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Show, with calculations, how the following data illustrate the law of multiple proportions: Compound 1: 47.5 mass \(\%\) sulfur and 52.5 mass \(\%\) chlorine Compound 2: 31.1 mass \(\%\) sulfur and 68.9 mass \(\%\) chlorine

Short Answer

Expert verified
The mass ratio of sulfur to chlorine in Compound 1 to Compound 2 is 2:1, illustrating the law of multiple proportions.

Step by step solution

01

Determine the mass ratios of sulfur to chlorine in each compound

For Compound 1: Calculate the mass ratio of sulfur to chlorine. Mass of sulfur = 47.5, mass of chlorine = 52.5. The ratio is \ \(\frac{47.5}{52.5} = 0.9048\). For Compound 2: Calculate the mass ratio of sulfur to chlorine. Mass of sulfur = 31.1, mass of chlorine = 68.9. The ratio is \ \(\frac{31.1}{68.9} = 0.4514 \).
02

Find the ratio between the mass ratios of the two compounds

Compare the mass ratio of sulfur to chlorine in Compound 1 to that in Compound 2. \ \(\frac{0.9048}{0.4514} = 2.004 \). This is approximately a simple whole number ratio of 2:1.
03

Conclude based on the law of multiple proportions

According to the law of multiple proportions, the ratio of the mass ratios of sulfur to chlorine between the two compounds is 2:1. This illustrates that these compounds are related and follow the law of multiple proportions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mass ratio
The mass ratio is a fundamental concept in chemistry that tells us how the mass of one element compares to another within a compound. It is calculated by dividing the mass percentage of one element by the mass percentage of another. This ratio helps us understand the relative amounts of each element in a compound.

In the given exercise, we find the mass ratios of sulfur to chlorine in two different compounds. For Compound 1, the mass of sulfur is 47.5% and chlorine is 52.5%. The mass ratio is \(\frac{47.5}{52.5}\) which equals 0.9048. For Compound 2, the mass of sulfur is 31.1% and chlorine is 68.9%. Here, the mass ratio is \(\frac{31.1}{68.9}\) which equals 0.4514.

To find how these compounds are related, we compare these mass ratios. We divide the mass ratio of Compound 1 by the mass ratio of Compound 2, which gives \(\frac{0.9048}{0.4514}\) that equals approximately 2.004. This ratio, simplified to a simple whole number, is essential to applying the law of multiple proportions.
chemical compounds
Chemical compounds are substances made up of two or more elements that are chemically bonded together. Each compound has a unique composition that is defined by the elements involved and their respective ratios. For example, in the given exercise, we have two compounds containing sulfur and chlorine.

Compound 1 has 47.5% sulfur and 52.5% chlorine, while Compound 2 has 31.1% sulfur and 68.9% chlorine. Despite having the same two elements, the different proportions of sulfur and chlorine result in distinct compounds. The law of multiple proportions helps us understand these differences by examining how the ratios of elements compare across different compounds.

This distinctive composition dictates the compound's properties and its behavior in chemical reactions, making the study of mass ratios and element proportions crucial in chemistry.
sulfur to chlorine ratio
The sulfur to chlorine ratio helps illustrate the law of multiple proportions, which states that when elements combine into more than one compound, the ratios of the masses of the second element that combine with a fixed mass of the first element can be expressed as a ratio of small whole numbers.

In our case study, the sulfur to chlorine ratios in Compound 1 and Compound 2 are calculated as follows:
  • For Compound 1: The ratio is \(\frac{47.5}{52.5} = 0.9048 \)
  • For Compound 2: The ratio is \(\frac{31.1}{68.9} = 0.4514 \)
To illustrate the law of multiple proportions, we compare these two ratios: we divide them \(\frac{0.9048}{0.4514} \), obtaining approximately 2.004. This translates to a simple whole number ratio of 2:1.

This confirms that sulfur and chlorine form two different compounds with different mass ratios that adhere to the law of multiple proportions. Understanding these ratios is crucial for predicting compound formation and analyzing chemical reactions.

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Most popular questions from this chapter

The anticancer drug Platinol (cisplatin), \(\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\), reacts with a cancer cell's DNA and interferes with its growth. (a) What is the mass \% of platinum (Pt) in Platinol? (b) If Pt costs \(\$ 51 / \mathrm{g},\) how many grams of Platinol can be made for \(\$ 1.00\) million (assume that the cost of Pt determines the cost of the drug)?

How is a structural formula similar to a molecular formula? How is it different?

Chlorine has two naturally occurring isotopes, \({ }^{35} \mathrm{Cl}\) (abundance \(76 \%\) ) and \({ }^{37} \mathrm{Cl}\) (abundance \(24 \%\) ), and it occurs as diatomic (two-atom) molecules. In a mass spectrum, peaks are seen for the molecule and for the separated atoms. (a) How many peaks are in the mass spectrum? (b) What is the \(m / e\) value for the heaviest particle and for the lightest particle?

Nitrogen monoxide (NO) is a bioactive molecule in blood. Low NO concentrations cause respiratory distress and the formation of blood clots. Doctors prescribe nitroglycerin, \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\), and isoamyl nitrate, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{ONO}_{2},\) to increase the blood level of NO. If each compound releases one molecule of NO per atom of \(\mathrm{N}\) it contains, calculate the mass percent of NO in each.

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