Problem 33 - Chemistry The Central Science

Question

Calculate the following quantities: (a) mass, in grams, of 0.105 mole of 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 0.105 mole of sucrose (C12H22O11) is approximately \(35.942g\). (b) There are 0.757 moles of Zn(NO3)2 in 143.50 g of this substance. (c) There are approximately \(6.022 \times 10^{17}\) molecules of CH3CH2OH in 1.0 x 10^{-6} mol. (d) There are approximately \(2.469 \times 10^{23}\) N atoms in 0.410 mol NH3.

Step 1: (a) - Calculate the molar mass of sucrose

To find the mass of 0.105 mole of sucrose, we first need to calculate the molar mass of sucrose. Sucrose (C12H22O11) has: - 12 Carbon atoms (C) - 22 Hydrogen atoms (H) - 11 Oxygen atoms (O) The molar masses of these elements are: - Carbon: 12.01 g/mol - Hydrogen: 1.01 g/mol - Oxygen: 16.00 g/mol So, the molar mass of sucrose is: \( Molar\_mass = 12*12.01 + 22*1.01 + 11*16.00 = 342.30 \,g/mol \)

Step 2: (a) - Calculate the mass of 0.105 mole of sucrose

Now we can calculate the mass of 0.105 mole of sucrose using its molar mass: \( Mass = moles * molar\_mass \) \( Mass = 0.105 * 342.30 \,g/mol = 35.942 \,g\) So the mass of 0.105 mole of sucrose is approximately \(35.942g\).

Step 3: (b) - Calculate the molar mass of Zn(NO3)2

To calculate the molar mass of Zn(NO3)2, we have: - 1 Zinc atom (Zn) - 2 Nitrogen atoms (N) in each NO3 group, total 4 Nitrogen atoms - 3 Oxygen atoms (O) in each NO3 group, total 6 Oxygen atoms The molar masses of these elements are: - Zinc: 65.38 g/mol - Nitrogen: 14.01 g/mol - Oxygen: 16.00 g/mol So, the molar mass of Zn(NO3)2 is: \( Molar\_mass = 65.38 + 4*14.01 + 6*16.00 = 189.39 \,g/mol\)

Step 4: (b) - Calculate the moles of Zn(NO3)2

Now we can calculate the moles of Zn(NO3)2 in 143.50 g of this substance: \( Moles = \frac{mass}{molar\_mass} \) \( Moles = \frac{143.50 \,g}{189.39 \,g/mol} = 0.757 \,mol\) So, there are 0.757 moles of Zn(NO3)2 in 143.50 g of this substance.

Step 5: (c) - Calculate the number of molecules of CH3CH2OH

To find the number of molecules in 1.0 x 10^{-6} mol CH3CH2OH, we will use Avogadro's number (6.022 x 10^{23}). \( Number\_of\_molecules = moles * Avogadro's\_number \) \( Number\_of\_molecules = 1.0 \times 10^{-6} \,mol * 6.022 \times 10^{23} \,molecules/mol \) \( Number\_of\_molecules = 6.022 \times 10^{17} \,molecules\) So there are approximately \(6.022 \times 10^{17}\) molecules of CH3CH2OH in 1.0 x 10^{-6} mol.

Step 6: (d) Calculate the number of N atoms in 0.410 mol NH3

To find the number of N atoms in 0.410 mol NH3, first, we calculate the number of NH3 molecules: \( Number\_of\_NH3\_molecules = moles * Avogadro's\_number \) \( Number\_of\_NH3\_molecules = 0.410 \,mol * 6.022 \times 10^{23} \,molecules/mol \) \( Number\_of\_NH3\_molecules = 2.469 \times 10^{23} \,molecules\) Since there is 1 N atom in each NH3 molecule: \( Number\_of\_N\_atoms = Number\_of\_NH3\_molecules = 2.469 \times 10^{23} \,atoms\) So there are approximately \(2.469 \times 10^{23}\) N atoms in 0.410 mol NH3.

Key Concepts

Molar Mass Calculation

Understanding how to calculate molar mass is fundamental in chemistry. It is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). To calculate the molar mass, you need to know the chemical formula of the substance and the atomic masses of its constituent elements, which can be found on the periodic table.

For example, in the exercise, the molar mass of sucrose (\(C_{12}H_{22}O_{11}\)) is determined by multiplying the number of each type of atom by its atomic mass (in g/mol) and then summing these values. Here's the breakdown:

• Carbon (\(C\)): 12 atoms \times 12.01 g/mol
• Hydrogen (\(H\)): 22 atoms \times 1.01 g/mol
• Oxygen (\(O\)): 11 atoms \times 16.00 g/mol

The molar mass of sucrose then becomes the sum of all these, resulting in 342.30 g/mol. Knowing the molar mass is essential as it enables you to convert between moles and grams, a vital step in stoichiometric calculations.

Avogadro's Number

Avogadro's number, approximately 6.022 x 10²³, is the number of units in one mole of any substance. It can refer to atoms, ions, or molecules, depending on the context. This constant is pivotal in converting moles to discrete particles and vice versa.

For instance, in step 5 of the exercise, to find out the number of molecules in 1.0 x 10^-6 moles of ethanol (\(CH_{3}CH_{2}OH\)), you multiply the moles by Avogadro's number: \[ Number\_of\_molecules = 1.0 \times 10^{-6} \,mol \times 6.022 \times 10^{23} \,molecules/mol \] yielding 6.022 x 10¹⁷ molecules. This concept is crucial for understanding how much of a substance is needed or produced in a chemical reaction.

Chemical Formula Stoichiometry

Chemical formula stoichiometry involves using the balanced equation of a chemical reaction to calculate the relative quantities of reactants and products. It is based on the principle of the conservation of mass and the mole concept. To perform these calculations, you'll need a balanced chemical equation and knowledge of the concepts of molar mass and Avogadro's number.

In the context of the given problems, once you determine the number of moles, you can calculate the number of molecules or atoms using stoichiometry. For example, in calculating the number of nitrogen atoms in 0.410 mole of ammonia (\(NH_{3}\)), you first find the number of molecules of ammonia and then use the fact that each molecule contains one nitrogen atom to determine the total number of nitrogen atoms: \[ Number\_of\_N\_atoms = Number\_of\_NH3\_molecules = Moles \times Avogadro's\_number \] Final stoichiometric calculations enable us to understand the detailed composition of substances involved in chemical reactions, crucial for laboratory preparations, and industrial processes.