Lecture 4: Chemical Reactions and Empirical Analysis

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Unit 1

Chapter 1

General Chemistry

Molecular Structure and Properties

Introduction

This lecture is part of the MEEP curriculum on General Chemistry. Information about Project MEEP and other General Chemistry lectures are available below.


Recommended Time: 1.5 Hours

Lecture Preview

  • What is a chemical reaction and how do we describe it?
  • How can we convert between different units in chemistry?
  • How do we balance a chemical equation and why is it important?
  • What types of quantitative analyses can we carry out from a chemical reaction and how are they useful?

Lecture Content (Part 1)

Chemical Reactions and Equations

Chemical reactions take place everywhere. To quantitatively analyze those reactions, we need a standardized system of describing a reaction. As a result, we developed the chemical equation.

In a typical chemical equation, there are 3 necessary components and some optional ones.

  1. Reactants (with coefficients, separated by +).
  2. Products (with coefficients, separated by +).
  3. An arrow – different arrows are used for different types of reactions.
    • \rightarrow (net forward reaction, the common one)
    • \rightleftarrows (reversible reaction)
    • \rightleftharpoons (equilibrium, Unit 3 content)

\leftrightarrow is another common arrow we use, but this isn’t used for chemical equations. This arrow is used for resonance structures, which we will discuss in Chapter 2.

  • The condition that the reaction will occur, such as “combustion”, a specific temperature, a specific acidity in the environment, light, heat, catalysts, etc.
  • The states of the products and reactants, especially when there are multiple states of matter present.
    • (s) for solid.
    • (l) for liquid.
    • (g) for gas.
    • (aq) for aqueous solutions.
    • \uparrow for gas generation in products when there is no gas in the reactants.
    • \downarrow for precipitation in products when there is no solid in the reactants.

In the end, a chemical equation quantitatively describes a chemical reaction, which is essentially a rearrangement of atoms in a specific way. For example, observe the reaction below:

CH_4 + 2\ O_2 \xrightarrow{\text{combustion}} CO_2 + 2\ H_2O

This describes a methane molecule interacting with 2 oxygen gas molecules when combusted to form a carbon dioxide molecule and 2 water molecules. The animation below demonstrates how the atoms rearrange from the reactants to form the products. We call this type of chemical equation a “molecular equation”

There are 4 slides to this animated diagram.

The Mole, Molar Mass, and Molarity

Most intuitively, the chemical equation describes how many molecules interact with each other. For example, observe the following equation:

B_2O_3\ (s) + 3\ Mg\ (s) \rightarrow 2\ B\ (s) + 3\ MgO\ (s)

The most intuitive explanation is that:

  • 1 molecule of boron oxide reacts with 3 atoms of magnesium and forms 2 atoms of boron and 3 molecules of magnesium oxide.

However, the coefficients can also describe ratios. As a result, another correct explanation of the equation above is:

\mathbf{B_2O_3}

reacts with

\mathbf{Mg}

and theoretically forms

\mathbf{B}

and

\mathbf{Mg}

1 part

3 parts

2 parts

3 parts

25 molecules

75 molecules

50 molecules

75 molecules

6.02\times 10^{23} molecules

18.06\times 10^{23} molecules

12.04\times 10^{23} molecules

18.06\times 10^{23} molecules

1 mole

3 moles

2 moles

3 moles

1.3 moles

3.9 moles

2.6 moles

3.9 moles

We now connect moles with chemical equations. However, we cannot easily measure the moles of a chemical. The easier method is to measure the mass of the chemical. We need a way to connect moles and mass.

Introducing the molar mass – a numerical equivalent of the atomic mass unit due to their definitions. The definition and formula for the molar mass are:

The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol).

(1)   \begin{equation*} M = \frac{m}{n} \end{equation*}

where:

  • M denotes the molar mass (g/mol)
  • m denotes the mass (g)
  • n denotes the amount of substance (mol)

The molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single C-12 atom weighs 12 amu (its atomic mass is 12 amu). A mole of C-12 weighs 12 g (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, C-12. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu.

Using molar mass, we can connect units like number of particles and mass together. For example:

  • Given that the molar mass for water is 18 g/mol, 1 mole of water will weigh around 18 grams while 27 grams of water corresponds to 1.5 moles of water.
  • Given that the molar mass for acetic acid is 60 g/mol, 3 moles of acetic acid will weigh 180 grams while 300 grams of acetic acid corresponds to 5 moles of acetic acid.

When it comes to solutions, we need a unit that represents the concentration of the solution so that we can convert between moles and volume. Introducing the molarity (a.k.a, molar concentration) – the concentration of a solution.

Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution.

(2)   \begin{equation*} c = \frac{n}{V} \end{equation*}

where:

  • c denotes the molarity (mol/L)
  • n denotes the amount of substance (mol)
  • V denotes the volume of the solution (L)

Unit Conversions in Chemistry

We already talked about unit conversions in Lecture 2, but this time, we will focus on unit conversions in chemistry.

We have the chemistry unit conversion trinity – molar mass (g/mol), molarity (mol/L), and density (g/mL) – to connect measurements like mass, moles, and volume. More specifically:

  • The chemical equation itself tells us how many moles of reactants are required for and how many products are theoretically produced by the reaction.
  • The molar mass helps us determine the mass of the solid required for or theoretically produced by the reaction.
  • The density helps us determine the volume of the liquid required for or theoretically produced by the reaction.
  • The molarity helps us determine the volume of solution required for the reaction if the reactant dissolved.
The chemistry unit conversion trinity – the molar mass, molarity, and density – connects units like the mole, mass, and volume. Avogadro’s Number is also included. Their symbols are represented in parentheses.

