Problem 33 - Chemistry: The Central Science

Question

Calculate the following quantities: (a) mass, in grams, of \(0.105\) moles sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) (b) moles of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) in \(143.50 \mathrm{~g}\) of this substance (c) number of molecules in \(1.0 \times 10^{-6} \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (d) number of \(\mathrm{N}\) atoms in \(0.410 \mathrm{~mol} \mathrm{NH}_{3}\)

Step-by-Step Solution

Answer

(a) The mass of sucrose is \(35.946 \mathrm{~g}\). (b) There are \(0.758 \mathrm{~mol}\) of \(\mathrm{Zn(NO}_{3})_{2}\). (c) There are \(6.022 \times 10^{17}\) molecules of \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\). (d) There are \(2.469 \times 10^{23}\) N atoms in \(\mathrm{NH}_{3}\).

Step 1: (a) Calculate the mass of sucrose

First, we need to find the molar mass of sucrose (C12H22O11). To do this, we look up the atomic masses of carbon, hydrogen, and oxygen, and then multiply them by the number of each atom in the sucrose molecule. Molar mass of sucrose = (12 Carbon atoms * Molar mass of Carbon) + (22 Hydrogen atoms * Molar mass of Hydrogen) + (11 Oxygen atoms * Molar mass of Oxygen) Using molar masses from the periodic table: Carbon = 12.01 g/mol, Hydrogen = 1.01 g/mol, and Oxygen = 16.00 g/mol. Molar mass of sucrose = (12 * 12.01 g/mol) + (22 * 1.01 g/mol) + (11 * 16.00 g/mol) = 144.12 g/mol + 22.22 g/mol + 176.00 g/mol = 342.34 g/mol Next, we can use the given moles of sucrose (0.105 mol) to calculate the mass in grams: Mass of sucrose = (Moles of sucrose) * (Molar mass of sucrose) = (0.105 mol) * (342.34 g/mol) = 35.946 g

Step 2: (b) Calculate moles of Zn(NO3)2

First, we need to find the molar mass of Zn(NO3)2 by adding the molar mass of zinc and the molar mass of two nitrate (NO3) ions. Using the periodic table, we find the molar masses of the elements: Zn = 65.38 g/mol N = 14.01 g/mol O = 16.00 g/mol Molar mass of Zn(NO3)2 = (Molar mass of Zn) + 2 * (Molar mass of N + 3 * Molar mass of O) = 65.38 g/mol + 2 * (14.01 g/mol + 3 * 16.00 g/mol) = 65.38 g/mol + 2 * (62.01 g/mol) = 65.38 g/mol + 124.02 g/mol = 189.40 g/mol Now, we can use the given mass of Zn(NO3)2 (143.50 g) to calculate the moles of Zn(NO3)2: Moles of Zn(NO3)2 = (Mass of Zn(NO3)2) / (Molar mass of Zn(NO3)2) = (143.50 g) / (189.40 g/mol) = 0.758 mol

Step 3: (c) Calculate the number of molecules of CH3CH2OH

To find the number of molecules, we'll use Avogadro's number (\(6.022 \times 10^{23}\)) and the moles of the substance: Number of molecules of CH3CH2OH = (Moles of CH3CH2OH) * (Avogadro's number) = (1.0 x 10^{-6} mol) * (6.022 x 10^{23}) = 6.022 x 10^{17} molecules

Step 4: (d) Calculate the number of N atoms in NH3

We have 0.410 mol of NH3 and there is one N atom in each NH3 molecule. Therefore, we have 0.410 mol N atoms. To find the number of N atoms, we'll use Avogadro's number (\(6.022 \times 10^{23}\)): Number of N atoms in NH3 = (Moles of N) * (Avogadro's number) = (0.410 mol N) * (6.022 x 10^{23}) = 2.469 x 10^{23} N atoms

Key Concepts

Mole to Mass Conversion

Understanding the concept of converting from moles to mass is fundamental in chemistry and can be seen as the bridge between the microscopic world of atoms and the macroscopic world we experience daily. Essentially, the mole is a counting unit for chemical substances based on the number of atoms, ions, or molecules present.

Mole to mass conversion involves using the molar mass of a compound, which is the mass of one mole of that substance. Molar mass acts as a conversion factor between the number of moles of a substance and its mass in grams. The process can be boiled down to a simple formula:
\[ \text{Mass (g)} = \text{Moles (mol)} \times \text{Molar Mass (g/mol)} \]
For example, if we have 0.105 moles of sucrose, and the molar mass is 342.34 g/mol, the calculation would be \( 0.105 \text{ mol} \times 342.34 \text{ g/mol} = 35.946 \text{ g} \).

In instructional materials, it's critical to provide clear examples and encourage students to practice with diverse chemical compounds to solidify their understanding of this concept.

Chemical Compound Molar Mass

The molar mass of a chemical compound is essentially the weight of one mole of that substance, and it is determined by summing up the atomic masses of each element present in the compound. Atomic masses are averaged weights of atoms, taking into account the different isotopes and their abundances.

Calculating the molar mass is straightforward but requires attention to detail. For a compound like sucrose \( \mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11} \), we must account for each carbon (C), hydrogen (H), and oxygen (O) atom. The respective atomic masses obtained from the periodic table are multiplied by the number of atoms of each kind in the formula. In the case of Zn(NO3)2, for instance, we factor in one zinc (Zn) atom and two nitrate ions, each consisting of one nitrogen (N) and three oxygen (O) atoms.

Step-by-Step Calculation:

The molar mass of Zn(NO3)2 is computed by adding together the mass of one zinc atom and the mass of two nitrate ions, including all the nitrogen and oxygen atoms within them. By clearly explaining these calculations with examples, we assist learners in moving from abstract concepts to practical applications.

Avogadro's Number Application

Avogadro's number, \(6.022 \times 10^{23}\), is a constant that represents the number of units in one mole of any substance and is pivotal in the realm of chemistry for converting between the amount of substance and the number of particles. This number allows chemists to count particles like atoms, molecules, and ions by weighing, which is a practical approach for laboratory results.

Its application is seen when we wish to determine the actual number of molecules or atoms in a given moles of a substance. For instance, to find out the number of molecules in \(1.0 \times 10^{-6} \text{ mol} \text{ CH}_{3}\text{CH}_{2}\text{OH}\), we multiply the moles by Avogadro's number, resulting in \(6.022 \times 10^{17}\) molecules.

Students must practice applying Avogadro's number to different contexts, such as calculating the number of atoms in a mole of a compound like ammonia (NH3), as this reinforces their understanding of the scale and proportion that this number represents in the microscopic world of chemistry.