Problem 166 - Introductory Chemistry Atoms First

Question

Consider the balanced chemical equation \(2 \mathrm{~A}+\mathrm{B} \rightarrow 2 \mathrm{C}+\mathrm{D}\) When \(8.0 \mathrm{~g}\) of A reacts completely with \(6.0 \mathrm{~g}\) of \(\mathrm{B}\), \(10.0 \mathrm{~g}\) of \(\mathrm{C}\) and \(4.0 \mathrm{~g}\) of \(\mathrm{D}\) are produced. Assuming the yield is \(100 \%\), (a) Which has a greater molar mass, A or C? (b) Which has a greater molar mass, A or B? (c) Which has a greater molar mass, A or D? (d) If the molar mass of \(\mathrm{A}\) is \(24.0 \mathrm{~g} / \mathrm{mol}\), determine the molar mass of \(\mathrm{B}, \mathrm{C}\), and \(\mathrm{D}\).

Step-by-Step Solution

Answer

(a) C has a greater molar mass than A. (b) B has a greater molar mass than A. (c) A and D have equal molar masses. (d) The molar masses of B, C, and D are 36.0 g/mol, 30.0 g/mol, and 24.0 g/mol, respectively.

Step 1: Calculate mole ratios A and B

To calculate the mole ratio between reactants A and B, we need to divide the mass of each reactant by its respective molar mass (we will denote these molar mass as M_A, M_B, M_C, and M_D for the four substances, respectively). Since we know the molar mass of A, we can calculate the number of moles of A. Then, we can determine the ratio of moles of A to moles of B using the balanced chemical equation. n_A = \(\frac{mass_A}{M_A}\) = \(\frac{8.0~g}{24.0~g/mol}\) = \(0.333~mol\) Since the balanced chemical equation states there is a 2-to-1 ratio for A to B, we can determine the number of moles of B: n_B = \(\frac{1}{2}\)n_A = \(\frac{1}{2}\)(0.333 mol) = \(0.167~mol\)

Step 2: Calculate the molar mass of B

Now that we know the number of moles of B, we can calculate the molar mass using the mass of B. M_B = \(\frac{mass_B}{n_B}\) = \(\frac{6.0~g}{0.167~mol}\) = \(36.0~g/mol\)

Step 3: Determine which has a greater molar mass: A or B

Since the molar mass of A is 24.0 g/mol and the molar mass of B is 36.0 g/mol, B has a greater molar mass than A.

Step 4: Calculate the number of moles of C and D

Using the balanced chemical equation, we see that the mole ratio of A:C is 2:2 = 1:1 and A:D is 2:1. Therefore, the number of moles of C must be equal to the number of moles of A, and the number of moles of D must be equal to half the number of moles of A: n_C = n_A = \(0.333~mol\) n_D = \(\frac{1}{2}\)n_A = \(0.167~mol\)

Step 5: Calculate the molar mass of C and D

We can calculate the molar masses of C and D using their masses and the calculated number of moles. M_C = \(\frac{mass_C}{n_C}\) = \(\frac{10.0~g}{0.333~mol}\) = \(30.0~g/mol\) M_D = \(\frac{mass_D}{n_D}\) = \(\frac{4.0~g}{0.167~mol}\) = \(24.0~g/mol\)

Step 6: Determine which has a greater molar mass: A, C, and D

Comparing the molar masses of A, C and D: - Molar mass of A: 24.0 g/mol - Molar mass of C: 30.0 g/mol - Molar mass of D: 24.0 g/mol So, C has a greater molar mass than A, and A and D have equal molar masses. The answers to the given questions are: (a) C has a greater molar mass than A. (b) B has a greater molar mass than A. (c) A and D have equal molar masses. (d) The molar masses of B, C, and D are 36.0 g/mol, 30.0 g/mol, and 24.0 g/mol, respectively.

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is essential in chemical reactions as it shows the proportion of reactants and products. In the equation: \[2 \mathrm{~A} + \mathrm{B} \rightarrow 2 \mathrm{C} + \mathrm{D}\]this means that two units of chemical A react with one unit of chemical B to form two units of chemical C and one unit of chemical D.
Balancing ensures that the same number of each type of atom appears on both the reactant and product sides of the equation.
For example, if there are two atoms of A on the left side, there should also be two atoms of A in the products. Keeping equations balanced is crucial for accurate stoichiometric calculations.

• It obeys the law of conservation of mass.
• Indicates how much reactant is needed or how much product is formed.

Understanding the balanced equation also allows you to predict how changing one variable can affect the entire reaction.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It acts like a bridge between the mass of a entity and the number of particles within that mass.
To determine molar mass, you need to add up the atomic masses of all the atoms in a molecule. For example:- For A, given: 24.0 g/mol
The molar mass allows us to convert easily between grams and moles, using the formula:\[ n = \frac{\text{mass}}{\text{molar mass}} \]where \( n \) is the number of moles.
For chemical conversions, this is essential as it enables us to calculate how much of a chemical we have or need in moles or grams, depending on what we have and what we want to find.

Mole Ratio

The mole ratio is derived from the coefficients of a balanced chemical equation, representing the relative number of moles of reactants and products involved. In our example equation:\[2 \mathrm{~A} + \mathrm{B} \rightarrow 2 \mathrm{C} + \mathrm{D}\]The ratio is:

• A to B: 2:1
• A to C: 2:2 (or 1:1)
• A to D: 2:1

This means, for every 2 moles of A used, 1 mole of B is needed, producing 2 moles of C and 1 mole of D.
The mole ratio is a critical concept in stoichiometry because it provides a direct relationship between the quantities of all substances in the reaction. This allows for precise calculations of required reactants and expected products in chemical reactions.

Chemical Reactions Analysis

Analyzing chemical reactions involves understanding how reactants transition to products, respecting conservation of mass and atom isolation. This analysis involves several strategic steps.

• Identifying reactants and products: Determine what starts the reaction and what ends it.
• Balancing the equation: Ensure the mass and number of each type of atom is conserved from reactants to products.
• Using stoichiometric concepts: Apply mole ratios to calculate the amount of each substance used or created.

When calculating, it's essential to ensure all measurements are accurately converted between grams and moles using the molar mass. This allows for precise deduction of outcomes in chemical assays.
The initial and final masses of reactants and products allow for checking the completeness of the reaction and verifying yield predictions, like in our exercise where it was assumed to be 100%.