Unit 4 Study Guide: Chemical Formulas and Equations

4a: List the rules for assigning most common electronic charge states (oxidation numbers) of compounds or elements.

Oxidation numbers are used to assign the most common electronic charge for a given element.

    • How do you assign oxidation numbers for a given element?

Review the section Determining Oxidation States, in Oxidation Numbers and Redox Reactions for a list of rules for assigning oxidation numbers.

Here are the general rules:

    1. The sum of oxidation states in a molecule must equal the molecule’s charge. The sum is zero for a neutral molecule.
    2. The atom that is more electronegative in a bond gets a negative oxidation state. The atom that is more electropositive gets a positive oxidation state.
    3. Certain elements always have the same oxidation state.

Review the table Determining Oxidation States, in Oxidation Numbers and Redox Reactions.


4b: Give oxidation numbers for each element in the formula of a compound.

Determining the oxidation number of elements in a formula gives us important information about the bonding and reactivity of the compound.

    • Determine oxidation states of all atoms in a compound.

Here, we use the rules listed in the previous learning outcome to assign oxidation numbers for each atom in the compound. It is important to apply these rules systematically and be sure to note any exceptions to the rules that may apply.

Review Oxidation Numbers and Redox Reactions.

For an example, let’s explore the nitrate ion NO3, which is in the text.

    • We know this compound is an ion with a -1 charge, so the overall oxidation numbers must add up to -1.
    • We know oxygen is more electronegative than nitrogen, so we know oxygen will have a negative oxidation number and nitrogen will have a positive one.
    • We know oxygen almost always has a -2 oxidation state (with a few exceptions).
    • Since this is not one of the exceptions, the oxidation state for each oxygen is -2.
    • Now we can determine the oxidation state of nitrogen.
    • There are three oxygen atoms, each with an oxidation state of -2.
    • So, the total oxidation state from the three oxygen atoms is -6.
    • To get the overall compound oxidation state to be -1, the oxidation state of nitrogen must be +5.


4c: Explain the significance of a chemical formula.

The chemical formula specifies the types of atoms in a chemical compound, and the number of each type of atom in the compound. The chemical formula defines the compound.

    • Given a chemical formula, determine the number of each type of atom in the compound.
    • Use the chemical formula to determine the molecular mass.
    • Use the chemical formula to determine the mole ratios of atoms in the compound.
    • Use the chemical formula to determine the percent composition of the compound.

For example, let’s look at ethanol, C2H6O, which is the type of alcohol found in alcoholic beverages. The formula tells us ethanol has two carbons, six hydrogens, and one oxygen.

When we know the chemical formula for a compound, we can determine its molecular mass, and molar mass.

Molecular mass, or molecular weight, is the mass of the compound in atomic mass units (amu). This is also called formula mass or formula weight. We can use the chemical formula to determine molecular mass by adding up the atomic masses of all atoms in the compound.

Review Example 4, in Chemical Formulas and their Arithmetic.

The chemical formula for a compound also allows us to calculate the mole ratios of elements for the compound. The atomic ratios in a formula are also the mole ratios of the atoms in the formula.

For example, in methane, CH4, there are four hydrogen atoms for every carbon atom. There are also four moles of hydrogen for every one mole of carbon.

Review Examples 5 and 6, in Chemical Formulas and their Arithmetic.

We can also use the chemical formula to determine the mass fraction or percent composition of the elements in a given compound.

Review Examples 8, 9, 10 and 11, in Chemical Formulas and their Arithmetic.


4d: Determine the formula of an ionic compound when given the name.

    • Know the names and charges of the ions listed in the text.
    • Use the names and charges of the ions to determine the formula of an ionic compound when given the name.

Before you review ionic compound formulas, review Section 4: Naming the Chemical Ions, in Naming Chemical Substances. It is essential to know the names and charges of the ions listed here to be able to name ionic compounds.

To determine the formula of an ionic compound from the name, you must have a strong command of the names and charges of single atom ions and polyatomic ions. It may be helpful to create flashcards of the ion symbols, with their charge and names, to learn them.

When given a chemical name for an ionic compound, the first name is the cation, or positive ion and the second name is the anion, or negative ion.

First, write the formula of the cation, including charge, and then write the formula of the anion, including charge. For cations that can have different charges, the charge will be written as a Roman numeral in parentheses. Then, balance the charges. In other words, make sure the positive charge equals the negative charge in the compound. To do this, you may need to alter the number of each type of ion.

For example, let’s look at the chemical copper (II) chloride. The cation is copper, and we are told it has a +2 charge by the (II). Therefore, the cation is: Cu2+. The anion is chloride, which is Cl. Now, balance the charges. There is a +2 charge from the cation and a -1 charge from the anion. Therefore, we need two chloride ions to get a -2 charge. The formula for copper (II) chloride is CuCl2.

Review more examples in salts in Section 5: Names of Ion-Derived Compounds, in Naming Chemical Substances.


4e: Name an ionic compound when given a formula.

    • Know the names and charges of the ions listed in the text.
    • Given a formula of an ionic compound, write the name.

Before you review ionic compound formulas, review Section 4: Naming the Chemical Ions, in Naming Chemical Substances. It is essential to know the names and charges of the ions listed here to be able to name ionic compounds.

Look at the formula to determine the name of an ionic compound. The first ion listed is the cation, or positive ion. The second ion listed is the anion, or negative ion. From the formula, write the name of the cation. Then, from the formula, write the name of the anion.

