Question

Give the name and calculate the molar mass for each of the following substances. a. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) b. \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(\mathrm{P}_{4} \mathrm{O}_{6}\) d. \(\mathrm{Bi}_{2} \mathrm{O}_{3}\) e. \(C S_{2}\) f. \(\mathrm{H}_{2} \mathrm{SO}_{3}\)

Short answer

The molar masses of the given substances are as follows: a. Cr₂O₃: 151.99 g/mol b. Cu(NO₃)₂: 187.56 g/mol c. P₄O₆: 123.90 g/mol d. Bi₂O₃: 465.96 g/mol e. CS₂: 76.14 g/mol f. H₂SO₃: 82.07 g/mol

Step-by-step solution

Identify the number of atoms of each element in the given substance

For each given substance, we will determine the number of atoms of each element included. a. Cr₂O₃ : There are 2 Chromium (Cr) atoms and 3 Oxygen (O) atoms. b. Cu(NO₃)₂ : There is 1 Copper (Cu) atom, 2 Nitrogen (N) atoms, and 6 Oxygen (O) atoms. c. P₄O₆ : There are 4 Phosphorus (P) atoms and 6 Oxygen (O) atoms. d. Bi₂O₃ : There are 2 Bismuth (Bi) atoms and 3 Oxygen (O) atoms. e. CS₂ : There is 1 Carbon (C) atom and 2 Sulfur (S) atoms. f. H₂SO₃ : There are 2 Hydrogen (H) atoms, 1 Sulfur (S) atom, and 3 Oxygen (O) atoms.

Calculate the molar mass of each substance

To calculate the molar mass of each substance, we multiply the atomic mass of each element by the number of its atoms in the substance, and then sum up the individual molar masses. a. Cr₂O₃ Molar mass = (2 × 51.996 g/mol Cr) + (3 × 16.00 g/mol O) = 151.99 g/mol b. Cu(NO₃)₂ Molar mass = (1 × 63.546 g/mol Cu) + (2 × 14.01 g/mol N) + (6 × 16.00 g/mol O) = 187.56 g/mol c. P₄O₆ Molar mass = (4 × 30.974 g/mol P) + (6 × 16.00 g/mol O) = 123.90 g/mol d. Bi₂O₃ Molar mass = (2 × 208.980 g/mol Bi) + (3 × 16.00 g/mol O) = 465.96 g/mol e. CS₂ Molar mass = (1 × 12.01 g/mol C) + (2 × 32.07 g/mol S) = 76.14 g/mol f. H₂SO₃ Molar mass = (2 × 1.008 g/mol H) + (1 × 32.07 g/mol S) + (3 × 16.00 g/mol O) = 82.07 g/mol

Key concepts

Chemical Compounds

Chemical compounds are substances consisting of two or more different elements that are chemically bonded together. These compounds are categorized based on the types of bonds they have, such as ionic or covalent bonds.
In this exercise, we examine a variety of chemical compounds like Chromium(III) oxide (\(\mathrm{Cr}_{2}\mathrm{O}_{3}\)) and Sulfur dioxide (\(\mathrm{H}_{2}\mathrm{SO}_{3}\)).
Understanding the composition of these compounds requires identifying the different elements present and determining how many atoms of each element are involved.
Some compounds have simple structures, such as \(\mathrm{CS}_{2}\), with only carbon and sulfur atoms.
Others, like \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\), have more complex structures involving polyatomic ions like nitrate.

Atomic Mass

The atomic mass of an element is a weighted average of the masses of its isotopes, measured in atomic mass units (amu).
It reflects the distribution of the element's isotopes and their respective abundances. Calculating atomic mass is crucial when determining the molar mass of compounds.
To ensure accurate calculations, it's important to use precise atomic mass values, such as 51.996 amu for Chromium (Cr) and 16.00 amu for Oxygen (O).
In practice, atomic masses are multiplied by the number of atoms of each element in the compound. For example:- Chromium in \(\mathrm{Cr}_{2}\mathrm{O}_{3}\): 2 atoms of Cr \(\times\) 51.996 amu.- Oxygen in nitrates: 6 atoms of O \(\times\) 16.00 amu in \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\).
Understanding atomic mass helps bridge the microscopic world of atoms with the macroscopic world of grams and moles.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It involves using the ratios of elements in compounds to determine their proportional compositions.
In this exercise, stoichiometry becomes particularly relevant when we calculate molar masses. By understanding stoichiometric relationships, we can predict how much of each element is needed or produced in a reaction.
In simpler terms, stoichiometry is about balancing the number of each type of atom on both sides of a chemical equation.
While the exercise involves individual compounds, the principles of stoichiometry remain the same:

• Recognize the proportion of each element.
• Use these proportions to calculate molar masses.
• Apply these concepts to solve real-world chemical problems, such as reaction yields and limiting reactants.

Mastering stoichiometry empowers you to decode the quantitative aspects of chemistry beyond just identifying compounds.

Elemental Composition

Understanding the elemental composition of a chemical compound means identifying which elements are present and in what quantities.
In our examples, this involves recognizing the individual contributions of elements like Oxygen, Sulfur, and Bismuth in compounds such as \(\mathrm{Bi}_{2}\mathrm{O}_{3}\) and \(\mathrm{H}_{2}\mathrm{SO}_{3}\).
Elemental composition is usually expressed in terms of the number of atoms of each element present in a compound.
For instance:

• \(\mathrm{CS}_{2}\) consists of 1 atom of Carbon and 2 atoms of Sulfur.
• \(\mathrm{P}_{4}\mathrm{O}_{6}\) has 4 atoms of Phosphorus and 6 atoms of Oxygen.

Appreciating elemental composition aids in predicting the chemical and physical properties of compounds, as these depend heavily on the types and quantities of atoms involved.