1.1 Atoms and Subatomic Particles

What are atoms?

Atoms (from the Greek word atomos, which means “uncuttable”) are tiny building blocks of matter. Specifically, an atom is the smallest quantity of matter that still retains the properties of matter. Elements are substances that cannot be broken down into simpler substances and are made up of atoms. Atoms comprise three fundamental subatomic particles, two of which are contained in a dense nucleus. The nucleus comprises electrically neutral neutrons and positively charged protons and is surrounded by a cloud of negatively charged electrons. The important properties of these subatomic particles are shown in Table 1.

Table 1 Mass and charge of subatomic particles. Note: formal unit of charge is coulomb (\(\pu{C}\)).

Particle	Mass	Mass	Charge	Charge
	\(\pu{g}\)	\(\pu{amu}\)	Absolute, \(\pu{C}\)	Relative
Electron	\(\pu{9.109e-28}\)	\(\pu{5.486e-4}\)	\(\pu{-1.6022e-19}\)	\(-1\)
Proton	\(\pu{1.673e-24}\)	\(1.007\)	\(\pu{1.6022e-19}\)	\(+1\)
Neutron	\(\pu{1.675e-24}\)	\(1.009\)	\(0\)	\(0\)

The mass of individual atoms is very low and is expressed in atomic mass units or \(\pu{amu}\). The \(\pu{amu}\) equals one-twelfth the mass of a carbon-12 atom.

\[ \pu{1 amu} = \pu{1.661e-24 g} = \pu{1.661e-27 kg} \]

Atoms are identified by the number of protons and neutrons they contain. The atomic number (\(Z\)) is the number of protons in the nucleus of each atom of an element. It also indicates the number of electrons in the atom - because for an atom to be neutral, it must contain the same number of protons and electrons. \(Z\), \(A\), and the number of neutrons are all positive integers (whole numbers).

Note

All atomic nuclei contain protons and neutrons except for the most common form of \(\ce{H}\), which has one proton and no neutrons.

The mass number (\(A\)) is the total number of neutrons and protons in an atom and is estimated as follows:

(1)\[\text{mass number}\ (A) = \text{# of protons}\ (Z) + \text{# of neutrons}\]

The accepted way to express this information in geochemical settings is as follows for a hypothetical element, \(X\) is \(^A_Z X\).

Example: Estimating subatomic particles

For the element fluorine (\(\ce{F}\)), \(A=19\) and \(Z=9\). Therefore, using Eq. (1), we can determine the number of neutrons in an atom of \(\ce{F}\) is \(19-9 = 10\). This information can be shown as \(\ce{^19_9{F}}\).

Atoms of any given element are not all identical. Instead, most elements have two or more isotopes. These elements have the same \(Z\) but different \(A\). Because \(Z\) does not change for isotopes of any element (if \(Z\) changed, it would be a different element!), \(Z\) can be dropped and only \(A\)} is often shown.

A small set of naturally occurring elements in the periodic table has only one stable isotope.

Example: Isotope notation

For example, \(\ce{H}\) has three isotopes: \(\ce{^1_1H}\) (protium), \(\ce{^2_1H}\) (deuterium), and \(\ce{^3_1H}\) (tritium) and can be shown as \(\ce{^1H}\), \(\ce{^2H}\), and \(\ce{^3H}\). Only \(\ce{H}\) isotopes have specific names; isotopes of all other elements are denoted by their mass numbers or names, as shown in the examples below:

1. Carbon-14 or \(\ce{C-14}\) or \(\ce{^{14}C}\)

2. Oxygen-18 or \(\ce{O-18}\) or \(\ce{^{18}O}\)

3. Uranium-238 or \(\ce{U-238}\) or \(\ce{^{238}U}\)

What is mass?

Atomic Mass

Atomic mass is the mass of an atom in amu. When you look up the atomic mass of most elements in a periodic table, those values are usually not whole numbers. The difference arises because most naturally occurring elements have more than one isotope. This means that when we determine an element’s atomic mass, we measure the average atomic mass of the naturally occurring mixture of isotopes. The average atomic mass of an element is a weighted average mass of all the isotopes present in a naturally occurring sample of that element. This equals the sum of each isotope’s mass multiplied by its fractional abundance.

