Chemical Calculations: The Mole Concept and Chemical Formulas
Chapter 9: Chemical calculations: The mole concept and chemical formulas
9.1 Law of Definite Proportions
Definition: In a pure compound, elements are always present in the same definite proportion by mass.
Key Point: The percentage of each element present in a given compound does not vary.
Example: ( NH_3 ) Samples
Sample 1: 4.63
Sample 2: 4.62
Mass Before Decomposition (g):
Sample 1: 1.840
Mass Nitrogen Produced (g): 1.513
Mass Hydrogen Produced (g): 0.327
Sample 2: 2.000
Mass Nitrogen Produced (g): 1.644
Mass Hydrogen Produced (g): 0.356
Ratio Calculation:
For Sample 1: ( \frac{1.513}{0.327} )
For Sample 2: ( \frac{1.644}{0.356} )
9.2 Calculation of Formula Masses
Formula Mass: The sum of atomic masses of all the atoms present in one formula unit of a substance, expressed in atomic mass units (amu).
For molecules, this is often referred to as molecular mass.
Calculation:
Add the atomic masses for all atoms in the formula, adjusting for multiple atoms of the same element.
Example of Formula Mass Calculation of Benzoic Acid ((C7H6O_2)):
Carbon (C): 7 x 12.01 amu = 84.07 amu
Hydrogen (H): 6 x 1.01 amu = 6.06 amu
Oxygen (O): 2 x 16.00 amu = 32.00 amu
Total = 84.07 + 6.06 + 32.00 = 122.13 amu
9.3 Significant Figures and Formula Mass
Use atomic masses rounded to the hundredths place (0.01).
Exception: Lead (Pb) is rounded only to the tenths place (0.1).
Note: Calculations for significant figures in formula mass assume pure addition problems, determined by addition rules.
Practice Calculations:
( Fe2O3 ): (55.85 + 55.85) + (16.00 + 16.00 + 16.00) = 159.70 amu
( CHF_2Cl ): 12.01 + 1.01 + (19.00 + 19.00) + 35.45 = 86.47 amu
( C9H8O_4 ): (9 x 12.01) + (8 x 1.01) + (4 x 16.00) = 180.17 amu
9.4 Mass Percent Composition of a Compound
Definition: Percent composition by mass represents the percent by mass of each element in the compound.
Example Calculation for ( H_2O ):
Formula Mass = (2 x 1.01) + 16.00 = 18.02 amu
% Oxygen (O) = (16.00 / 18.02) x 100 ≈ 88.79%
% Hydrogen (H) = (2.02 / 18.02) x 100 ≈ 11.21%
Percent by Mass Calculation:
Formula: % by mass of an element = ( \frac{mass \ of \ element}{mass \ of \ compound} \times 100 )
Calculation Steps from Experimental Data:
Calculate the formula mass of the compound.
Divide the mass of each element by the formula mass and multiply by 100 to find the mass percent.
Example Calculation for Acetaminophen (( C8H9O_2N )):
Formula mass = (8 x 12.01) + (9 x 1.01) + (2 x 16.00) + 14.01 = 151.18 amu
Percent composition:
C: ( \frac{96.08}{151.18} \times 100 \approx 63.55% )
H: ( \frac{9.09}{151.18} \times 100 \approx 6.01% )
O: ( \frac{32.00}{151.18} \times 100 \approx 21.17% )
N: ( \frac{14.01}{151.18} \times 100 \approx 9.267% )
Practice Problems:
Calculate the percent by mass composition for ( C2H2 )
Calculate the percent by mass composition for ( C6H6 )
Observations on the two formulas.
Solutions for Practice:
( C2H2 ):
Formula mass = (12.01 + 12.01) + (1.01 + 1.01) = 26.04 amu
C: ( \frac{24.02}{26.04} \times 100 \approx 92.24% )
H: ( \frac{2.02}{26.04} \times 100 \approx 7.76% )
( C6H6 ):
Formula mass = (6 x 12.01) + (6 x 1.01) = 78.12 amu
C: ( \frac{72.06}{78.12} \times 100 \approx 92.24% )
H: ( \frac{6.06}{78.12} \times 100 \approx 7.76% )
9.5 The Mole: Chemist's Counting Unit
Definition: In chemistry, the mole is a counting unit used to quantify the amount of a substance in discrete units, not to be confused with animals or molecules.
Other Counting Units:
One dozen = 12 items, two dozen = 24 items, etc.
Mole Definition: 1 mole = ( 6.022 × 10^{23} ) objects (Avogadro's number).
Large Scale Counting:
The amounts of atoms or molecules in chemistry often lead to very large numbers, necessitating a large counting unit: the mole.
