Problem 36 - Chemistry: The Central Science in SI Units, Global Edition

Question

Calculate the following quantities: (a) mass, in grams, of \(1.50 \times 10^{-2} \mathrm{~mol} \mathrm{CdS}\) (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 \(1.50 \times 10^{-2}\) mol CdS is 2.167 g. (b) The number of moles of NH4Cl in 86.6 g is 1.62 mol. (c) The number of molecules in \(8.447 \times 10^{-2}\) mol C6H6 is \(5.08 \times 10^{22}\) molecules. (d) The number of O atoms in \(6.25 \times 10^{-3}\) mol Al(NO3)3 is \(3.39 \times 10^{22}\) O atoms.

Step 1: (a) Calculate the mass of \(1.50 \times 10^{-2}\) mol CdS

To calculate the mass of \(1.50 \times 10^{-2}\) mol of CdS (cadmium sulfide), we'll first determine the molar mass of CdS. The molar mass of CdS = (molar mass of Cd) + (molar mass of S) = \(112.41 \mathrm{~g/mol}\) + \(32.07 \mathrm{~g/mol}\) = \(144.48 \mathrm{~g/mol}\) Now, we use the equation \(m = n \times M\) to calculate the mass of CdS. mass = \((1.50 \times 10^{-2} \mathrm{~mol~CdS}) \times (144.48 \mathrm{~g/mol}) = 2.167 \mathrm{~g~CdS}\)

Step 2: (b) Calculate the number of moles of NH4Cl in 86.6 g of this substance.

Now, we will determine the number of moles of NH4Cl (ammonium chloride) in 86.6 g of the substance. Molar mass of NH4Cl = (molar mass of N) + 4 × (molar mass of H) + (molar mass of Cl) = \(14.01 \mathrm{~g/mol}\) + 4 × \(1.01 \mathrm{~g/mol}\) + \(35.45 \mathrm{~g/mol}\) = \(53.49 \mathrm{~g/mol}\) Using the formula \(n = m/M\), we can compute the number of moles of NH4Cl. number of moles = \(\frac{86.6 \mathrm{~g}}{53.49 \mathrm{~g/mol}} = 1.62 \mathrm{~mol~NH4Cl}\)

Step 3: (c) Calculate the number of molecules in \(8.447 \times 10^{-2}\) mol C6H6.

To find the number of molecules in \(8.447 \times 10^{-2}\) mol of C6H6 (benzene), we use the formula \(n_\text{particles} = n \times N_A\). Number of molecules = \((8.447 \times 10^{-2} \mathrm{~mol~C6H6}) \times (6.022 \times 10^{23} \, \text{molecules/mol})\) = \(5.08 \times 10^{22} \mathrm{~molecules~C6H6}\)

Step 4: (d) Calculate the number of O atoms in \(6.25 \times 10^{-3}\) mol Al(NO3)3.

First, we'll find the number of moles of O atoms in \(6.25 \times 10^{-3}\) mol of Al(NO3)3. For each Al(NO3)3 molecule, there are 3 times 3 O atoms (9 O atoms) in each NO3 group. number of moles of O atoms = \((6.25 \times 10^{-3} \mathrm{~mol~Al(NO3)3}) \times (9 \, \text{O atoms per Al(NO3)3 molecule})\) = \(5.625 \times 10^{-2} \mathrm{~mol~O}\) Now we use the formula \(n_\text{particles} = n \times N_A\) to find the number of O atoms. Number of O atoms = \((5.625 \times 10^{-2} \mathrm{~mol~O}) \times (6.022 \times 10^{23} \, \text{atoms/mol})\) = \(3.39 \times 10^{22} \mathrm{~O~atoms}\)

Key Concepts

Moles and Molecules

The concept of moles and molecules is foundational in chemistry. A **mole** is a unit that measures quantity of particles, typically atoms or molecules, and it is represented by Avogadro's number, \(6.022 \times 10^{23}\). This means one mole of any substance contains exactly \(6.022 \times 10^{23}\) molecules or atoms. For practical purposes, the mole allows chemists to convert between atoms, molecules, and grams.

For example, if you have 0.08447 mol of benzene (\(C_6H_6\)), using Avogadro's number, we can find the total molecules in this sample as follows:

• Given that the number of molecules \(n_{particles} = n \times N_A\)
• Substitute, \(n_{particles} = 0.08447 \times 6.022 \times 10^{23}\)

This calculation will yield the total number of molecules, giving us a concrete idea of the molecular scale and abundance involved. Understanding moles and molecules streamlines measurements and computations in chemistry by bridging macroscopic and molecular scales.

Chemical Reactions

Chemical reactions are processes that convert reactants into products through the breaking and forming of bonds. Understanding chemical reactions requires knowledge of how quantities relate in reactions, usually expressed in chemical equations. Each component in an equation is expressed in moles, providing a way to convey the proportions of each substance involved.

For instance, to determine how many oxygen atoms are present in a compound like \(Al(NO_3)_3\), recognize that each molecule contains nine oxygen atoms. Once you know how much of \(Al(NO_3)_3\) you have, you can determine the total number of oxygen atoms by multiplying this quantity by the number of oxygen atoms in the formula.

• Identify the number of moles (e.g., \(6.25 \times 10^{-3}\) mol of \(Al(NO_3)_3\))
• Multiply by the number of O atoms per formula unit (9 in this case)
• Compute total atoms using \(n \times N_A\)

Thus, chemical equations not only represent the reactants and products in a reaction but help to indicate the stoichiometric relationships necessary for complete conversions.

Stoichiometry

Stoichiometry is the field in chemistry that studies the quantitative relationships between substances in chemical reactions. It helps predict the amount of products that result from given reactants and is an essential tool in chemistry for practical and theoretical applications. It relies on mole ratios from balanced chemical equations.

Consider a reaction where we calculated the mass of \(CdS\) from moles. Begin by determining a compound's molar mass using atomic masses from the periodic table, then use this to convert moles to grams. For \(1.50 \times 10^{-2}\) moles of \(CdS\), calculate its mass using:

• Find the molar mass of \(CdS\) (e.g., 144.48 g/mol)
• Utilize the equation \(mass = n \times M\)
• Substitute values: \(mass = (1.50 \times 10^{-2}) \times 144.48\)

This approach highlights the utility of stoichiometry in linking reactants to precise quantities of products, establishing it as a core competency for chemists in predicting and verifying chemical phenomena.