Question

Name each compound and calculate its formula mass. (a) \(\mathrm{PtO}_{2}\) (b) \(\mathrm{N}_{2} \mathrm{O}_{5}\) (c) \(\mathrm{Al}\left(\mathrm{ClO}_{3}\right)_{3}\) (d) \(\mathrm{PBr}_{5}\)

Short answer

(a) Platinum(IV) oxide, 227.08 amu; (b) Dinitrogen pentoxide, 108.02 amu; (c) Aluminum chlorate, 277.335 amu; (d) Phosphorus pentabromide, 429.97 amu.

Step-by-step solution

Naming the Compounds

Use appropriate nomenclature rules for each compound. For (a), the compound is called platinum(IV) oxide, as Pt is in the +4 oxidation state and O is oxide. For (b), the name is dinitrogen pentoxide, where 'di-' and 'penta-' denote the number of N and O atoms, respectively. For (c), the name is aluminum chlorate, where chlorate is the anion \( \(ClO_{3}\)^{-} \). For (d), the compound is called phosphorus pentabromide; 'penta-' indicates five bromine atoms bonded to phosphorus.

Calculating Formula Mass for \( \mathrm{PtO}_{2} \)

Calculate the atomic masses of all atoms and multiply by their respective counts in the molecule. Pt has an atomic mass of approximately 195.08 amu, and O has an atomic mass of approximately 16.00 amu. For \( \mathrm{PtO}_{2} \), the formula mass is \( 195.08 \, amu + 2 \times 16.00 \, amu = 227.08 \, amu \)

Calculating Formula Mass for \( \mathrm{N}_{2} \mathrm{O}_{5} \)

N has an atomic mass of approximately 14.01 amu, and O has an atomic mass of approximately 16.00 amu. For \( \mathrm{N}_{2} \mathrm{O}_{5} \), the formula mass is \( 2 \times 14.01 \, amu + 5 \times 16.00 \, amu = 108.02 \, amu \)

Calculating Formula Mass for \( \mathrm{Al}\left(\mathrm{ClO}_{3}\right)_{3} \)

Al has an atomic mass of approximately 26.98 amu, Cl has an atomic mass of approximately 35.45 amu, and O has an atomic mass of 16.00 amu. For \( \mathrm{Al}\left(\mathrm{ClO}_{3}\right)_{3} \), the formula mass is \( 26.98 \, amu + 3 \times \left( 35.45 \, amu + 3 \times 16.00 \, amu \right) = 277.335 \, amu \)

Calculating Formula Mass for \( \mathrm{PBr}_{5} \)

P has an atomic mass of approximately 30.97 amu, and Br has an atomic mass of approximately 79.90 amu. For \( \mathrm{PBr}_{5} \), the formula mass is \( 30.97 \, amu + 5 \times 79.90 \, amu = 429.97 \, amu \)

Key concepts

Formula Mass Calculation

Understanding formula mass is essential for anyone studying chemistry. Simply put, it's the sum of the atomic masses of all the atoms in a chemical formula. To calculate the formula mass, you should:

• First, identify each element present in the compound.
• Then, find the atomic mass of each element using the periodic table.
• Next, multiply the atomic mass of each element by the number of atoms of that element in the compound.
• Finally, sum all these values to get the total formula mass of the compound.

For instance, in the compound PtO₂, platinum (Pt) has an atomic mass of about 195.08 amu, and oxygen (O) is approximately 16.00 amu. The formula mass is calculated by adding the atomic mass of Pt with twice the atomic mass of O, resulting in a formula mass of 227.08 amu. This process is followed for each compound to determine its respective formula mass.

Nomenclature Rules

The process of naming chemical compounds is governed by a set of rules known as nomenclature rules. These ensure that each compound has a unique and universally accepted name. Some fundamental rules include:

• Use of prefixes (mono-, di-, tri-, etc.) to indicate the number of atoms of each element in the compound.
• For ionic compounds, the cation (positive ion) is named first, followed by the anion (negative ion).
• When a metal can have more than one oxidation state, the specific state is indicated in the name with Roman numerals.
• In covalent compounds, the more electronegative element is named last and gets an 'ide' suffix.

Applying these rules, for example, the compound N₂O₅ is named dinitrogen pentoxide, where 'di-' signifies two nitrogen atoms, and 'penta-' indicates five oxygen atoms.

Oxidation States

The oxidation state, often called oxidation number, is a crucial concept that describes the degree of oxidation of an atom in a chemical compound. It is an indicator of the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic.

Here's how to determine the oxidation states:

• For an elemental substance, the oxidation state is 0.
• The sum of oxidation states in a neutral compound must equal zero, whereas in an ion it should match the ion's charge.
• Some elements have more stable oxidation states that are commonly seen in compounds, such as oxygen generally having an oxidation state of -2.
• Using known oxidation states can help determine the less obvious ones in a compound.

For instance, when naming PtO₂ as platinum(IV) oxide, we understand that platinum has an oxidation state of +4 since oxygen's known oxidation state is -2 and there are two oxygen atoms, balancing the total charge to zero.

Chemical Compounds Naming

Naming chemical compounds involves combining the rules of nomenclature with the knowledge of oxidation states to give each compound a unique name. Compounds are named systematically to avoid ambiguity no matter who uses those names or where they are used.

For Ionic Compounds:

• Start with the name of the metal (cation) and then the name of the non-metal (anion) with an 'ide' suffix.
• If the metal has more than one possible charge, use Roman numerals to indicate its oxidation state.

For Molecular (Covalent) Compounds:

• Use prefixes to denote the number of atoms of each element.
• The more electropositive element is named first using the element's name, followed by the more electronegative element with an 'ide' ending.

For example, the compound Al(ClO₃)₃ is named aluminum chlorate, following the rules for the naming of ionic compounds with polyatomic ions. The prefix 'chlor-' stems from the element chlorine, and '-ate' indicates that it's an oxyanion with oxygen in the formula.