Quantitative Composition of Compounds

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 ARE FIXED. 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 DEFINITE PROPORTIONS.

Certain SOLIDS are exceptions of the Law of Constant Composition: 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 CHEMICAL FORMULA.

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 we know 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 the same 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=112

12g612C

mol612C

×1mol6

12C6.0221×1023atoms6

12C

⎛

⎝ ⎜ ⎞

⎠ ⎟=1.66054 ×10−24 g/ atom6

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(1.008g/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

Cl2 HCl H Cl

6.022 x 1023

molecules

6.022 x 1023

molecules

6.022 x 1023

atoms

6.022 x 1023

atoms

70.90 g Cl2 36.46g HCl 1.008g H 35.45g Cl

1 mol Cl2 1 mol HCl 1 at-gr H 1 at-gr Cl

1 molar mass Cl2 1 molar mass

HCl

1 molar mass

H

1 molar mass Cl

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

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 chlorine. What is the % composition of the compound?

13.43 gAl +53.18 gCl=66.61 gAlCl3

13.43 gAl

66.61 g AlCl 3

⎛

⎝ ⎜

⎞

⎠ ⎟×100 =20.16% Al

53.18 gCl66.61 g AlCl3

⎛

⎝ ⎜

⎞

⎠ ⎟×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 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

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 g

⎛ ⎝ ⎜ ⎞

⎠ ⎟ =6.661 mol C

Molecular Formula from Empirical Formula

H : 20.0 gH×1 mol H atoms

1.008 g

⎛ ⎝ ⎜ ⎞

⎠ ⎟ =19.84 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 H 6