Concept of the Mole:
1 mole contains ( 6.022 × 10^{23} ) particles (atoms, molecules, ions).
Example of calculating objects in a molar quantity: For 1.20 moles of carbon monoxide (CO): 1.20 moles CO x ( 6.022 × 10^{23} ) molecules/1 mole CO = 7.23 x 10^{23} CO molecules.
9.6 The Mass of a Mole
Concept: The mass of a mole varies depending on the chemical substance.
Molar Mass Definition: The molar mass of an element is defined as the mass in grams numerically equal to the atomic mass when the element is present in atomic form.
Relationship Example:
Mass of 1 carbon atom = 12.01 amu means mass of 1 mole of carbon atoms = 12.01 g (molar mass).
Summarized table of Atomic Mass and Molar Mass:
H: 1.008 amu = 1.008 g
C: 12.01 amu = 12.01 g
O: 16.00 amu = 16.00 g
Molar Mass for Diatomic Elements:
Example: 1 mole of Cl atoms = 35.45 g Cl = 6.022 x 10^{23} Cl atoms.
1 mole of ( Cl2 ) molecules = 70.90 g ( Cl2 ) = 6.022 x 10^{23} ( Cl_2 ) molecules.
Molar Mass of Compounds:
Molar mass of a compound is numerically equal to the formula mass expressed in grams. By using the atomic masses of constituent atoms, the following masses can be deduced:
Mass of 1 H2O molecule = 18.02 amu = 18.02 g.
Mass of 1 NH3 molecule = 17.04 amu = 17.04 g.
Mass of 1 BaCl2 molecule = 208.23 amu = 208.23 g.
9.7 Significant Figures with Avogadro’s Number
The value of Avogadro’s number is 6.0221415 x 10^{23}, but typically only four significant figures (6.022 x 10^{23}) are required for practical use.
9.8 Relationship Between AMU and Gram Units
Conversion Factor: 6.022 x 10^{23} amu = 1.000 g.
Example:
For 104.0 amu, the mass in grams is calculated as follows:
104.0 amu x (1 g / 6.022 x 10^{23} amu) = 1.727 x 10^{-22} g.
9.9 Mole and Chemical Formulas
The chemical formula indicates the number of moles of atoms of each element present in one mole of a compound.
Example Calculation with ( C2H6 ):
For 1 mole of ( C2H6 ) (6.022 x 10^{23} molecules):
Number of C atoms = 2 moles ( C ) atoms.
Number of H atoms = 6 moles ( H ) atoms.
9.10 The Mole and Chemical Calculations
The mole is essential in calculating:
The number of particles of a substance.
The number of moles of a substance.
The number of grams of a substance.
Interrelationships among these quantities:
Number of particles and moles.
Grams of substance and moles.
Moles of compound and moles of elements in the compound.
9.11 Purity of Samples
Most samples contain impurities.
Percent Purity Definition: Percent by mass of a specified substance in an impure sample.
Example Calculation with HNO3:
Given 32.00 g sample is 96.20% pure.
Calculate:
(a) Mass of HNO3 = 32.00 g x 0.9620 = 30.78 g
(b) Impurity mass = 32.00 - 30.78 = 1.22 g.
9.12 Empirical and Molecular Formulas
Empirical Formula: Simplest formula giving the smallest whole number ratio of atoms in a compound.
Molecular Formula: Actual formula representing the number of atoms present in a formula unit of a compound.
Relationship: Molecular formula = whole number multiple ( n ) x empirical formula.
Example of Empirical and Molecular Formulas:
Empirical Formula of C2H4 is CH2.
Molecular Formula of C6H6 is CH.
Hydrogen chloride as Cl and HCI both represent the same molecular and empirical formulas respectively.
9.13 Determination of Empirical Formulas
Usually calculated from experimental data (e.g. % composition).
Steps for Calculation:
Assume a starting quantity (often 100.0 g).
Convert grams of each element to moles.
Divide by smallest mole value.
Use integers as subscripts for final empirical formula.
Example Calculation for Freon-12:
Given % compositions of C, Cl, and F.
Calculate moles from mass and express in ratio to derive empirical formula ( CCl2F2 ).
9.14 Determination of Molecular Formulas
The molecular formula can be calculated from the empirical formula and known molar mass.
Example: Given empirical formula ( CH ) and molecular formula mass of 78.12 amu:
Calculate ( n ) = ( 78.12 / 13.03 = 6.000 ).
Hence, the molecular formula is ( (CH)6 = C6H_6 ).
Practice Exercises for Finding Molecular Formulas:
Practice problems provided for determining molecular formulas based on given empirical formulas and respective molecular masses.