While doing unit conversions, remember to use the right units. Otherwise, our calculations can be magnitudes off.

Let’s go through some examples.

Question: What is the molar concentration of water?

Hint:

  • The molar mass of water is around 18.0 g/mol.
  • The density of water is 1.00 g/mL.

Response:

To obtain the molarity of water, we use our equation:

(3)   \begin{equation*}c=\frac{n}{V}\end{equation*}

Let’s assume we have 1 mole of water to make our calculations easier. Since this is a theoretical value, it does not follow sig fig rules.

(4)   \begin{equation*}n=1 \text{ mol}\end{equation*}

However, we don’t know the volume of water. Thankfully, we can convert moles of water to the mass of water and then to the volume (because molar mass and density of water are provided). So, we have:

(5)   \begin{equation*}V=\frac{m}{\rho}=\frac{nM}{\rho}=\frac{1\ \text{mol}\times 18.0\ \text{g/mol}}{1.00\ \text{g/mL}\times \frac{1000\ \text{mL}}{1\ \text{L}}}=0.0180\ \text{L}\end{equation*}

Notice that we want our volume to be in liters because that is what molarity is defined as – moles per liter. Now that we have our mole and volume, we can calculate the molar concentration of water.

(6)   \begin{equation*}c=\frac{n}{V}=\frac{1\ \text{mol}}{0.0180\ \text{L}}=55.6\ \text{mol/L}\end{equation*}

Answer: The molarity of water is 55.6 mol/L.

The animation below visualizes how we can solve the problem.

Visualization of how the molarity of water can be calculated. There are 7 slides in this animation.

Here is another example:

Another Example: Use the resource below and answer the question.

From https://www.nutritionvalue.org/

This is a nutritional table for a bottle of beverage. It is known that:

  • The bottle has a volume of 20.0 fluid ounces.

For simplicity, let’s assume that:

  • There is only 1 type of sugar contained in this beverage.
  • The sugar is sucrose.

Here are some important Values:

  • 1 fluid ounce = 29.5 mL
  • Sucrose has a molar mass of 342 g/mol.

Question 1: What is the density of the beverage?
Question 2: What is the molarity of the sugar in this beverage?

Hint:

  • Question 1 uses this equation: \rho = \frac{m}{V}
  • Question 2 uses this equation: c = \frac{n}{V}
  • As a convention, we will use g/mL for the density unit and mol/L for the molarity unit.

Response:

From the equations, we need the following values.

1. The mass of the beverage

2. The volume of the beverage

3. The mole of the sugar

Let’s calculate one by one.

  • Firstly, the mass of the beverage can be found on the nutritional label – 600 grams.

(7)   \begin{equation*} m = 600\text{ g}\end{equation*}

  • Secondly, the volume of the beverage is 20.0 fluid ounces. We need to convert it into mL.

(8)   \begin{equation*} V = 20.0\text{ fl oz} \times \frac{29.5\text{ mL}}{1\text{ fl oz}}=591\text{ mL}\end{equation*}

  • Thirdly, the mole of the sucrose. From the nutritional label, we know that it contains 64 grams of sugar. Thus, we can get the number of moles.

(9)   \begin{equation*} n = \frac{m}{M} = \frac{64\text{ g}}{342\text{ g/mol}} = 0.19\text{ mol}\end{equation*}

Now, we can solve the question. For the density of the beverage, we have:

(10)   \begin{equation*} \rho = \frac{m}{V} = \frac{600\text{ g}}{591\text{ mL}} = 1.02\text{ g/mL}\end{equation*}

For the molarity of the sugar, we have:

(11)   \begin{equation*} c = \frac{n}{V} = \frac{0.19\text{ mol}}{591\text{ mL} \times \frac{1\text{ L}}{1000\text{ mL}}} = 0.32\text{ mol/L}\end{equation*}

Notice that we cheekily converted mL into L in the equation above.

Answer: The density of the beverage is 1.02 g/mL, and the molarity of the sugar is 0.32 mol/L.

More practice will be available in the worksheet. Practice makes perfect!

Types of Chemical Reactions

Not surprisingly, there are a plethora of chemical reaction types. A chemical reaction can have multiple types. We will introduce some types here, especially the ones we will encounter in future chapters.

  1. Precipitation reactions
  2. Acid-Base reactions
  3. Oxidation-Reduction reactions

Precipitation Reactions

Suppose we mix 2 aqueous solutions together, and suddenly, we see solids forming. Those solids are called “precipitates”, and the reaction is called “precipitation”. Those reactions are used widely in industry for production of a number of commodity and specialty chemicals. They also play a central role in many chemical analysis techniques.

If we want to generalize what a precipitation reaction looks like, the following chemical equation may help. Notice the right arrow in the equation – precipitation reactions usually are highly favorable, resulting in an irreversible reaction.

AB(aq) + CD(aq)\rightarrow AC(s)\downarrow + BD(aq)

The precipitation is noted by the downward arrow.

The ability of a substance to dissolve in water (or any solvent) is quantitatively expressed as its solubility. The higher the solubility, the more the substance can dissolve in water. Quantitative analysis of solubility will be discussed in Unit 3.

