Atoms
The formula for water is H2O. What is its molar mass?
2 2100g/mol 2016 g/mol1O = 1(16.00g/mol) = 16.00 g/mol
102 g/mol = molar mass of water
Formula Mass
Meaning
Molar Mass
Mole - Mass Conversions
The Mole
1 mol = 6.022 x 1023 items
Cl2
6.022 x 1023 molecules
HCl H Cl
36.46g HCl 1.008g H
1 mol HCl 1 at-gr H
35.45g Cl
36.46g HCl1 molar mass H
1 mol Cl2
70.90 g Cl2
1 at-gr H
1 molar mass HCl1 molar mass Cl2
6.022 x 1023 molecules 6.022 x 1023 atoms 6.022 x 1023 atoms
Meaning of Mole
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
Molar Mass
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
Mole to Mass Conversion
% Composition from Formula
Empirical Formula from % Composition
Molecular Formula from Empirical Formula
% Composition from Experimental Data
Formulas
Percent composition of K2CrO7?
molar mass K2CrO7 = (78.20 + 52.00 + 112.00)g/mol = 242.20g/mol
78.20242.20
%K = x 100 = 32.29%
112.00242.20
%O = x 100 = 46.24%
Note that percents must add to 100
52.00242.20
%Cr = x 100 = 21.47%
Mass % from Formula
Calculate mass of compound formed
Divide mass of each element by total mass of compound and multiply by 100.
Aluminum chloride is formed by reacting 13.43 g aluminum with 53.18 g chlorineWhat is the % composition of the compound?
13.43 gAl +53.1 gCl=66.61 gAlCl3
13.43 gAl
66.61 g AlCl 3
⎛
⎝ ⎜
⎞
⎠ ⎟×100 =20.16% Al
53.1 gCl66.61 g AlCl3
⎛
⎝ ⎜
⎞
⎠ ⎟×100 =79.4%Cl
% Composition from Exp. Data
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 %
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 %
A sample of acetic acid (C, H, O atoms) weighing 1.000 g burns to 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 ×1 mol CO244.01 g CO2
×1molC
1 mol CO2×12.01gC1molC
=0.394gC
Empirical 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
Empirical Formula from Analytical Data
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 g
⎛ ⎝ ⎜ ⎞
⎠ ⎟ =6.661 mol C
Molecular Formula from Empirical Formula
H : 20.0 gH×1 mol H atoms
1.00 g ⎛ ⎝ ⎜ ⎞
⎠ ⎟ =19.4 mol H
Divide each value by smaller number of moles
H : 19.84
6.661=2.97 ~3.00 C:
6.6616.661
=1.00
Empirical Formula: CH3
n= molar mass
mass of empirical formula=
30.0g15.01g
~2
Molecular Formula: (CH3 )2 = C2 H6