Problem 13 - Chemistry The Central Science In Si Units

Question

A chemist finds that \(30.82 \mathrm{~g}\) of nitrogen will react with 17.60 , 35.20,70.40 , or \(88.00 \mathrm{~g}\) of oxygen to form four different compounds. (a) Calculate the mass of oxygen per gram of nitrogen in each compound. (b) How do the numbers in part (a) support Dalton's atomic theory?

Step-by-Step Solution

Answer

(a) 0.571, 1.142, 2.284, 2.854 g/g (b) Ratios show simple whole-number relationships, supporting Dalton's theory.

Step 1: Calculate Oxygen per Gram of Nitrogen in Compound 1

For the first compound: - Given mass of oxygen = \(17.60\, \text{g}\);- Mass of nitrogen = \(30.82\, \text{g}\).Calculate the mass of oxygen per gram of nitrogen:\[\text{Mass of Oxygen per gram of Nitrogen} = \frac{17.60\, \text{g}}{30.82\, \text{g}} = 0.571\, \text{g/g}\]

Step 2: Calculate Oxygen per Gram of Nitrogen in Compound 2

For the second compound:- Given mass of oxygen = \(35.20\, \text{g}\);- Mass of nitrogen = \(30.82\, \text{g}\).Calculate the mass of oxygen per gram of nitrogen:\[\text{Mass of Oxygen per gram of Nitrogen} = \frac{35.20\, \text{g}}{30.82\, \text{g}} = 1.142\, \text{g/g}\]

Step 3: Calculate Oxygen per Gram of Nitrogen in Compound 3

For the third compound:- Given mass of oxygen = \(70.40\, \text{g}\);- Mass of nitrogen = \(30.82\, \text{g}\).Calculate the mass of oxygen per gram of nitrogen:\[\text{Mass of Oxygen per gram of Nitrogen} = \frac{70.40\, \text{g}}{30.82\, \text{g}} = 2.284\, \text{g/g}\]

Step 4: Calculate Oxygen per Gram of Nitrogen in Compound 4

For the fourth compound:- Given mass of oxygen = \(88.00\, \text{g}\);- Mass of nitrogen = \(30.82\, \text{g}\).Calculate the mass of oxygen per gram of nitrogen:\[\text{Mass of Oxygen per gram of Nitrogen} = \frac{88.00\, \text{g}}{30.82\, \text{g}} = 2.854\, \text{g/g}\]

Step 5: Comparing the Ratios

List the calculated ratios:- Compound 1: \(0.571\, \text{g/g}\)- Compound 2: \(1.142\, \text{g/g}\)- Compound 3: \(2.284\, \text{g/g}\)- Compound 4: \(2.854\, \text{g/g}\)Compare these ratios—note that each ratio is a simple multiple of the smallest (0.571) by 2, 4, and 5, respectively, supporting the simple whole-number concept in Dalton's theory.

Step 6: Relate to Dalton's Atomic Theory

Dalton's atomic theory suggests that elements combine in simple, whole number ratios to form compounds. The calculated ratios of oxygen per gram of nitrogen (0.571, 1.142, 2.284, 2.854) demonstrate simple multiple relationships (1, 2, 4, 5). This supports Dalton's postulate that compounds form from atoms in small whole-number ratios.

Key Concepts

Mass Ratios

When dealing with chemical reactions, it is crucial to understand how elements combine with each other. One key aspect of this is mass ratios, which tell us how many grams of one element combine with a specific number of grams of another. In this exercise, the chemist finds how many grams of oxygen combine with a constant amount of nitrogen (30.82 g) to form different compounds.

To calculate the mass of oxygen per gram of nitrogen, you divide the mass of oxygen by the mass of nitrogen. This calculation helps us comprehend the quantity of oxygen needed for every gram of nitrogen in each specific reaction.

• For instance, in the first compound, it is 0.571 g of oxygen per gram of nitrogen.
• In the second compound, it increases to 1.142 g.
• The third goes up further to 2.284 g.
• Finally, the fourth compound has 2.854 g per gram of nitrogen.

This step-by-step calculation process is crucial as it lays the groundwork for understanding the formation of compounds, leading to our next topic.

Compound Formation

Compound formation is a fundamental concept in chemistry that involves elements coming together to form new substances. These substances have properties different from the individual elements that compose them. In the scenario given, different amounts of oxygen react with a fixed mass of nitrogen to produce four unique compounds.

The way nitrogen and oxygen unite in varying proportions showcases the versatile nature of compound formation. Despite the same elements being involved, differing mass ratios lead to distinctive compounds because of their unique atomic structures and configurations. This is a real-world illustration of Dalton’s theory which posits that atoms rearrange and recombine in fixed ratios to form new substances.

Understanding compound formation helps explain the wide variety of materials found in the natural world. The different chemicals that can be manufactured by playing with simple mass ratios emphasize that even slight changes in the composition can lead to vastly different outcomes.

Simple Whole-Number Ratios

Dalton's Atomic Theory postulates that elements combine in simple whole-number ratios to form compounds. This stems from the idea that atoms are indivisible and combine in fixed, predictable ways. Referring back to our exercise, the mass ratios calculated (0.571, 1.142, 2.284, and 2.854) between oxygen and nitrogen support this principle through their simple relationships.

If we examine these ratios closer, each is a multiple of the smallest ratio (0.571 g/g), multiplying correspondingly by 1, 2, 4, and 5. This illustrates the whole-number principle endorsed by Dalton. By understanding these simple ratios:

• 1 part: 0.571 tells us the baseline, the smallest ratio of combination.
• 2 parts: 1.142 is simply 2 times the baseline ratio.
• 4 parts: 2.284 is 4 times the smallest.
• 5 parts: 2.854 is 5 times the base.

This predictability in ratios reaffirms why small, whole-number multiples are integral to the consistency and predictability of chemical reactions and supports the validity of Dalton’s atomic theory even today.