Let’s look at some example reactions.

Example 1: Clogged Water Pipes

Water pipes get clogged because precipitates form inside the pipes. Precipitates form because we usually run hard water in our pipes, which contain ions like Ca^{2+} and Mg^{2+}. Those ions can reacting with other ions to form precipitates. Those reactions include but are not limited to:

Ca^{2+} + CO_3^{2-}\rightarrow CaCO_3 \downarrow (limestone clogging)

Mg^{2+} + 2\ OH^- \rightarrow Mg(OH)_2 \downarrow

Notice that the chemical equation above does not look like the ones we introduced earlier. This is because this is called a “net ionic equation” – an equation focusing on ion interactions rather than molecular interactions. We use the net ionic equation for this example because we aren’t sure exactly what molecules are involved. Consider the following logic:

  1. CaCl_2(aq) and CaNO_3(aq) can both exist in hard water, and both are highly soluble in water as well.
  2. In a typical limestone clogging scenario, we don’t know if it’s CaCl_2 or CaNO_3 reacting.
  3. However, we do know that Ca^{2+} will reacting regardless of which molecule it belongs in.
  4. So, we will use the net ionic equation to minimize confusion.

Example 2: Kidney Stones

Kidney stones are hard deposits made of minerals and salts that form inside your kidneys. They can be very painful. They are a product of precipitation reactions and supersaturation.

  • Supersaturation means that a solution contains more solute than the solubility allows. The excess solute then exists as a solid rather in an aqueous state.

A type of kidney stones – calcium stones, are formed from calcium (Ca^{2+}) and oxalate (C_2O_4^{2-}). Because calcium oxalate has a low solubility in water, it precipitates when formed.

Ca^{2+} + C_2O_4^{2-} \rightarrow CaC_2O_4 \downarrow

Quite predictably, kidney stones are usually caused by high blood calcium and oxalate concentrations. Those high concentrations can be achieved by (not an exhaustive list at all):

  • Not drinking enough fluids.
  • Eating too much meat and other protein-rich foods.
  • Having certain medical conditions such as diabetes, obesity, and osteoporosis.

More information is available at the end of this lecture.

Example 3: Gout

Gout is a form of arthritis – a chronic condition that causes inflammation of the joints, tissues around the joints, or other connective tissues. It is characterized by the deposition of urate crystals in joints and tissues.

Urate C_5H_3N_4O_3^- (structurally and chemically distinct from urea CO(NH_2)_2) is produced as an end product of purine metabolism. When paired with sodium ions in the blood, the monosodium urate crystals can deposit around the joints.

Na^+ + C_5H_3N_4O_3^- \rightarrow NaC_5H_3N_4O_3 \downarrow

Gout has an increased risk of expression in certain diets, including but not limited to:

  • Drinks and sweets with high amounts of sugar or high fructose corn syrup.
  • Alcohol.
  • Organ, game, and red meat and certain seafood.

More information is available at the end of this lecture.

Acid-Base Reactions

Acid-base reactions will be a huge part of general chemistry because of how important it is in our body and also because of how difficult it could be in future chapters. For now, let’s ease in to this concept of a reaction.

Here are some important terms to remember.

  • A proton is chemically equivalent to a hydrogen ion H^+ since the hydrogen atom contains only 1 proton and 1 electron.
  • An acid is a substance that donates a proton.
  • A base is a substance that accepts a proton.
  • Protonation is the gaining of proton(s).
  • Deprotonation is the losing of proton(s).

In a way, acid-base reactions are simply proton transfers. Protons leave acids for bases.

We can generalize the acid-base reaction into the following chemical equation:

AH + B \rightarrow A^- + HB^+

where:

  • AH stands for “acid”
  • B stands for “base”
  • H stands for a hydrogen atom

However, in common acid-base reactions, other ions are involved, so this generalized equation won’t fit perfectly in the examples we discuss below.

The most textbook example of an acid-base reaction would be hydrochloric acid reacting with sodium hydroxide, forming sodium chloride (table salt) and water.

HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)

  • Acid: HCl
  • Base: NaOH

Other textbook examples include:

H_2SO_4 + 2\ KOH \rightarrow K_2SO_4 + 2\ H_2O

  • Acid: H_2SO_4
  • Base: KOH

2\ HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2\ H_2O

  • Acid: HNO_3
  • Base: Ca(OH)_2

Let’s look at some actually interesting examples.

Example 1: Antacids and Stomach Acid

Too much stomach acid (HCl) can lead to a number of problems, such as acid reflex, stomach ulcers, gastritis (inflammation of the stomach lining). To combat excess stomach acid, we consume antacids.

Antacids are not the same as ant acid (known formally as formic acid). Surprisingly, antacids are not acids at all. Antacids are substances which neutralize stomach acidity and are used to relieve heartburn, indigestion, or an upset stomach. Marketed antacids contain salts of aluminum Al, calcium Ca, magnesium Mg, or sodium Na. Below are some examples of those salts.

1. Al(OH)_3

2. CaCO_3

3. Mg(OH)_2

4. NaHCO_3

When digested, antacids participate in acid-base reactions to neutralize our stomach acid.

