Problem 34 - Chemistry The Central Science

Question

Calculate the following quantities: (a) mass, in grams, of \(1.50 \times 10^{-2} \mathrm{~mol}\) of CdS (b) number of moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(86.6 \mathrm{~g}\) of this substance (c) number of molecules in \(8.447 \times 10^{-2} \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{6}\) (d) number of \(\mathrm{O}\) atoms in \(6.25 \times 10^{-3} \mathrm{~mol} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\)

Step-by-Step Solution

Answer

(a) The mass of CdS is approximately \(2.17 \mathrm{~g}\). (b) There are approximately \(1.62 \mathrm{~mol}\) of NH4Cl. (c) There are approximately \(5.08 \times 10^{22}\) molecules of C6H6. (d) There are approximately \(1.13 \times 10^{23}\) O atoms in Al(NO3)3.

Step 1: (a) Calculate the mass of CdS

To calculate the mass of CdS, we need to use its molar mass. The molar mass of a compound is the total mass of all the elements in the compound added together. For CdS, this will be the sum of the molar mass of cadmium (Cd) and sulfur (S). From the periodic table, we have: Molar mass of Cd = 112.41 g/mol Molar mass of S = 32.07 g/mol The molar mass of CdS = 112.41 + 32.07 = 144.48 g/mol Now, multiply the moles of CdS by the molar mass of CdS to find the mass: Mass of CdS = 1.50 × 10^{-2} mol × 144.48 g/mol ≈ 2.17 g

Step 2: (b) Calculate the moles of NH4Cl

To find the moles of NH4Cl given the mass, we need to use the molar mass of NH4Cl. The molar mass of NH4Cl is the sum of the molar masses of its elements: nitrogen (N), hydrogen (H), and chlorine (Cl). From the periodic table, we have: Molar mass of N = 14.01 g/mol Molar mass of H = 1.01 g/mol Molar mass of Cl = 35.45 g/mol The molar mass of NH4Cl = 14.01 + 4 × 1.01 + 35.45 ≈ 53.5 g/mol Now, divide the mass of NH4Cl by the molar mass to find the moles: Moles of NH4Cl = 86.6 g / 53.5 g/mol ≈ 1.62 mol

Step 3: (c) Calculate the number of molecules of C6H6

To calculate the number of molecules of C6H6, we will use Avogadro's number, which is approximately 6.022 × 10^23. This is the number of units (molecules, atoms, etc.) in one mole. Multiply the moles of C6H6 by Avogadro's number to find the number of molecules: Number of molecules of C6H6 = 8.447 × 10^{-2} mol × 6.022 × 10^23/mol ≈ 5.08 × 10^22 molecules

Step 4: (d) Calculate the number of O atoms in Al(NO3)3

First, we need to find the number of moles of O atoms in one mole of Al(NO3)3. In one molecule of Al(NO3)3, there are three molecules of NO3, and in each molecule of NO3, there is one O atom. Therefore, there are three O atoms in one molecule of Al(NO3)3. Now we can calculate the number of O atoms in 6.25 × 10^{-3} mol of Al(NO3)3 using Avogadro's number: Number of O atoms = 6.25 × 10^{-3} mol × 6.022 × 10^23 O atoms/mol × 3 ≈ 1.13 × 10^23 O atoms

Key Concepts

Molar Mass Calculation

Molar mass is a foundational concept in stoichiometry. It refers to the mass of one mole of a substance, typically in grams per mole (g/mol). The molar mass can be calculated by adding the atomic masses of each element in the molecule, which are sourced from the periodic table.

For instance, when calculating the molar mass of water (H2O), we add the atomic mass of hydrogen (H) multiplied by 2, since there are two hydrogen atoms, to the atomic mass of oxygen (O):

• Molar mass of H = 1.01 g/mol
• Molar mass of O = 16.00 g/mol
• Molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol

Understanding molar mass is crucial as it bridges the gap between the macroscopic world of grams and the microscopic world of molecules and atoms.

Avogadro's Number

Avogadro's number is a constant that represents the number of atoms, ions, or molecules in one mole of any substance. It is approximately 6.022 x 10^23 units per mole. This immense number allows chemists to count individual entities by weighing large numbers of atoms or molecules in a way that can be measured with standard laboratory equipment.

For example, when you have 1 mole of carbon atoms, you essentially have 6.022 x 10^23 carbon atoms. This concept simplifies the transition from the world of individual particles to macroscopic quantities in grams, enabling the use of a laboratory balance for measurements.

Moles to Mass Conversion

Converting moles to mass and vice versa is vital in stoichiometry for it allows the quantification of substances involved in chemical reactions. To convert moles to mass, multiply the number of moles by the molar mass of the substance:

• Mass (g) = Moles (mol) x Molar Mass (g/mol)

Conversely, to convert mass to moles, divide the mass of the substance by its molar mass:

• Moles (mol) = Mass (g) / Molar Mass (g/mol)

These conversions are necessary for creating balanced chemical equations, determining reactant quantities, and calculating product yields.

Molecular Formulas

Molecular formulas convey the exact number and type of atoms in a single molecule of a compound. They are critical to understanding the composition of molecules and predicting the stoichiometry of reactions. For example, the molecular formula for glucose is C6H12O6, indicating that each molecule of glucose contains 6 carbon, 12 hydrogen, and 6 oxygen atoms.

Molecular formulas can also be derived from empirical formulas, which give the simplest whole-number ratio of atoms in a compound. When empirical and molecular formulas differ, the molecular formula is a multiple of the empirical formula. Knowledge of molecular formulas supports calculations of molar masses and helps predict the outcome of chemical reactions.