Lecture 3. Moles and molar mass

Tuesday 23 January 2024

The mole, Avogadro's number, and molar mass. Converting between moles and number of atoms. Converting between mass and chemical amount (moles). Converting between mass and number of atoms. Introduction to electromagnetic radiation and properties of waves.

Reading: Tro NJ.Chemistry: Structure and Properties (3rd ed.) - Ch.1, §1.10 (pp.60-65). Ch.2, §2.1-2.2 (pp.79-89t)

---

Summary

At the outset, we described chemistry as providing a bridge between the macroscopic world and the nanoscale realm. Here we introduce the concept of the mole, which serves as the chemist's tool to translate atomic and molecular quantities into convenient macroscopic quantities. The use of the mole and the calculation of molar masses is illustrated in the examples below.

The mole and the "Weighing is counting" principle

How many carbon atoms does it take to make up a macroscopic amount of carbon? More specifically, how many atoms of ¹²C does it take to make up a sample of carbon that weighs exactly 12 g? The answer to this question is constitutes the exact definition of a mole:

A mole is defined as the number of ¹²C atoms in exactly 12 g of a pure sample of this isotope. This number, called Avogadro's number (N_A), serves as a conversion factor between numbers of atoms, molecules, or any nanoscale particles and macroscopic chemical amounts of these ultrasmall entities. The mole (unit abbreviation: mol) is the SI base unit for chemical amount.

As a well determined physical constant, the value of Avogadro's number is known to a high precision:

N_A  =  6.0221418 × 10²³ mol⁻¹

Avogadros' number, which is the number of particles in a mole, is an essential concept and tool in our study of chemistry. It allows us to use atomic or molecular masses in amu as conversion factors with the units g/mol. Hence we can convert masses of elements and compounds into moles (i.e. a known number of atoms or molecules) and vice-versa. This conversion is of fundamental importance in chemistry, and you will use it repeatedly in problem-solving and laboratory work. You should be able to perform such conversions routinely and readily.

Two important concepts:

• The mole - such a large number that a mole's worth of atoms makes up a macroscopic amount. Furthermore, the value of the mole is specially chosen so that a mole of atoms of any element has a mass in grams equal to the atomic weight of the element.
• Atomic mass and formula mass as conversion factors (units = g/mol)

Weighing is counting - since substances are composed of molecules and atoms, any amount of a pure element or compound will consist of a finite number of identical molecules. When we go into the lab and weigh out the formula mass, in grams, of a molecular compound, we have just counted out Avogadro's number of molecules of that compound, i.e. 1 mole. Similarly, we can count out any fraction or multiple of a mole by weighing out in grams the same fraction or multiple of the formula mass.

Using Avogadro's number (N_A)

Example 1: Convert the quantity 5.8 × 10²⁴ aluminum atoms to moles of aluminum.

Solution: Use (1/N_A) as the conversion factor for number of Al atoms to mol Al (i.e. we can associate with 1/N_A the units mol Al / Al atoms):

(5.8 × 10²⁴ Al atoms)( 1/N_A mol Al / Al atoms)   =  (5.8 × 10²⁴ Al atoms) / (6.022 × 10²³ Al atoms/mol Al)

= 9.6 mol Al

This makes sense, since we can see from the magnitude of N_A that 5.8 × 10²⁴ aluminum atoms is almost 10 times the former.

Example 2: How many atoms of copper are in 4.878 × 10⁻² mol of copper?

Solution: Use N_A as the conversion factor for mol Cu to number of Cu atoms:

(4.878 ×10⁻² mol Cu)( N_A Cu atoms / mol Cu)   =  (4.878 ×10⁻² mol Cu)(6.02214 × 10²³ Cu atoms / mol Cu)

= 2.938 × 10²² Cu atoms

Note that we chose to use a value for N_A with 2 more significant figures than the value for mol Cu given.

Using molar mass

Example: How many atoms of copper are in a pure sample of copper of mass 3.10 g?

Solution: We use the molar mass of copper to convert mass (in g) Cu to mol Cu:

(3.10 g Cu) / (63.55 g Cu / mol Cu)

This yields mol Cu which is converted to number of Cu atoms by using as the conversion factor, as in the previous example. Putting this all together, we have

(3.10 g Cu) × ( 1 / 63.55 g Cu / mol Cu) × (6.022 × 10²³ Cu atoms/mol Cu) = 2.94 × 10²² Cu atoms.

Light and electromagnetic radiation

The secrets of modern atomic theory have been revealed through careful investigation of the interactions between matter and energy, in the form of electromagnetic (EM) radiation. Visible light is only one part of the EM spectrum, which encompasses radiation from long-wavelength (low frequency) radio waves to short wavelength (high frequency) X-rays. EM radiation propagates through space as wave, as a wave , and is thus characterized by wavelength, frequency, amplitude and speed. The speed of light (represented as c) is a well determined physical constant,

c  =  2.9979 × 10⁸ m/s

and the governing relationship between wavelength and frequency for EM radiation

speed of wave  =  (frequency) × (wavelength)     or      speed  =  ν λ

One of our principal objectives is to use this relationship to convert between frequency and wavelength for any given value of either of these. We may also find it necessary to use decimal multipliers for unit conversion, as frequencies and wavelengths of EM radiation vary over many orders of magnitude.

Example: What is the frequency (ν) of visible light with wavelength λ = 535 nm?

Solution: We'll use c =  2.99792458 × 10⁸ m·s⁻¹ and a rearranged form of the relation between c, ν, and λ:

ν = c / λ

Since wavelength is given to us in nm, the conversion from nm to m must be included in our calculation:

ν = (2.99792458 × 10⁸ m·s⁻¹) / (535 nm)(10⁻⁹ m/nm) = 5.60 × 10¹⁴ s⁻¹

In this context, the unit s⁻¹ gets a special designation, hertz (Hz), so we will write the answer as 5.60 × 10¹⁴ Hz.

Where some of the interactions between light and matter are concerned, the wave description fails in its ability to account for certain observations such as black body radiation and the photoelectric effect. A quite different model is provided by quantum theory, which postulates a quantized nature of light. Ultimately, we learn that EM radiation manifests both wave and particle properties, while matter (in the nano- and picoscale realms of atoms and subatomic particles) also manifests properties of waves.