Al(OH)_3 + 3\ HCl \rightarrow AlCl_3 + 3\ H_2O

CaCO_3 + 2\ HCl \rightarrow CaCl_2 + H_2O + CO_2\uparrow

Mg(OH)_2 + 2\ HCl \rightarrow MgCl_2 + 2\ H_2O

NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2\uparrow

Of those examples:

  • Acid: HCl
  • Base: Al(OH)_3, CaCO_3, Mg(OH)_2, NaHCO_3

Notice that in 2 of those reactions, we have CO_2 produced as a gas. This is because carbonate CO_3^{2-} and bicarbonate HCO_3^- ions, after protonation (gaining H^+), form carbonic acid H_2CO_3, which immediately dissociates into water H_2O and carbon dioxide CO_2. So, simply speaking:

CO_3^{2-} + 2\ H^+ \rightarrow H_2CO_3 \xrightarrow{\text{immediately}} H_2O + CO_2\uparrow
HCO_3^- + H^+ \rightarrow H_2CO_3 \xrightarrow{\text{immediately}} H_2O + CO_2\uparrow

Since the dissociation of H_2CO_3 is immediate, we won’t write H_2CO_3 as a product but H_2O + CO_2\uparrow. Also, because CO_2 is produced, we get a bit gassy after consuming some antacids, specifically the ones that contain carbonate CO_3^{2-} or bicarbonate HCO_3^-.

Example 2: Chlorine Bleach and Cleaners

Safety Warning: Do NOT mix bleach and cleaners!

Never mix bleach with other cleaners. The chemical reactions can produce hazardous gases or compounds, posing serious health risks. Always use cleaning agents separately and in well-ventilated areas. Let’s look at an example.

  • Most bleach is mostly composed of sodium hypochlorite NaOCl.
  • A common acid-based cleaner is hydrochloric acid HCl.

Mixing those 2 compounds will yield the following reaction.

NaOCl + 2\ HCl \rightarrow NaCl + H_2O + Cl_2\uparrow

  • Acid: HCl
  • Base: NaOCl

NaCl and H_2O are harmless. However, chlorine gas Cl_2 is produced in this reaction as well. Cl_2 – greenish-yellow in color – can cause coughing, difficulty breathing, and chemical burns to the respiratory tract and mucous membranes. In serious cases, it can be lethal.

There are other acid-based cleaners that do not use HCl. For example, sulfuric acid H_2SO_4, phosphoric acid H_3PO_4, and even vinegar CH_3COOH if we’re on a saving spree. However, the net ionic reaction is the same.

2\ OCl^- + 4\ H^+ \rightarrow 2\ H_2O + Cl_2\uparrow

Even with non-acid-based cleaners, we can still get very harmful substances from mixing those cleaners with bleach. The reactions are not acid-base, but it’s still nice to be aware of them. For example:

  • Ammonia-based cleaners: NaOCl + NH_3 \rightarrow NH_2Cl + NaOH. Chloramines (NH_2Cl, NHCl_2, NCl_3) are irritating to the eyes, nose, throat, and lungs.
  • Alcohol-based cleaners (like isopropyl alcohol C_3H_7OH): 3\ NaOCl + C_3H_7OH \rightarrow CHCl_3 + 3\ NaOH + C_2H_4O. Chloroform CHCl_3 is a sedative at high concentrations and can damage the liver and kidneys.

Example 3: Blood Buffers

Our blood needs to maintain a constant pH level – a quantity that measures the acidity of a solution. If our blood gets too acidic or too basic (the opposite of being acidic), our cells can get damaged. So, our blood has a system to regulate our blood pH levels.

This is called the bicarbonate buffer system of the blood because it relies on bicarbonate HCO_3^- and carbonic acid H_2CO_3. Simply speaking, they do the following:

  • When exposed to acidic substances (generally referred as H_3O^+):
    • Chemical equation: H_3O^+ + HCO_3^- \rightarrow H_2O + H_2CO_3
    • Acid: H_3O^+
    • Base: HCO_3^-
  • When exposed to basic substances (generally referred as OH^-):
    • Chemical equation: OH^- + H_2CO_3 \rightarrow H_2O + HCO_3^-
    • Acid: H_2CO_3
    • Base: OH^-

Notice when acidic or basic substances are introduced into our bloodstream, our bicarbonate buffer system can neutralize those substances to reduce the impact of those substances on our blood pH levels (again, pH is a quantity describing the acidity of a solution).

Also notice that, in the chemical equations above, we didn’t dissociate H_2CO_3 into H_2O and CO_2. This is because the environment of the blood favors a bit more H_2CO_3 compared to that of our previous examples.

We will revisit our blood buffering system again in Unit 3 and in Biochemistry.

Oxidation-Reduction Reactions

We endearingly call those reactions redox reactions. This type of reaction is also very important, and we will also discuss this reaction in detail in the future.

Redox reactions involve 1 thing – changes in oxidation state.

Here are some terminologies:

  • Oxidation: an increase in oxidation state.
  • Reduction: a decrease in oxidation state.
  • A substance is oxidized when its oxidation state is increased.
  • A substance is reduced when its oxidation state is decreased.
  • A substance is an oxidizing agent when it oxidizes others.
  • A substance is a reducing agent when it reduces others.

We’ll look at examples soon.

Similar to how donating and accepting protons in acid-base reactions can be boiled down to the concept of “proton transfers”, changes in oxidation state in redox reactions can be boiled down to the concept of “electron transfers” and “atom transfers (not just protons, but entire atoms)”. We will discuss this in detail in Unit 2. For now, let’s just ease ourselves into this concept.

