Question

Calculate the following quantities (a) mass, in grams, of \(5.76 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{CdS}\) (b) number of moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(112.6 \mathrm{~g}\) of this substance (c) number of molecules in \(1.305 \times 10^{-2} \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{6}\) (d) number of \(\mathrm{O}\) atoms in \(4.88 \times 10^{-3} \mathrm{~mol} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\)

Short answer

(a) The mass of \(5.76 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{CdS}\) is \(0.832~\mathrm{g}\). (b) There are 2.106 moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(112.6 \mathrm{~g}\) of this substance. (c) There are \(7.853\times10^{21}\) molecules in \(1.305 \times 10^{-2} \mathrm{ ~mol} \mathrm{C}_{6} \mathrm{H}_{6}\). (d) There are \(8.810\times10^{22}\) O atoms in \(4.88 \times 10^{-3} \mathrm{ ~mol} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\).

Step-by-step solution

(a) Calculate the molar mass of CdS

First, we need to find the molar mass of cadmium sulfide (CdS). From the periodic table, we have that the atomic masses of the elements are: - Molecular mass of Cd (cadmium) = 112.4 g/mol - Molecular mass of S (sulfur) = 32.1 g/mol The molar mass of CdS is the sum of the atomic masses of cadmium and sulfur: \( M_{CdS} = 112.4 + 32.1 = 144.5 \mathrm{ ~g/mol} \)

(a) Determine the mass of CdS

Now we can determine the mass by multiplying the number of moles by the molar mass: \[ mass = moles~of~CdS \times molar~mass~of~CdS \] \[ mass = 5.76\times10^{-3}~\mathrm{mol} \times 144.5~\mathrm{g/mol} \] \[ mass = 0.832~\mathrm{g} \] For (a), the mass of \(5.76 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{CdS}\) is \(0.832~\mathrm{g}\).

(b) Determine the number of moles of NH4Cl

First, we calculate the molar mass of ammonium chloride (NH4Cl): - Molecular mass of N (nitrogen) = 14.0 g/mol - Molecular mass of H (hydrogen) = 1.0 g/mol - Molecular mass of Cl (chlorine) = 35.5 g/mol \( M_{NH4Cl} = 14.0 + 4(1.0) + 35.5 = 53.5 \mathrm{ ~g/mol} \) Now, we determine the number of moles by dividing the mass by the molar mass: \[ moles = \frac{mass}{molar~mass} = \frac{112.6 \mathrm{~g}}{53.5 \mathrm{ ~g/mol}} = 2.106 \mathrm{ ~mol} \] For (b), there are 2.106 moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(112.6 \mathrm{~g}\) of this substance.

(c) Determine the number of molecules in C6H6

To find the number of molecules, we multiply the number of moles by Avogadro's number: \[ molecules = moles \times Avogadro's~number \] \[ molecules = 1.305\times10^{-2}~\mathrm{mol} \times 6.022\times10^{23}~\mathrm{molecules/mol} \] \[ molecules = 7.853\times10^{21}~\mathrm{molecules} \] For (c), there are \(7.853\times10^{21}\) molecules in \(1.305 \times 10^{-2} \mathrm{ ~mol} \mathrm{C}_{6} \mathrm{H}_{6}\).

(d) Determine the number of O atoms in Al(NO3)3

To find the number of oxygen atoms, we first find the number of \(\mathrm{NO}_{3}\) ions in the compound by multiplying the number of moles by Avogadro's number: \[ moles~of~NO_{3}~ions = moles~ of~Al\left(\mathrm{NO}_{3}\right)_{3} \times 3 \] \[ moles~of~NO_{3}~ions = 4.88\times10^{-3}~\mathrm{mol} \times 3 \] Now, we multiply the \(moles~of~NO_{3}~ions\) by Avogadro's number and then, since there is one O atom in each \(\mathrm{NO}_{3}\) ion, we multiply by three to obtain the number of O atoms: \[ O~atoms = moles~of~NO_{3}~ions \times Avogadro's~number \times 3 \] \[ O~atoms = (4.88\times10^{-3} \times 3) \times 6.022\times10^{23} \] \[ O~atoms = 8.810\times10^{22}~\mathrm{O~atoms} \] For (d), there are \(8.810\times10^{22}\) O atoms in \(4.88 \times 10^{-3} \mathrm{ ~mol} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\).

Key concepts

Molar Mass

Understanding the concept of molar mass is essential for performing stoichiometry calculations. The molar mass is the mass (in grams) of one mole of a substance. A mole is defined based on Avogadro's number, which is approximately 6.022 x 10²³ entities (such as atoms, molecules, or ions), and represents a standard quantity for counting particles at the atomic scale.

Think of molar mass as a bridge between the microscopic world of atoms and the macroscopic world we can measure. Each element's molar mass is given on the periodic table and units are expressed as grams per mole (g/mol). For instance, the molar mass of carbon is 12.01 g/mol, indicating that one mole of carbon atoms weighs 12.01 grams.

To find the molar mass of a compound, like cadmium sulfide (CdS), we add the molar masses of the individual elements. This value can be used for important conversions, such as converting moles to grams or vice versa, which is crucial in chemical calculations like those in this exercise.

Avogadro's Number

Avogadro's number, 6.022 x 10²³, is a fundamental constant in chemistry known as the number of atoms, molecules, or other particles in one mole of a substance. It is named after the Italian scientist Amedeo Avogadro.

Using Avogadro's number allows chemists to count particles in a substance by measuring its mole quantity. When we say that one mole of carbon contains 6.022 x 10²³ carbon atoms, we're applying Avogadro's number. In the example of benzene (C₆H₆), knowing that we have 1.305 x 10^-2 moles of benzene, we can determine the number of benzene molecules present using Avogadro's number, providing a link between measurable amounts of substances and the number of particles they contain.

Moles to Mass Conversion

In chemistry, moles to mass conversion is a fundamental skill for quantifying the amount of substance present in a sample. By using the molar mass of a substance, we can convert moles to mass and vice versa, aligning the amount of substance with a tangible weight.

To convert moles to grams, you multiply the number of moles by the molar mass of the substance. Conversely, to convert grams to moles, you divide the mass of the substance by its molar mass. The formulas look like this:

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

Understanding this conversion allows students to bridge the theoretical understanding of chemical compounds with practical laboratory experiments, increasing the palpability of otherwise abstract concepts.

Molecular Formula

The molecular formula provides an explicit way of denoting the number and type of atoms in a molecule. It is a representation that displays the actual number of atoms of each element in one molecule or formula unit of a substance. For instance, the molecular formula of glucose is C₆H₁₂O₆, which shows that a glucose molecule consists of 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.

In the context of the exercise involving aluminum nitrate, Al(NO₃)₃, the molecular formula reveals not only the types of atoms present but also the ratio with which they combine. This detailed information is crucial for understanding the compound's composition, and when paired with stoichiometry calculations, it allows for the precise determination of particles like oxygen atoms in a given amount of the compound.