If the cation can have different charges, we must write the charge of the anion in the formula name. From the formula, determine the total negative charge from the anions. Then, determine the charge of the cation needed to balance out the total negative charge from the anions. Write the charge of the cation in Roman numerals in parentheses after the name of the cation.

For example, let’s look at Fe2S3. The cation is iron and the anion is sulfide. Iron can have different charges, so we will need to write the cation charge in the name. The sulfide ion has a -2 charge. However, there are 3 sulfide ions in this compound. Therefore, the total anion charge is -6. The cation charge must balance this with a total +6 charge. There are 2 iron ions in this compound, so each iron ion must have a charge of +3. The name for this compound is iron (III) sulfide.

Review more examples in salts in Section 5: Names of Ion-Derived Compounds, in Naming Chemical Substances.


4f: Name binary molecular compounds using prefixes.

Many molecular compounds are binary, which means they consist of two types of atoms.

    • Know the prefixes for binary molecular compounds.
    • Given a formula, write the correct binary molecular compound name.
    • Given a name, write the correct binary molecular compound formula.

See the chart of numerical prefixes in Section 3: Naming the Binary Compounds, in Naming Chemical Substances. Consider making flashcards to help you memorize the numerical prefixes and their numerical values.

To write the name of a binary molecular compound from the formula, you need to know the names of the atoms involved. Write the names of the elements in the order they appear in the formula. The second element should end in ide rather than the element name. Sometimes these are polyatomic ions rather than elements. Then, add the numerical prefixes from the chart to the names of the elements.

For example, we can name P4S3. The first element is phosphorus and the second element is sulfur. We change the second element to sulfide. Now, we add the numerical prefixes. There are four phosphorus atoms so it is tetraphosphorus. There are three sulfur atoms so it is trisulfide. The name of the molecule is tetraphosphorus trisulfide.

Given a name of a binary compound we can determine the formula. Consider dinitrogen tetroxide. Here, we use the prefixes and element names to determine the chemical formula. The first element is nitrogen, and from the di– prefix we know there are two nitrogens. The second element is oxygen, and from the tetra– prefix we know there are four oxygens. Therefore, the formula is N2O4.

Review more examples in Section 3: Naming the Binary Moleculesin Naming Chemical Substances.


4g: Balance a chemical equation.

Chemical equations express the net change of composition that occur during a chemical change. Understanding how chemists write chemical reactions, is an important part of the language of chemistry.

    • Identify reactants and products in a chemical reaction.
    • Explain why chemical reactions must be balanced.
    • Balance a given unbalanced chemical equation.

In a chemical reaction, reactants are transformed into products. It is convention in chemistry that we write the reactants on the left side of the equation and the products on the right.

We write an arrow going from reactants to products to signify the chemical change:

Reactants → Products

We need to balance a chemical equation to comply with the Law of Conservation of Mass, which states that matter (mass) in a chemical reaction must be conserved. This means you cannot make or lose mass during a chemical reaction. Balancing a chemical equation ensures the amount of reactants equals the amount of products. You need to ensure an equal number of each type of atom appears on both sides of the equation (reactant and product).

Review balancing a chemical equation in Problem Examples 1 and 2, in Chemical Equations and Calculations.

See the blue box on the left of Examples 1 and 2 to watch YouTube videos that show how to balance chemical equations in Chemical Equations and Calculations.

Let’s examine the reaction in Problem Example 1: the combustion of propane, C3H8. The unbalanced chemical equation for this reaction is:

C3H8+ O2 → H2O + CO2

To begin, you should tally up the number of each type of atom on each side of the equation.

    • Reactant side: three carbons (C), eight hydrogens (H), two oxygens (O);
    • Product side: one carbon (C), two hydrogens (H), and three oxygens (O).

Then, add whole number coefficients to the molecules to ensure the number of each type of atom on each side of the equation is equal. Note you can only alter the number of molecules—you cannot change the formulas of the molecules by changing the number of individual atoms in the molecule.

Let’s start by balancing carbon. To balance carbon, put a coefficient of three in front of CO2 in the products. Then re-tally the atom count.

C3H8+ O2 → H2O + 3CO2

Reactants: 3 C, 8 H, 2 O; Products, 3 C, 2 H, 7 O

*Note that since CO2 has two oxygen atoms, the three CO2 molecules have six oxygen atoms. There is also an oxygen in the water in the products.

Next, repeat the process of balancing a different atom and calculating a new atom tally. Continue until the number of each type of atom on the reactant side is the same as the product side.

Secondly, let’s balance hydrogen. Since eight hydrogen atoms are on the reactant side and two are on the product side, you should put a coefficient of four in front of the water to make eight hydrogens on the product side.

C3H8+ O2 → 4H2O + 3CO2

Reactants: 3 C, 8 H, 2 O; Product: 3 C, 8 H, 10 O

Finally, let’s balance oxygen. Since two oxygen atoms are on the reactant side and 10 are on the product side, you should put a coefficient of five in front of the oxygen on the reactant side to balance.

C3H8+ 5O2 → 4H2O + 3CO2

Reactants: 3 C, 8 H, 10 O; Products: 3 C, 8 H, 10 O

The equation is balanced.


Unit 4 Vocabulary

      • Balancing a chemical equation
      • Binary molecular compound
      • Chemical equation
      • Chemical formula
      • Formula mass/formula weight
      • Law of Conservation of Mass
      • Molar mass
      • Molecular mass/molecular weight
      • Oxidation number
      • Polyatomic ion
      • Product
      • Reactant
Last modified: Wednesday, July 17, 2019, 6:06 PM