Let’s look at some textbook examples. The diagrams below the reactions are redox tables. We will use them throughout this course. In our redox tables:

  • The chemical reaction is written as intended.
  • The oxidation state of an individual element is listed above said element.
  • The change in oxidation state of an individual element is listed below said element.
  • The roles the elements play in the redox reactions are below the changes in oxidation state.
    • “RED” stands for “reduced”.
    • “OX” stands for “oxidized”.
    • “RED agent” stands for “reducing agent”.
    • “OX agent” stands for “oxidizing agent”.

2\ Mg + O_2 \xrightarrow{\text{combustion}} 2\ MgO

Zn + CuSO_4 \rightarrow ZnSO_4 + Cu

Here is a harder textbook example:

Textbook Example: Dichromate and Iron (II) Ions

Observe the following reaction, which involves a dichromate ion Cr_2O_7^{2-} reacting with iron (II) ions under acidic conditions – implied by protons in the reactants.

Cr_2O_7^{2-} + 6\ Fe^{2+} + 14\ H^+ \rightarrow 2\ Cr^{3+} + 6\ Fe^{3+} + 7\ H_2O

From our previously established rule, we can calculate the oxidation state for Cr in Cr_2O_7^{2-}.

\text{Oxidation State of Cr in }Cr_2O_7^{2-} = +6

Using this information, we can fill in the blanks in our redox table. Arrows are added for clarity.

Hopefully, the textbook examples above gave us a good enough gist of what redox reactions are like. But, before we get into the more relatable examples, there’s something else we need to know about redox reactions.

  • Oxidation (increased oxidation state) also means the atom losing electrons.
  • Reduction (decreased oxidation state) also means the atom gaining electrons.

So, in the example above with the dichromate ions, chromium Cr was reduced because its oxidation state decreased, meaning that it gained electrons. The iron (II) ions were oxidized because its oxidation state increased, meaning that it lost electrons. In a way, the electrons were transferred from the iron (II) ions to the dichromate, forming iron (III) ions, chromium (III) ions, and water.

Now, with that out of the way, let’s look at some more relatable examples.

Example 1: Cellular Respiration

When we breathe, we are providing our cells with oxygen O_2 such that the following reaction can be carried out.

C_6H_{12}O_6(aq) + 6\ O_2(g) \xrightarrow{\text{Glycolysis, CAC, ETC}} 6\ CO_2(g) + 6\ H_2O(l)

  • C_6H_{12}O_6 is glucose.
  • Glycolysis, CAC (Citric acid cycle), and ETC (Electron transport chain) are the metabolic pathways involved in the chemical reaction above. We will talk about this in detail in biochemistry.

We will encounter this equation in different forms in biology and biochemistry because it serves as a fundamental biochemical concept – its metabolic logic can be found in other metabolic pathways, which can be highly useful in pharmaceutical development.

For this example, we will only look at it in the most overarching perspective – no details, just the entire reaction, from a redox point of view.

Let’s calculate the oxidation state of carbon in glucose together since it’s a big compound.

\text{Oxidation State of C in }C_6H_{12}O_6 = 0

Let’s write out the redox table.

So, in this example, from an electron-transport point of view, we can describe the following.

  • The carbon in glucose C_6H_{12}O_6 is oxidized, meaning that it lost electrons.
  • The oxygen in oxygen gas O_2 is reduced, meaning that it gained electrons.
  • In a way, electrons are transferred from glucose to oxygen, and through the transfer, glucose becomes CO_2 and O_2 becomes H_2O.

That is why we sometimes might hear words such as “oxygen is final electron acceptor in cellular respiration”.

Example 2: Rust

Iron can reacting with oxygen and water. This causes the iron to rust. Rusting involves a lot of chemical reactions between iron metal Fe, oxygen O_2, and water H_2O. Let’s look at an example.

In a general sense, rusting can be summarized into the following equation.

4\ Fe + 3\ O_2 + 6\ H_2O \rightarrow 4\ Fe(OH)_3

Let’s fill in the redox table.

After all this, Fe(OH)_3 participates in various other reactions to form a combination of FeO, Fe_2O_3, Fe(OH)_2, Fe(OH)_3, and FeO(OH), which we call rust.

In this example, we can also describe the following:

  • The iron metal Fe is oxidized, meaning that it lost electrons.
  • The oxygen gas O_2 is reduced, meaning that it gained electrons.
  • In a way, Fe transferred electrons to O_2, and alongisde H_2O, formed Fe(OH)_3 in the process.

Rust can be dangerous in terms of structural integrity of objects made with pure iron and in terms of our health as well. Interestingly, rust does not cause tetanus. However, the tetanus bacteria that is housed in rust cause tetanus. Because rusty objects are often found in dirt, it’s easy to house harmful bacteria inside those objects.

In the end, any open skin wound is susceptible to tetanus as long as the wound is exposed to tetanus bacteria. Fortunately, it can be prevented by vaccines.

More information about tetanus can be found at the end of the lecture.

Example 3: Reactive Oxygen Species

Reactive oxygen species (ROS) are highly reactive chemicals formed from diatomic oxygen O_2, water H_2O, and hydrogen peroxide H_2O_2.

