Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In a series of experiments, a chemist prepared three different compounds that contain only iodine and fluorine and determined the mass of each element in each compound: $$ \begin{array}{ccc} \hline \text { Compound } & \text { Mass of lodine (g) } & \text { Mass of Fluorine (g) } \\ \hline 1 & 4.75 & 3.56 \\ 2 & 7.64 & 3.43 \\ 3 & 9.41 & 9.86 \\ \hline \end{array} $$ (a) Calculate the mass of fluorine per gram of iodine in each compound. (b) How do the numbers in part (a) support the atomic theory?

Short Answer

Expert verified
The mass ratios of fluorine to iodine for the three compounds are approximately \(\frac{3}{4}\), \(\frac{1}{2}\), and \(\frac{1}{1}\), which are simple whole number ratios. These ratios support the atomic theory, as it demonstrates that iodine and fluorine combine in fixed, simple, whole number ratios to form different compounds.

Step by step solution

01

Find the mass ratio of Fluorine to Iodine for each compound.

We can find the mass ratio by dividing the mass of fluorine by the mass of iodine in each compound: 1. For Compound 1: Ratio = \(\frac{3.56}{4.75}\) 2. For Compound 2: Ratio = \(\frac{3.43}{7.64}\) 3. For Compound 3: Ratio = \(\frac{9.86}{9.41}\)
02

Calculate the mass ratios.

Now, let's calculate the mass ratios for each compound: 1. For Compound 1: Ratio = \(\frac{3.56}{4.75}\) = 0.749 2. For Compound 2: Ratio = \(\frac{3.43}{7.64}\) = 0.449 3. For Compound 3: Ratio = \(\frac{9.86}{9.41}\) = 1.048 These numbers represent the mass of fluorine per gram of iodine in each compound.
03

Analyze the ratios for their relationship to atomic theory.

The atomic theory suggests that elements combine in simple, whole number ratios to form compounds. Let's analyze the ratios we calculated in step 2: 1. Compound 1: Ratio = 0.749 ≈ \(\frac{3}{4}\) (a simple whole number ratio) 2. Compound 2: Ratio = 0.449 ≈ \(\frac{1}{2}\) (a simple whole number ratio) 3. Compound 3: Ratio = 1.048 ≈ \(\frac{1}{1}\) (a simple whole number ratio) Observing these ratios, we can see that they approximate simple, whole number ratios. This supports the atomic theory, as it shows that iodine and fluorine are combining in fixed, simple, whole number ratios to form different compounds.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Atomic Theory
Atomic theory is a fundamental concept in chemistry. It posits that matter is made up of tiny, indivisible particles known as atoms. Atoms are the basic units of chemical elements and combine in simple whole-number ratios to form compounds. This theory was first proposed by John Dalton in the early 19th century. It laid the foundation for modern chemistry. One of its key principles is that atoms of different elements can combine in varied fixed ratios to form different compounds, demonstrating their distinct combinations and properties.

In the context of the given exercise, observing how the mass ratios of fluorine to iodine in three compounds are simple and constant provides strong evidence in support of atomic theory. It confirms that these elements are aggregating in predictable, measurable ratios that align perfectly with Dalton’s ideas. This consistency highlights the inherent structure and laws guiding chemical reactions and formations.
Fluorine and Iodine Compounds
Fluorine and iodine are halogens, a group of highly reactive non-metals. Fluorine is the most electronegative element, readily forming compounds by a strong attraction to electrons. Meanwhile, iodine is less electronegative but still reactive, often forming binary compounds such as those with fluorine.

In the experiment described, three different fluorine and iodine compounds were analyzed to determine the mass of each element present. These experiments shed light on how these elements interact with each other in various proportions.
  • Fluorine compounds typically manifest in forming powerful bonds due to fluorine's high reactivity.
  • Iodine, though less reactive than fluorine, still forms stable compounds and often shows varied stoichiometric ratios in its compounds.
Understanding these properties helps chemists predict how fluorine and iodine will combine, their potential uses, and how they might behave in different chemical reactions.
Mass Ratio Calculation
Mass ratio calculation is a process used to quantify the relative amounts of different elements in a compound. By finding the mass ratio of one element to another, chemists can determine how compounds are composed.

For each iodine and fluorine compound, this exercise required finding the mass ratio of fluorine to iodine. This was done by dividing the mass of fluorine by the mass of iodine.
  • In Compound 1: The mass ratio calculated was 0.749.
  • In Compound 2: The mass ratio calculated was 0.449.
  • In Compound 3: The mass ratio calculated was 1.048.
These calculations reflect how much fluorine exists for each gram of iodine in the compounds. By consistently applying this mass ratio method, chemists can better understand the composition and stoichiometry of compounds, which contributes to broader chemical analysis and applications.
Law of Multiple Proportions
The Law of Multiple Proportions is closely related to Dalton’s atomic theory. It states that if two elements form more than one compound together, the ratios of the masses of the second element which combine with a fixed mass of the first element are simple whole numbers.

This law is a critical principle in chemistry, and it further supports the idea that chemical reactions are predictable and governed by specific rules. For the compounds of fluorine and iodine given in the exercise, the experimentally determined ratios of mass:
  • 0.749 for Compound 1, simplifying to approximately \(\frac{3}{4}\).
  • 0.449 for Compound 2, simplifying to about \(\frac{1}{2}\).
  • 1.048 for Compound 3, simplifying to \(\frac{1}{1}\).
Each of these can be expressed as a simple ratio of whole numbers, just as the law predicts. This adds further evidence that the compound formations are not arbitrary but instead follow these natural laws. The law of multiple proportions is essential for understanding compound formation and chemical stoichiometry principles.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The first atoms of seaborgium (Sg) were identified in 1974 . The longest-lived isotope of Sg has a mass number of \(266 .\) (a) How many protons, electrons, and neutrons are in an \({ }^{266} \mathrm{Sg}\) atom? (b) Atoms of Sg are very unstable, and it is therefore difficult to study this element's properties. Based on the position of Sg in the periodic table, what element should it most closely resemble in its chemical properties?

The elements of group \(4 \mathrm{~A}\) show an interesting change in properties moving down the group. Give the name and chemical symbol of each element in the group and label it as a nonmetal, metalloid, or metal.

The compound cyclohexane is an alkane in which six carbon atoms form a ring. The partial structural formula of the compound is as follows: (a) Complete the structural formula for cyclohexane. (b) Is the molecular formula for cyclohexane the same as that for \(n\) -hexane, in which the carbon atoms are in a straight line? If possible, comment on the source of any differences. (c) Propose a structural formula for cyclohexanol, the alcohol derived from cyclohexane.

Identify the element represented by each of the following symbols and give the number of protons and neutrons in each: (a) \(33^{7} X\), (b) \(\frac{127}{53} \mathrm{X}\), (c) \(\frac{152}{63} \mathrm{X}\) (d) \({ }_{83}^{209} \mathrm{X}\).

What can we tell about a compound when we know the empirical formula? What additional information is conveyed by the molecular formula? By the structural formula? Explain in each case.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free