(2)\[\text{average mass} = ∑_i (\text{fractional abundance} \times \text{isotopic mass})_i\]

Example: Calculating average atomic masses

Naturally occurring \(\ce{C}\) has two major stable isotopes - \(\ce{^{12} C}\) (\(99\%\) of all stable \(\ce{C}\)) and \(\ce{^{13} C}\) (\(1\%\) of all stable \(\ce{C}\)).

The atomic mass of \(\ce{^{12} C}\) and \(\ce{^{13} C}\) is \(\pu{12 amu}\) and \(\pu{13 amu}\), respectively. Using Eq. (2), we can calculate the overall atomic mass of naturally occurring \(\ce{C}\):

\[ = \frac{99}{100} \cdot \pu{12 amu} + \frac{1}{100}\cdot \pu{13 amu} = \pu{12.01 amu} \]

Mole

Atoms are so tiny that even the smallest quantity of matter contains many atoms. Knowing the exact number of atoms involved in a chemical reaction is useful. For convenience, chemists use a unit of measurement called the mole to denote these large numbers of atoms. A mole is defined as the amount of a substance that contains as many elementary entities (e.g., atoms, ions, etc.) as there are atoms in exactly \(\pu{0.012 kg}\) (\(\pu{12 g}\)) of carbon-12. The number of (experimentally determined) atoms in exactly \(\pu{12 g}\) of carbon-12 is known as Avogadro’s Number (\(N_A\)).

Note

After Italian scientist Amedeo Avogadro (1776-1856). See this link to read about Avogadro: (https://en.wikipedia.org/wiki/Amedeo_Avogadro}.

The currently accepted value of Avogadro’s number is \(\pu{6.0221418e23}\), although we usually round this number to \(\pu{6.022e23}\).

How big is a mole? (Not the animal, the other one.) - TED-Ed.

Molar Mass

We know the number of atoms present in a mole of any substance, but how much does this mole weigh? We need to see the weight of each atom of that substance and add all these masses. An easier way to deal with this is the molar mass concept.

The molar mass of a substance is the mass in grams of one mole of a substance. Atomic mass has the units of atomic mass units or amu.

Example: Calculating molar masses:

Carbon-12 or \(\ce{^{12} C}\) (most abundant stable isotope of \(\ce{C}\)) is used the standard example – mass of one mole of \(\ce{^{12} C}\) is exactly \(\pu{12 g}\), i.e., molar mass of \(\ce{^{12} C}\) is equal to its atomic mass in grams.

The overall molar mass of naturally occurring \(\ce{C}\)

\[ = \frac{99}{100} \cdot \pu{12 g} + \frac{1}{100}\cdot \pu{13 g} = \pu{12.01 g} \]

So, how do we convert from moles to molar masses to the number of atoms? Set up conversion factors such that units are explicitly shown and are consistent.

Example: Conversion of moles to grams

\(\ce{Ca}\) is the most abundant metal in the human body. A typical human body contains roughly \(\pu{30 moles}\) of \(\ce{Ca}\). How much is this amount in kg?

First, convert moles of \(\ce{Ca}\) to the number of atoms of \(\ce{Ca}\) as follows:

\[ 30\ \cancel{\text{moles Ca}} \left(\dfrac{\pu{6.022e23 atoms Ca}}{1\ \cancel{\text{mole Ca}}}\right) = \pu{1.807e25 atoms Ca} \]

Then, convert the number of atoms of \(\ce{Ca}\) to kg of \(\ce{Ca}\) as follows:

\[ 30\ \cancel{\text{moles Ca}} \left( \dfrac{\pu{40.078 g Ca}}{1\ \cancel{\text{mole Ca}}}\right) = \pu{1202.3 g Ca} \]

Therefore, \(\pu{1202.3 g}\) of Ca exists in an average human body.

Example: Conversion of grams to atoms

Copper (\(\ce{Cu}\)) is commonly used to fabricate electrical wire. How many \(\ce{Cu}\) atoms are in \(\pu{5.00 g}\) of a \(\ce{Cu}\) wire?