Recall our cellular respiration example above (the metabolism of C_6H_{12}O_6). During part of the process, H_2O_2 (an ROS) is formed as a byproduct. Alongside oxygen and water, it can form other ROS such as:

Radical hydroxyl OH\cdot

Hydroperoxide O_2H

Superoxide O_2^-

ROS are intrinsic to cellular functioning, and are present at low and stationary levels in normal cells. However, when the levels are not controlled, it can cause irreversible damage to DNA as they oxidize and modify some cellular components and prevent them from performing their original functions.

Our cells have an enzyme called catalase that dissociates H_2O_2 into O_2 and H_2O as a mechanism to maintain low ROS levels to prevent cell damage.

2\ H_2O_2 \xrightarrow{\text{catalase}} 2\ H_2O + O_2

This reaction is interesting, and we will see why in our redox table.

Apparently, when H_2O_2 dissociates into water and oxygen, it is both oxidized and reduced. It may seem weird at first, but since the oxidation states change, it is still a redox reaction.

From an electron transfer perspective, we can imagine the following.

  • 1 oxygen atom in H_2O_2 is oxidized, meaning that it lost electrons.
  • 1 oxygen atom in H_2O_2 is reduced, meaning that it gain electrons.
  • In a way, 1 oxygen atom in H_2O_2 transferred electrons to another oxygen atom in the same molecule, forming H_2O and O_2 in the process (given the correct stoichiometry).

Break Time: 10 Minutes

Take a short break! This lecture is content-heavy.

In the meantime, enjoy this composition by Ben Johnston (1926-2019) – an American contemporary music composer, known for his use of just intonation. This excerpt below is his final (4th) movement of his final (10th) string quartet performed by the Kepler Quartet. His string quartets feature extensive uses of microtones (tones that fit between half notes/semitones) so if something feels “out of tune”, it’s actually not.

If the last few minutes of this excerpt sounds familiar, it’s because it’s from Londonderry Air – an Irish folk tune – from which its motifs can be heard in multiple other songs.


Lecture Content (Part 2)

Balancing Chemical Equations

Now that we have a more conceptual understanding of chemical reactions and equations, let’s learn how to balance an equation. Balancing equations is important because we want to satisfy the laws of physics – namely, the conservation of mass and charge between the reactants and the products. Moreover, having balanced equations can give us accurate empirical analyses and understanding of reactions. We probably don’t want to end up calculating a wrong dosage of medication.

Conservation of Mass

The number of atoms involved in the reactants should be the same in the products. If there are 4 oxygen atoms and 8 hydrogen atoms in the reactants, we should count 4 oxygen atoms and 8 hydrogen atoms in the products.

Conservation of Charge

The net charge in the reactants should be the same in the products. If the net charge of reactants is 2-, then the net charge of the products should be 2- as well.

  • Unbalanced equation: Fe + O_2 + H_2O \rightarrow Fe(OH)_3
    • Reason that it’s unbalanced: \text{2 }H\text{ at reactant} \ne \text{3 }H\text{ at product}
  • Balanced equation: 4\ Fe + 3\ O_2 + 6\ H_2O \rightarrow 4\ Fe(OH)_3
    • Reason that it’s balanced:
      • \text{4 }Fe\text{ at reactant} = \text{4 }Fe\text{ at product}
      • \text{12 }O\text{ at reactant} = \text{12 }O\text{ at product}
      • \text{12 }H\text{ at reactant} = \text{12 }H\text{ at product}
      • Net charge of reactants = Net charge of products (0=0)

Truth be told, there are already enough online resources available to balance equations for us. One site is WebQC (Web Quintessential Chemistry). If we scroll down, we can see that they also teach us how to balance equations.

https://www.webqc.org/balance.php

The reason we might want to know how to balance equations is not because we have to do it by hand in the future (even though it will be tested) but because it helps improve our own (not ChatGPT’s or others’) understanding of chemistry in general. Also, forming a skill in one area often helps us in unexpected ways in seemingly unrelated areas in life.

Among multiple methods to balance equations, we’ll talk about 2 methods: the algebraic method and the oxidation state method.

The Algebraic Method

It’s essentially an algorithm. With the right programming and coding knowledge, we can technically create our own chemical equation balancing app. Let’s see how we can do this with an example.

Introducing the chemical equation we have been using for a while: Iron rusting in its unbalanced form.

Fe + O_2 + H_2O \rightarrow Fe(OH)_3

  • Step 1: Assign the coefficients to each molecule involved in the reaction (bolded for convenience).

(12)   \begin{equation*}\mathbf{c_1}\ Fe + \mathbf{c_2}\ O_2 + \mathbf{c_3}\ H_2O \longrightarrow \mathbf{c_4}\ Fe(OH)_3\end{equation*}

  • Step 2: Write down a system of equations that satisfies the law of conservation of mass. In this example:
    • \text{Number of }Fe\text{ atoms in reactants} = \text{Number of }Fe\text{ atoms in products}
    • \text{Number of }O\text{ atoms in reactants} = \text{Number of }O\text{ atoms in products}
    • \text{Number of }H\text{ atoms in reactants} = \text{Number of }H\text{ atoms in products}

(13)   \begin{equation*}\left\{ \begin{array}{thing}1 \times c_1 &=& 1 \times c_4 \quad & \text{(# of Fe atoms)}\\ 2 \times c_2 + 1 \times c_3 &=& 3\times c_4 \quad & \text{(# of O atoms)}\\ 2 \times c_3 &=& 3 \times c_4 \quad & \text{(# of H atoms)} \end{array} \right.\end{equation*}

  • Step 3: Arbitrarily set c_1 = 1 and solve the equation.

