chapter 2 p2

Atoms & Molecules

Introduction

This section introduces fundamental concepts in general chemistry, focusing on the nature of atoms, isotopes, and molecules, and how these relate to chemical calculations. Understanding these concepts is essential for performing quantitative work in chemistry.

Average Atomic Mass

Isotopes and Atomic Mass

Most elements occur in nature as mixtures of isotopes, which are atoms of the same element with different numbers of neutrons and thus different masses. The average atomic mass of an element is a weighted average of the masses of its naturally occurring isotopes.

• Isotope: Atoms of the same element with different numbers of neutrons.

• Average Atomic Mass: The value listed on the periodic table, reflecting the weighted average of all naturally occurring isotopes.

Example: Carbon Isotopes

• 12C: 98.89% abundance, mass = 12.000 amu (exactly)

• 13C: 1.11% abundance, mass = 13.0034 amu

• 14C: Trace (<0.01%), mass = 14.0032 amu

The average atomic mass of carbon as listed on the periodic table is 12.011 amu.

Calculating Average Atomic Mass

The average atomic mass is calculated using the following formula:

$\text{Average Atomic Mass} = \sum (\text{fractional abundance} \times \text{isotopic mass})$

Example Calculation for Carbon:

• $(0.9889 \times 12.000) + (0.0111 \times 13.0034) = 12.011$ amu

For stoichiometric calculations, we treat natural carbon as if it consists of atoms with an average mass of 12.01 amu, even though no single atom has this mass. This allows us to count atoms by weighing samples.

Practice Problem: Identifying an Element by Isotopic Mass

A sample consists of 62.60% of an isotope with mass 186.956 amu and 37.40% of an isotope with mass 184.953 amu. Calculate the average atomic mass and identify the element.

• Average atomic mass = $(0.6260 \times 186.956) + (0.3740 \times 184.953)$

• Average atomic mass ≈ 186.207 amu

• Element with atomic mass ≈ 186.2 is Osmium (Os)

Chemical Stoichiometry

Definition and Importance

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in chemical reactions. It allows chemists to predict the amounts of substances consumed and produced in a given reaction.

• Involves counting by weighing: using the average mass of objects (atoms, molecules) to determine their number in a sample.

• Atoms and molecules behave as though they are all identical for calculation purposes.

Counting by Weighing: Example

Example: A pile of marbles weighs 394.80 g; 10 marbles weigh 13.10 g. How many marbles are in the pile?

• Average mass per marble = $\frac{13.10\ g}{10} = 1.31\ g$

• Number of marbles = $\frac{394.80\ g}{1.31\ g} \approx 301$ marbles

This method is analogous to counting atoms or molecules by weighing a sample and using the average mass.

The Mole and Avogadro's Number

Definition of the Mole

The mole is the SI unit for the amount of substance. One mole contains exactly $6.022 \times 10^{23}$ entities (Avogadro's number, $N_A$), which is the number of atoms in exactly 12 grams of carbon-12.

• Avogadro's Number ($N_A$): $6.022 \times 10^{23}$

• 1 mole of anything = $6.022 \times 10^{23}$ units of that thing

• 12.00 g of $^{12}$C contains $6.022 \times 10^{23}$ atoms

• The mass of one mole of atoms of an element (in grams) is numerically equal to its average atomic mass (in amu)

Examples:

• 1 atom of carbon: 12.01 amu

• 1 mole of carbon atoms: 12.01 g

Problem Solving in Stoichiometry

General Approach

Effective problem solving in chemistry involves a systematic approach:

1. Where are we going? Analyze the problem and decide on the final goal.

2. How do we get there? Work backwards from the final goal to decide where to start.

3. Check Does the answer make sense? Is it reasonable?

Practice Problems

• Example 1: Californium is an element that does not occur naturally and is used in neutron detectors. Calculate the mass in grams of a sample of 6 atoms of californium. Additional info: Use the atomic mass of californium (approx. 251 amu) and Avogadro's number for calculation.

• Example 2: Calculate the number of iron atoms in a 4.48 mole sample of iron.

  • Number of atoms = $4.48 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol}$

• Example 3: Calculate the number of moles in a 125 g sample of silver.

  • Moles = $\frac{125\ g}{107.87\ g/mol}$ (where 107.87 g/mol is the molar mass of silver)

Summary Table: Key Quantities and Relationships

Quantity	Unit	Relationship
Atomic Mass	amu	Mass of a single atom
Molar Mass	g/mol	Mass of 1 mole of atoms/molecules
Number of Particles	atoms, molecules, etc.	Number = moles × Avogadro's number
Mole	mol	1 mol = $6.022 \times 10^{23}$ entities