The number of \(\ce{Cu}\) atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of \(\ce{Cu}\), and then using Avogadro’s number (\(N_A\)) to convert this molar amount to the number of \(\ce{Cu}\) atoms:

\[ \pu{5.00 g Cu} \left(\dfrac{\pu{1 mol Cu}}{\pu{63.55 g}}\right) \left(\dfrac{\pu{6.022e23 atoms}}{\pu{1 mol}}\right) = \pu{4.74e22 atoms Cu} \]

Chemical Formulas

A chemical formula denotes the composition of any chemical substance. In a molecular formula, an exact number of atoms of each element in a molecule is shown.

Molecular formulas

\(\ce{H2}\) is the molecular formula for hydrogen, \(\ce{O2}\) is that for oxygen, \(\ce{CO2}\) is that for carbon dioxide, and \(\ce{H2O}\) is that of water. The subscript number indicates the number of atoms of an element present in the molecule. There is no subscript for \(\ce{O}\) in \(\ce{H2O}\) because there is only one oxygen atom in a molecule of water, and \(1\) is never used as a subscript in a chemical formula.

Structural Formulas

Structural formulas show the elemental composition but also the general arrangement of the atoms within the molecule. See different representations in Fig. 2 and Fig. 3.

Fig. 2 A methane molecule can be represented as (a) a molecular formula, (b) a structural formula, (c) a ball-and-stick model, and (d) a space-filling model. Carbon and hydrogen atoms are represented by black and white spheres, respectively. Image source: 2.4 Chemical Formulas - Chemistry: Atoms First | OpenStax

Fig. 3 Molecular structure of caffeine molecule. The structure is shown as a ball-and-stick model, with the color of each ball representing an atom of a specific element. Typically, black balls are \(\ce{C}\), blue balls are \(\ce{N}\), white balls are \(\ce{H}\), and the red balls are \(\ce{O}\). Source: Caffeine - Wikipedia

Empirical Formulas

Empirical (“from experience” or “experimentally”) formulas tell which elements are present in a molecule and in what whole-number ratio relative to each other. This is usually determined experimentally. It is the most straightforward chemical formula possible for any molecule. Usually, it is written by reducing the subscripts of molecular formulas to the smallest possible whole number. But molecular formulas are actual formulas of molecules.

Example of empirical formulas

Below are some examples of molecular and empirical formulas.

Compound	Molecular Formula	Empirical formula
Water	\(\ce{H2O}\)	\(\ce{H2O}\)
Hydrogen Peroxide	\(\ce{H2O2}\)	\(\ce{HO}\)
Propane	\(\ce{C3H8}\)	\(\ce{C3H8}\)
Benzene	\(\ce{C6H6}\)	\(\ce{CH}\)

In the above example, water and propane already have the simplest formula possible with a whole number of subscripts. Therefore, both molecular and empirical formulas are the same.

Molecular and Formula Masses

Molecular mass is the sum of the atomic masses (in amu) of the atoms that make up the molecule. Often, the molecular mass is referred to as molecular weight.

Example: Molecular mass

Calculate the molecular mass of \(\ce{H2O}\).

\[\begin{split} \begin{aligned} \text{Molecular mass of}\, \ce{H2O} &= 2(\text{Atomic mass of H}) + 1(\text{Atomic mass of O})\\ &= 2(\pu{1.008 amu}) + (\pu{16.00 amu})\\ &= \pu{18.02 amu} \end{aligned} \end{split}\]

Since ionic compounds are not whole molecules, their masses are based on their empirical formulas and are called formula mass or formula weight.

Examples of molecular and formula masses

Compound	Chemical Formula	Molecular Mass, \(\pu{amu}\)	Empirical Mass, \(\pu{amu}\)
Water	\(\ce{H2O}\)	\(18.02\)	\(18.02\)
Hydrogen Peroxide	\(\ce{H2O2}\)	\(34.01\)	\(17.01\)
Propane	\(\ce{C3H8}\)	\(44.10\)	\(44.10\)
Benzene	\(\ce{C6H6}\)	\(78.11\)	\(13.02\)

Percent Composition

If we know a compound’s molecular or empirical formula, we can easily estimate the percent composition by mass of each element in that compound. The percent composition of each element is calculated as follows:

(3)\[\text{% mass of an element} = \dfrac{n\times\text{atomic mass of element}}{\text{molecular/formula mass of compound}}\times 100\]

where \(n\) is the number of atoms of the element in the compound, a list or table of the percent by mass of each element in a compound is known as the compound’s percent composition by mass.