(14)   \begin{equation*}\left\{ \begin{array}{thing}1 \times c_1 &=& 1 \times c_4\\2 \times c_2 + 1 \times c_3 &=& 3\times c_4\\2 \times c_3 &=& 3 \times c_4\end{array} \right.\quad\xrightarrow{\text{setting }c_1=1}\left\{ \begin{array}{thing}c_1 &=& 1\\c_2 &=& \frac{3}{4}\\c_3 &=& \frac{3}{2}\\c_4 &=& 1\end{array}\right.\end{equation*}

  • Step 4: If necessary, multiply the solutions altogether such that they are all integers.

(15)   \begin{equation*}\left\{ \begin{array}{thing}c_1 &=& 1\\c_2 &=& \frac{3}{4}\\c_3 &=& \frac{3}{2}\\c_4 &=& 1\end{array}\right.\quad\xrightarrow{\times4 \text{ on all terms}}\left\{ \begin{array}{thing}c_1 &=& 4\\c_2 &=& 3\\c_3 &=& 6\\c_4 &=& 4\end{array}\right.\end{equation*}

  • Step 5: Balance the equation.

4\ Fe + 3\ O_2 + 6\ H_2O \rightarrow 4\ Fe(OH)_3

  • Step 6 (Optional): Check our work by ensuring the adherence of the laws of conservation of mass and charge.
    • \text{4 }Fe\text{ at reactant} = \text{4 }Fe\text{ at product}
    • \text{12 }O\text{ at reactant} = \text{12 }O\text{ at product}
    • \text{12 }H\text{ at reactant} = \text{12 }H\text{ at product}
    • Net charge of reactants = Net charge of products (0=0)

Those are the 6 steps of balancing a chemical equation using the algebraic method. As we can see, it is highly algorithmic. It also directly reflects the law of conservation of mass. Here’s another example.

Example: Balance the following equation using the algebraic method.

C_8H_{18} + O_2 \xrightarrow{\text{combustion}} CO_2 + H_2O

This is the combustion of octane C_8H_{18}. The animation below details how we can balance this equation using the algebraic method. It’s fast but it’s intended to familiarize us with the flow of the algorithm rather than the details. More practice will be available in the lecture worksheet at the end.

There are 7 slides to this animation.

2\ C_8H_{18} + 25\ O_2 \xrightarrow{\text{combustion}} 16\ CO_2 + 18\ H_2O

The Oxidation State Method

Because a change in oxidation state signals a transfer in electrons, we want to make sure that the number of electrons donated is the same as the number of electrons accepted. In short, the essence of this method is ensuring the law of the conservation of charge is satisfied, which goes hand in hand with the algebraic method that ensures the law of conservation of mass is satisfied.

\text{total } e^- \text{ lost from oxidized substances} = \text{total } e^- \text{ gained from reduced substances}

Let’s use the same example as the algebraic method.

Fe + O_2 + H_2O \rightarrow Fe(OH)_3

  • Step 1: Create the redox table for this reaction.
  • Step 2: Calculate the transfer of electrons per molecule (instead of per atom) for the ones that have their oxidation states changed. We assume that oxidation state changes directly reflect electron transfer.
  • Step 3: Find the least common multiple (lcm) of the numbers in the electron transfer. This gives us the most simplified number of total electrons transferred during this reaction that satisfy the law of conservation of charge.

\mathbf{lcm}(-3,4) = 12

  • Step 4: Calculate the number of molecules needed for each substance to donate or accept that number of electrons. It is denoted by “multiplier” in the redox table.
  • Step 5: The multipliers (bolded) correspond to the coefficients of the respective substances. Using these, we can figure out the rest of the coefficients.

4\ Fe + 3\ O_2 + 6\ H_2O \rightarrow 4\ Fe(OH)_3

  • Step 6 (Optional): Check our work by ensuring the adherence of the laws of conservation of mass and charge.
    • \text{4 }Fe\text{ at reactant} = \text{4 }Fe\text{ at product}
    • \text{12 }O\text{ at reactant} = \text{12 }O\text{ at product}
    • \text{12 }H\text{ at reactant} = \text{12 }H\text{ at product}
    • Net charge of reactants = Net charge of products (0=0)

Just like the algebraic method, there are 6 steps to this method. Using this method, we also can notice how the electrons transfer between molecules. They will be quite important once we dive deeper into redox reactions.

Here is an example.

Example: Balance the same example equation using the oxidation state method.

C_8H_{18} + O_2 \xrightarrow{\text{combustion}} CO_2 + H_2O

Similarly, the animation below details how we can balance this equation using the oxidation state method. It’s fast but it’s intended to familiarize us with the flow of the process rather than the details. More practice will be available in the lecture worksheet at the end.

There are 7 slides to this animation.

2\ C_8H_{18} + 25\ O_2 \xrightarrow{\text{combustion}} 16\ CO_2 + 18\ H_2O

To summarize, the algebraic method is the algorithm – we’ll most likely get it right all the time, but it will take us some time with all the equation solving. The oxidation state method, on the other hand, is more of a shortcut – we might make a bit more mistakes with all the electron transfer calculations, but we save a good amount of time.

In the end, use whatever method that works the best.

Analyzing Chemical Reactions

We can analyze a reaction through meticulous observations – are there bubbles forming? does the color change? do we smell anything? etc. In this lecture, we will analyze chemical reactions from a calculation point of view. In other words, how can a reaction be analyzed through its chemical equation?

