Question

Calculate the number of molecules present in each of the following samples. a. \(3.45 \mathrm{g}\) of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) b. 3.45 mol of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. \(25.0 \mathrm{g}\) of \(\mathrm{ICl}_{5}\) d. \(1.00 \mathrm{g}\) of \(\mathrm{B}_{2} \mathrm{H}_{6}\) e. \(1.05 \mathrm{mmol}\) of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\)

Short answer

The number of molecules in each sample is as follows: a. \(1.15 \times 10^{22}\) molecules of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) b. \(2.08 \times 10^{24}\) molecules of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. \(4.94 \times 10^{22}\) molecules of \(\mathrm{ICl}_{5}\) d. \(2.17 \times 10^{22}\) molecules of \(\mathrm{B}_{2} \mathrm{H}_{6}\) e. \(6.32 \times 10^{20}\) molecules of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\)

Step-by-step solution

Calculate the molar mass of the substance

\( M(C_6 H_{12} O_6) = 6 \times 12.01 + 12 \times 1.01 + 6 \times 16.00 = 72.06 + 12.12 + 96.00 = 180.18 \mathrm{g/mol} \)

Convert the given mass to moles

\( \mathrm{moles} = \frac{\mathrm{mass}}{\mathrm{molar\ mass}} = \frac{3.45 \mathrm{g}}{180.18 \mathrm{g/mol}} = 0.01914 \mathrm{mol} \)

Use Avogadro's number to convert moles to molecules

\( \mathrm{molecules} = \mathrm{moles} \times \mathrm{Avogadro's\ number} = 0.01914 \mathrm{mol} \times 6.022 \times 10^{23}\ \mathrm{molecules/mol} = 1.15 \times 10^{22} \ \mathrm{molecules} \) b. 3.45 mol of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

No need for molar mass calculation

Since we are given the moles, there is no need to calculate the molar mass this time.

Use Avogadro's number to convert moles to molecules

\( \mathrm{molecules} = \mathrm{moles} \times \mathrm{Avogadro's\ number} = 3.45 \mathrm{mol} \times 6.022 \times 10^{23}\ \mathrm{molecules/mol} = 2.08 \times 10^{24} \ \mathrm{molecules} \) c. \(25.0 \textrm{g}\) of \(\mathrm{ICl}_{5}\)

Calculate the molar mass of the substance

\( M(ICl_5) = (1 \times 126.90) + (5 \times 35.45) = 126.90 + 177.25 = 304.15 \mathrm{g/mol} \)

Convert the given mass to moles

\( \mathrm{moles} = \frac{\mathrm{mass}}{\mathrm{molar\ mass}} = \frac{25.0 \mathrm{g}}{304.15 \mathrm{g/mol}} = 0.08216 \mathrm{mol} \)

Use Avogadro's number to convert moles to molecules

\( \mathrm{molecules} = \mathrm{moles} \times \mathrm{Avogadro's\ number} = 0.08216 \mathrm{mol} \times 6.022 \times 10^{23}\ \mathrm{molecules/mol} = 4.94 \times 10^{22} \ \mathrm{molecules} \) d. \(1.00 \textrm{g}\) of \(\mathrm{B}_{2} \mathrm{H}_{6}\)

Calculate the molar mass of the substance

\( M(B_2H_6) = (2 \times 10.81) + (6 \times 1.01) = 21.62 + 6.06 = 27.68 \mathrm{g/mol} \)

Convert the given mass to moles

\( \mathrm{moles} = \frac{\mathrm{mass}}{\mathrm{molar\ mass}} = \frac{1.00 \mathrm{g}}{27.68 \mathrm{g/mol}} = 0.03613 \mathrm{mol} \)

Use Avogadro's number to convert moles to molecules

\( \mathrm{molecules} = \mathrm{moles} \times \mathrm{Avogadro's\ number} = 0.03613 \mathrm{mol} \times 6.022 \times 10^{23}\ \mathrm{molecules/mol} = 2.17 \times 10^{22} \ \mathrm{molecules} \) e. \(1.05 \mathrm{mmol}\) of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\)

Convert mmol to mol

\( 1.05 \mathrm{mmol} = 1.05 \times 10^{-3} \mathrm{mol} \)

Use Avogadro's number to convert moles to molecules

\( \mathrm{molecules} = \mathrm{moles} \times \mathrm{Avogadro's\ number} = 1.05 \times 10^{-3} \mathrm{mol} \times 6.022 \times 10^{23}\ \mathrm{molecules/mol} = 6.32 \times 10^{20} \ \mathrm{molecules} \)

Key concepts

Avogadro's Number

Avogadro's Number is a fundamental concept in chemistry that represents the number of atoms, ions, or molecules in one mole of a substance. This number is a constant and is approximately \( 6.022 \times 10^{23} \). It is named after the Italian scientist Amedeo Avogadro. Understanding Avogadro's Number is crucial for converting between the macroscopic scale of substances that we can measure and the microscopic scale of atoms and molecules.

Avogadro's Number allows chemists to relate the amount of substance to the number of particles involved. This link is essential because it connects the mass of a substance with the number of molecules, which is fundamental for understanding reactions at a molecular level.

When you have the number of moles of a substance, multiplying by Avogadro's Number will give you the total number of molecules in the sample. For example, if you have 2 moles of water, the calculation would be:

• \( 2 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 1.2044 \times 10^{24} \text{ molecules} \)

By using Avogadro's Number, chemists can scale up small-scale measurements to real-world applications.

Molar Mass Calculation

Molar mass is a key concept when dealing with molecules and their conversions. It represents the mass of one mole of a given substance and is expressed in grams per mole (g/mol). To calculate the molar mass, you add up the atomic masses of all atoms in the molecular formula of a compound.

For instance, let's calculate the molar mass of glucose \( C_6H_{12}O_6 \):

• Carbon (C): \( 12.01 \ \text{g/mol} \times 6 = 72.06 \ \text{g/mol} \)
• Hydrogen (H): \( 1.01 \ \text{g/mol} \times 12 = 12.12 \ \text{g/mol} \)
• Oxygen (O): \( 16.00 \ \text{g/mol} \times 6 = 96.00 \ \text{g/mol} \)
• Total Molar Mass of Glucose: \( 180.18 \ \text{g/mol} \)

Calculating molar masses is essential for converting mass to moles, which is the stepping stone to finding the number of molecules. Understanding and calculating molar masses enables us to precisely measure quantities for chemical reactions, ensuring accurate and consistent results.

Moles to Molecules Conversion

Converting moles to molecules is a common calculation in chemistry, facilitating the transition from bulk materials to microscopic entities. This conversion requires two main factors: the quantity of moles you have and Avogadro's Number.

To convert from moles to molecules, utilize the following formula:
\[\text{Number of Molecules} = \text{Number of Moles} \times 6.022 \times 10^{23} \ \text{molecules/mol}\]Let's take a practical example with 0.01914 moles of a compound. The conversion goes like this:

• \( 0.01914 \ ext{moles} \times 6.022 \times 10^{23} \ ext{molecules/mol} = 1.15 \times 10^{22} \ ext{molecules} \)

Understanding this conversion is crucial for chemical reactions, allowing you to predict and measure the quantities of substances involved. It's a vital calculation that supports much of the work done in labs, research, and industry. Knowing how to convert moles to molecules ensures that measurements reflect the actual number of particles involved, leading to more accurate scientific conclusions.