Question

This problem is designed to incorporate several concepts and techniques into one situation. You have gone back in time and are working with Dalton on a table of relative masses. Following are his data. \(0.602 \mathrm{g}\) gas A reacts with \(0.295 \mathrm{g}\) gas \(\mathrm{B}\) \(0.172 \mathrm{g}\) gas \(\mathrm{B}\) reacts with \(0.401 \mathrm{g}\) gas \(\mathrm{C}\) \(0.320 \mathrm{g}\) gas \(\mathrm{A}\) reacts with \(0.374 \mathrm{g}\) gas \(\mathrm{C}\) a. Assuming simplest formulas \((\mathrm{AB}, \mathrm{BC}, \text { and } \mathrm{AC}),\) construct a table of relative masses for Dalton. b. Knowing some history of chemistry, you tell Dalton that if he determines the volumes of the gases reacted at constant temperature and pressure, he need not assume simplest formulas. You collect the following data: 6 volumes gas \(A+1\) volume gas \(B \rightarrow 4\) volumes product 1 volume gas \(\mathrm{B}+4\) volumes gas \(\mathrm{C} \rightarrow 4\) volumes product 3 volumes gas \(A+2\) volumes gas \(C \rightarrow 6\) volumes product Write the simplest balanced equations, and find the actual relative masses of the elements. Explain your reasoning.

Short answer

In summary, we have found the actual relative masses of the elements A, B, and C to be 1, 1/6, and 2/3, respectively. Using these masses, we've written the simplest balanced equations for the reactions between these elements as: 1. \(6A + (1/6)B \rightarrow 4\) Product 2. \((1/6)B + 4(2/3)C \rightarrow 4\) Product 3. \(3A + 2(2/3)C \rightarrow 6\) Product

Step-by-step solution

Calculate Mass Ratios#from the Given Mass Data

First, let's calculate the mass ratios from the given mass data: 1. For the reaction between A and B: Ratio = mass of A / mass of B = \(0.602 \mathrm{g} / 0.295 \mathrm{g} = 2.04\) 2. For the reaction between B and C: Ratio = mass of B / mass of C = \(0.172 \mathrm{g} / 0.401 \mathrm{g} = 0.43\) 3. For the reaction between A and C: Ratio = mass of A / mass of C = \(0.320 \mathrm{g} / 0.374 \mathrm{g} = 0.86\) Now that we have the mass ratios, we can construct a table of relative masses for Dalton in the next step.

Construct a Table of Relative Masses for Dalton

Based on the mass ratios calculated in Step 1, we can construct the following table of relative masses for Dalton: | Elements | Mass Ratio | |-------|------------| | A : B | 2.04 | | B : C | 0.43 | | A : C | 0.86 |

Use Volume Data to Write Simplest Balanced Equations#for the Reactions

Now let's use the volume data to write the simplest balanced equations for the reactions: 1. For the reaction between A and B: \(6A + B \rightarrow 4\) Product 2. For the reaction between B and C: \(B + 4C \rightarrow 4\) Product 3. For the reaction between A and C: \(3A + 2C \rightarrow 6\) Product

Find the Actual Relative Masses of the Elements

Using the balanced equations from Step 3, we can find the actual relative masses of the elements: 1. From the first equation, we get: A / B = 6 / 1 So, B = A / 6 2. From the second equation, we get: B / C = 1 / 4 So, C = 4 * B = 4 * (A / 6) = (2 / 3) * A Now we can determine the actual relative masses of the elements: A = 1 (assuming A's mass as a reference) B = 1 / 6 C = 2 / 3 Based on these actual relative masses, we can rewrite the balanced equations: 1. For the reaction between A and B: \(6A + (1/6)B \rightarrow 4\) Product 2. For the reaction between B and C: \((1/6)B + 4(2/3)C \rightarrow 4\) Product 3. For the reaction between A and C: \(3A + 2(2/3)C \rightarrow 6\) Product

Key concepts

Chemical Reactions

Chemical reactions are the processes in which substances, known as reactants, transform into different substances, called products. These transformations involve the breaking and forming of chemical bonds, characterized by changes in energy and often changes in physical states. For example, in the exercise, we have gases A, B, and C reacting in various combinations to form products.

Understanding the nature of chemical reactions is crucial because it allows us to predict the outcomes and design reactions for various applications, from industrial synthesis to biological processes. In our exercise, we are considering reactions that likely follow the law of constant composition, meaning the same reactants will always react in the same fixed ratios by mass to produce a given compound.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It's based on the conservation of mass and the concept that atoms are indivisible in chemical reactions. Stoichiometry allows chemists to predict the amounts of substances consumed and produced in a given reaction.

In the exercise, stoichiometry helps students to translate the mass of reactants into a table of relative atomic masses. By understanding the ratios of volumes reacted, students can construct balanced chemical equations that reflect the stoichiometric relationships. It also facilitates the calculation of actual relative masses, a critical step in understanding the composition of compounds and their interactions in chemical reactions.

Law of Multiple Proportions

The law of multiple proportions is a principle of chemistry stating that when elements combine, they do so in ratios of small whole numbers (if you keep one element's mass constant). This is essential when elements can form more than one compound together, like carbon and oxygen forming both carbon monoxide and carbon dioxide.

In the exercise, when we calculate mass ratios and determine the simplest formulas for the products (AB, BC, AC), we're applying the law of multiple proportions. This law guides us when interpreting the volume data and constructing the relative masses table. It helps us understand why 6 volumes of gas A can react with 1 volume of gas B to produce 4 volumes of a product, illuminating the underlying fixed ratio relationships between the elements involved.