This is called stoichiometry – the relationships among the weights of reactants and products before, during, and following chemical reactions.

Limiting Reactants and Stoichiometry

Observe the following reaction (from our bleach and cleaner mix example).

NaOCl + 2\ HCl \rightarrow NaCl + H_2O + Cl_2\uparrow

What we can obtain from this chemical equation is that:

  • 1 part of NaOCl reacts with 2 parts of HCl to form 1 part of NaCl, 1 part of H_2O, and 1 part of Cl_2. The coefficients are called stoichiometric factors.

So, let’s consider the following logic.

  1. Suppose we mix \text{3 moles of } NaOCl and \text{3 moles of }HCl together.
  2. If we want all \text{3 moles of } NaOCl react, we actually need \text{6 moles of }HCl, but we don’t.
  3. In this example, HCl is our limiting reactant, and NaOCl is our excess reactant.
  4. As a result, what will happen is \text{1.5 moles of } NaOCl and \text{3 moles of }HCl will react together, leaving \text{1.5 moles of } NaOCl behind.

Similar to how we have the redox table, we can create a stoichiometric table, where we record relevant data of the reactants and products to make calculations more streamlined and convenient.

Using our example (3 moles of NaOCl and HCl each), we can fill in our stoichiometric table like so:

  • Step 1: Fill in the “Given” row with whatever is given to us in the prompt.
    • IF: the given values in the prompt is “mass” or “volume”, fill in the corresponding fields and convert them into moles. In those cases, relevant physical properties (molarity and density) should be given.
  • Step 2: Assume that all of the first substance will react, using the stoichiometric factors, fill in the “Expected” row.
  • Step 3: Compare the values between the “Given” and “Expected” row.
    • IF: No value in “Expected” is bigger than “Given”, then the first substance is the limiting reagent.
    • IF: One value in “Expected” is bigger than “Given”, then that corresponding substance is the limiting reagent.
    • IF: More than one value in “Expected” is bigger than “Given”, repeat Step 2, but assume the next substance will react completely.
  • Step 4: Fill in the “Reagent” row based on Step 3. “L” denotes the limiting reagent, while “E” denotes excess reagent(s).
The comparison is boxed (Expected 6 > 3 Given), indicating that it is a limiting reagent.

Knowing those, we can fill in the “Necessary stoichiometric calculations” field, which contains the rows “Moles”, “MM”, and “Mass”.

  • Step 5: Search up the molar masses of the substances on the Internet and fill those in.
  • Step 6: Fill in the “Moles” row using the limiting reagent as a reference.
  • Step 7: Calculate the mass using the formula m = nM.
The arrow indicates that the limiting reagent is used as a reference. Because this is a purely theoretical scenario, sig fig rules are not applied.

The “Conditional stoichiometric calculations” field needs to be filled if the substance is a liquid (density and volume) or solution (molarity and volume). If the substance is a gas, then we will use another set of calculations (the gas laws will be explained in Chapter 3).

Percent Yield

In theory, every reaction will proceed completely. However, in practice, a lot of complicated reasons will make our reaction yield lower than we would expect from pure calculations. To see how much we actually yield compared to our calculations, we have the percent yield.

(16)   \begin{equation*} % \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100% \end{equation*}

The yield can be the amount of substance (moles), mass, or volume. Simply take the one that makes the most sense as long as the actual yield and theoretical yield as expressed in the same unit.

Let’s look at an example.

Example: Synthesis of Sodium Benzoate.

Sodium benzoate C_6H_5COONa is used as a preservative and medicine. It can be used as the following.

  • Food and beverage preservatives.
  • Medication preservatives (mostly in cough syrup).
  • Treatment for urea cycle disorder.
  • Potential add-on treatment for schizophrenia.

Sodium benzoate is quite simple to synthesize in lab – we put benzoic acid C_6H_5COOH and sodium hydroxide NaOH together, and that’s it.

C_6H_5COOH + NaOH \rightarrow C_6H_5COONa + H_2O

Question: Suppose we have 198.2 grams of benzoic acid and excess NaOH. After the reaction, we yield 165.4 grams of sodium benzoate. What is our percent yield?

Response:

We can use our stoichiometric table for this. Since we have excess NaOH, benzoic acid will be our limiting reagent. The following animation describes how we calculate our percent yield using our stoichiometric table.

There are 5 slides to this animation.

Answer: The percent yield for sodium benzoate is 70.68%.


Assignments

1. Preview Questions

  • What is a chemical reaction and how do we describe it?
  • How can we convert between different units in chemistry?
  • How do we balance a chemical equation and why is it important?
  • What types of quantitative analyses can we carry out from a chemical reaction and how are they useful?

2. Lecture Worksheet (TBD)

The lecture worksheet is available as a pdf file below. Remember, practice makes perfect!

3. Further Reading

Kidney Stones: https://my.clevelandclinic.org/health/diseases/15604-kidney-stones
Gout: https://my.clevelandclinic.org/health/diseases/4755-gout
Tetanus: https://www.mayoclinic.org/diseases-conditions/tetanus/symptoms-causes/syc-20351625

You finished the lecture! It’s a long lecture, so let’s take a break for today – it takes time for our brain to fully absorb new materials. Don’t forget to review!


Image Attributions and Citations

Some images (with dots as background) are original creations using canva.com

Lecture Schedule