Problem 23 - Chemistry: The Central Science in SI Units, Global Edition

Question

Determine the formula weights of each of the following compounds: (a) lead (IV) chloride; (b) copper(II) oxide; (c) iodic acid, \(\mathrm{HIO}_{3} ;(\mathbf{d})\) sodium perchlorate, \(\mathrm{NaClO}_{4} ;\) (e) indium nitride, (f) phosphorus pentoxide, \(\mathrm{P}_{4} \mathrm{O}_{10} ;(\mathbf{g})\) boron trichloride.

Step-by-Step Solution

Answer

The formula weights for the given compounds are: (a) Lead(IV) chloride: \(349 \, g/mol\) (b) Copper(II) oxide: \(79.546 \, g/mol\) (c) Iodic acid: \(175.912 \, g/mol\) (d) Sodium perchlorate: \(122.44 \, g/mol\) (e) Indium nitride: \(128.825 \, g/mol\) (f) Phosphorus pentoxide: \(283.896 \, g/mol\) (g) Boron trichloride: \(117.17 \, g/mol\)

Step 1: (1) Lead(IV) chloride#

: First, we will write the chemical formula for lead(IV) chloride. The Roman numeral (IV) indicates the oxidation state of lead, which is +4. Chlorine has an oxidation state of -1. Thus, we need to balance the charges to get PbCl4. Now, we will calculate the formula weight: \(PbCl_4 = 1 \times (Pb) + 4 \times (Cl) \\ = 1 \times 207.2 + 4 \times 35.45 \\ = 207.2 + 141.8 \\ = 349 \, g/mol\)

Step 2: (2) Copper(II) oxide#

: First, we will write the chemical formula for copper(II) oxide. The Roman numeral (II) indicates the oxidation state of copper, which is +2. Oxygen has an oxidation state of -2. Thus, we have CuO. Now, we will calculate the formula weight: \(CuO = 1 \times (Cu) + 1 \times (O) \\ = 1 \times 63.546 + 1 \times 16 \\ = 63.546 + 16 \\ = 79.546 \, g/mol\)

Step 3: (3) Iodic acid (\(HIO_3\))#

: The chemical formula for iodic acid is given as \(HIO_3\). We need to calculate its formula weight: \(HIO_3 = 1 \times (H) + 1 \times (I) + 3 \times (O) \\ = 1 \times 1.008 + 1 \times 126.904 + 3 \times 16 \\ = 1.008 + 126.904 + 48 \\ = 175.912 \, g/mol\)

Step 4: (4) Sodium perchlorate (\(NaClO_4\))#

: The chemical formula for sodium perchlorate is given as \(NaClO_4\). We need to calculate its formula weight: \(NaClO_4 = 1 \times (Na) + 1 \times (Cl) + 4 \times (O) \\ = 1 \times 22.99 + 1 \times 35.45 + 4 \times 16 \\ = 22.99 + 35.45 + 64 \\ = 122.44 \, g/mol\)

Step 5: (5) Indium nitride#

: First, we write the chemical formula for indium nitride. Indium typically has an oxidation state of +3 and nitrogen has an oxidation state of -3, so we have InN. Now, we will calculate the formula weight: \(InN = 1 \times (In) + 1 \times (N) \\ = 1 \times 114.818 + 1 \times 14.007 \\ = 114.818 + 14.007 \\ = 128.825 \, g/mol\)

Step 6: (6) Phosphorus pentoxide (\(P_4O_{10}\))#

: The chemical formula for phosphorus pentoxide is given as \(P_4O_{10}\). We need to calculate its formula weight: \(P_4O_{10} = 4 \times (P) + 10 \times (O) \\ = 4 \times 30.974 + 10 \times 16 \\ = 123.896 + 160 \\ = 283.896 \, g/mol\)

Step 7: (7) Boron trichloride#

: First, we write the chemical formula for boron trichloride. Boron typically has an oxidation state of +3 and chlorine has an oxidation state of -1, so we have BCl3. Now, we will calculate the formula weight: \(BCl_3 = 1 \times (B) + 3 \times (Cl) \\ = 1 \times 10.81 + 3 \times 35.45 \\ = 10.81 + 106.35 \\ = 117.17 \, g/mol\) Formula weights for the given compounds are as follows: (a) Lead(IV) chloride: 349 g/mol (b) Copper(II) oxide: 79.546 g/mol (c) Iodic acid: 175.912 g/mol (d) Sodium perchlorate: 122.44 g/mol (e) Indium nitride: 128.825 g/mol (f) Phosphorus pentoxide: 283.896 g/mol (g) Boron trichloride: 117.17 g/mol

Key Concepts

Oxidation States

Oxidation states, also known as oxidation numbers, play a crucial role in understanding chemical reactions and formulas. An oxidation state is a theoretical charge that an atom would have if all bonds were completely ionic. This helps chemists keep track of how electrons are transferred in reactions. In many compounds, especially transition metals, you will see Roman numerals in names like lead(IV) chloride. These indicate the oxidation state of the metal. For instance, the (IV) in lead(IV) chloride means lead has an oxidation state of +4. Knowing this helps determine the chemical formula of a compound because the sum of the oxidation states in a neutral compound must be zero.

• For example, lead(IV) chloride, with lead at +4 and each chlorine at -1, combines to form PbCl₄.
• In copper(II) oxide, the copper's +2 counteracts the oxygen's -2, resulting in the formula CuO.

Understanding oxidation states is key in deducing the correct chemical formulas needed for precise formula weight calculations.

Chemical Formulas

Chemical formulas provide a shorthand way of presenting the elements and their proportions in a compound. They are essential for calculating formula weights, which give us the mass of the compound. A chemical formula like PbCl₄ provides a lot of information:

• It shows the types and numbers of atoms in the compound.
• It reflects the stoichiometry of the compound, or how atoms combine based on their oxidation states.

To write a chemical formula correctly, you must first know the oxidation states of the involved elements. This allows you to determine the necessary number of each atom to balance the charges. For example: - In PbCl₄, lead has a +4 charge, and each chlorine has a -1 charge, so four chlorine atoms are needed to neutralize one lead atom. - For phosphorus pentoxide, P₄O₁₀, the formula tells us that four phosphorus atoms combine with ten oxygen atoms. Understanding and writing chemical formulas are fundamental skills in studying chemical reactions and properties.

Molar Mass

Molar mass, often synonymous with formula weight, is the mass in grams of one mole of a substance. It is calculated by summing the atomic weights of all atoms in a compound's chemical formula.For example, when finding the molar mass of copper(II) oxide (CuO), we use the atomic weights of copper (63.546 g/mol) and oxygen (16 g/mol).

• Formula: CuO
• Molar Mass Calculation:\[ CuO = 63.546 + 16 = 79.546 \, g/mol \]

The molar mass is critical in chemistry for converting between moles and grams, a common task in lab work and chemical equations. Here's how molar mass connects to other concepts:- It reinforces the importance of correctly applying oxidation states to derive accurate chemical formulas.- Precise chemical formulas ensure you calculate the right molar mass, crucial when measuring out reagents or predicting reaction yields.In conclusion, knowing how to calculate and interpret molar mass is vital for anyone studying chemistry. It links the microscopic world of atoms to measurable quantities in the lab.