Example: Calculating percent composition of \(\ce{H2O}\)

Atomic masses of \(\ce{H}\) and \(\ce{O}\) are \(\pu{1.008 amu}\) and \(\pu{16.00 amu}\) respectively, the molecular mass of \(\ce{H2O}\) is \(\pu{18.02 amu}\).

Percent composition is calculated as follows:

\[\begin{split} \begin{aligned} \ce{H} &= \dfrac{2 (\pu{1.008 amu H})}{(\pu{18.02 amu \ce{H2O}})} \times 100 = \pu{11.19 \%} \\ \ce{O} &= \dfrac{1 (\pu{16.00 amu O})}{(\pu{18.02 amu \ce{H2O}})} \times 100 = \pu{88.81 \%} \end{aligned} \end{split}\]

The sum of the percentages adds up to \(\pu{100 \%}\).

Determining Chemical Formulas

Percent composition is often determined experimentally and precisely using very sensitive analytical instrumentation. This information can be used to determine the empirical formula of a substance of interest.

Rules for determining empirical formulas

1. Determine the weight percent of each element in the compound

2. Determine weight in grams of each element per 100 g of compound

3. Divide the weight of each element in grams by atomic weight to determine the number of moles

4. Using the element with the least number of moles, determine the relative number of moles of all other elements – these integers represent the formula for the compound

It is best to set this up as a table. See the example below.

Example: Determining chemical formulas

During a chemical reaction, a precipitate forms. The percent mass was determined experimentally as follows: \(\ce{Ca} = \pu{39.74 \%}\), \(\ce{P} = \pu{18.42 \%}\), \(\ce{O} = \pu{38.07 \%}\), and \(\ce{F} = \pu{3.77 \%}\). Determine the chemical formula for this compound.

First, let’s set this information up as a table and calculate the moles of each element.

Element	Wt., %	Wt./100 g	At. Wt.	Moles	Rel. Moles
\(\ce{Ca}\)	\(39.74\)	\(39.74\)	\(40.08\)	\(0.992\)	\(5\)
\(\ce{P}\)	\(18.42\)	\(18.42\)	\(30.97\)	\(0.595\)	\(3\)
\(\ce{O}\)	\(38.07\)	\(38.07\)	\(16.00\)	\(2.379\)	\(12\)
\(\ce{F}\)	\(3.77\)	\(3.77\)	\(19.00\)	\(0.198\)	\(1\)

Based on this information, the formula of the compound is \(\ce{Ca5P3O12F}\). If we re-arranged this formula, we get \(\ce{Ca5(PO4)3F}\) (fluorapatite).

In the above example, since the percent mass was determined experimentally, the resulting formula is considered an empirical formula or the simplest whole-number ratio of atoms that make up a compound. If we know or determine the molecular weight of this compound by other means, we could compare the formula and molecular weights and then determine the exact molecular formula.

Example: Determining the molecular formula

If the empirical formula of a compound is \(\ce{CH3}\) and its molecular weight is \(\pu{30 g}\), the molecular formula can be determined as follows.

First, calculate the formula weight of \(\ce{CH3}\)

\[\begin{split} \begin{aligned} \ce{CH3} &= 1(\pu{12.01 g mol-1 C}) + 3(\pu{1.008 g mol-1 H}) \\ &= \pu{15.03 g mol-1}\, \ce{CH3} \end{aligned} \end{split}\]

Since the molecular weight is twice the formula weight, the molecular formula is \(\ce{C2H6}\).

Solutions and Concentrations

Concentration is a significant factor influencing reactions in aqueous solutions. It is the amount of solute dissolved in a given solvent or solution.

Molarity, or molar concentration, \(\pu{M}\), is defined as the number of moles of solute per liter of solution and can also be written as \(\pu{mol L-1}\).