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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: $$ \begin{array}{lcc} \hline \text { Compound } & \text { Mass of Iodine (g) } & \text { Mass of Fluorine (g) } \\ \hline 1 & 4.75 & 3.56 \\ 2 & 7.64 & 3.43 \\ 3 & 9.41 & 9.86 \\ \hline \end{array} $$ (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 per gram of iodine in the compounds are approximately 0.749 g/g, 0.449 g/g, and 1.048 g/g. These values support the atomic theory because they show the relationship between the masses of fluorine and iodine in each compound formed, in accordance with the law of multiple proportions. The mass ratios are in simple fractions or whole numbers, which indicates that elements combine in fixed whole number ratios, as the atomic theory suggests.

Step by step solution

01

(a) Calculate Mass Ratio

To calculate the mass ratio of fluorine per gram of iodine in each compound, divide the mass of fluorine given in the problem by the mass of iodine. For Compound 1: $$ \text{Mass ratio} = \frac{\text{Mass of Fluorine}}{\text{Mass of Iodine}} = \frac{3.56\,\text{g}}{4.75\,\text{g}} \approx 0.749\,\text{g/g} $$ For Compound 2: $$ \text{Mass ratio} = \frac{\text{Mass of Fluorine}}{\text{Mass of Iodine}} = \frac{3.43\,\text{g}}{7.64\,\text{g}} \approx 0.449\,\text{g/g} $$ For Compound 3: $$ \text{Mass ratio} = \frac{\text{Mass of Fluorine}}{\text{Mass of Iodine}} = \frac{9.86\,\text{g}}{9.41\,\text{g}} \approx 1.048\,\text{g/g} $$
02

(b) Relation to Atomic Theory

The numbers calculated in part (a) show the relationship between the masses of fluorine and iodine in each compound formed. The compounds were formed by combining the individual atoms of iodine and fluorine together in different whole number ratios. This supports the atomic theory by showcasing the law of multiple proportions, which states that when two elements combine to form more than one compound, the masses of one element that combines with a fixed mass of the other element are in a ratio of small whole numbers. Here, we observe that the mass ratios of fluorine to iodine are in ratios close to either whole numbers or simple fractions (e.g., \(\frac{3}{4}\), \(\frac{1}{2}\), and 1). This indicates that elements combine in fixed whole number ratios, as the atomic theory suggests. In conclusion, the calculated mass ratios of fluorine to iodine in each compound demonstrate the law of multiple proportions, which serves as an important foundation for the atomic theory.

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Key Concepts

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

Mass Ratios
In chemistry, mass ratios are a fundamental concept used to understand how elements combine to form compounds. Mass ratios are calculated by comparing the mass of one element within the compound to the mass of another. In the context of our example, we divide the mass of fluorine by the mass of iodine for each compound. This gives us the mass ratio of fluorine per gram of iodine, helping us see how these elements interact on an atomic level.
  • For Compound 1: The mass ratio is approximately 0.749, meaning there are 0.749 grams of fluorine for every gram of iodine.
  • For Compound 2: The mass ratio is around 0.449, indicating a lesser amount of fluorine compared to iodine.
  • For Compound 3: The ratio is about 1.048, showing a slightly larger mass of fluorine relative to iodine.
These ratios illustrate how different quantities of fluorine are needed for combining with iodine to form various compounds. This demonstrates a key part of the law of multiple proportions, which arises from Dalton's atomic theory where elements mix in whole number ratios to yield chemical compounds.
Atomic Theory
Atomic theory is a cornerstone of chemistry, establishing that matter is composed of distinct units called atoms. This theory has evolved over centuries to explain the nature and behavior of atoms. John Dalton further developed this theory in the early 19th century, proposing that:
  • Each element is made of tiny, indivisible particles called atoms.
  • Atoms of the same element are identical in mass and properties.
  • Compounds are formed when atoms of different elements combine in fixed, whole number ratios.
The second part of our exercise highlights the atomic theory by examining the mass ratios of compound 1, 2, and 3. When the chemist observed that these mass ratios can be related back to simple whole number ratios, it supported Dalton's concept that compounds are composed of atoms bonded in consistent proportions, reflecting the concept of atoms themselves being discrete and quantifiable entities.
Chemical Compounds
Chemical compounds are substances formed when two or more elements are chemically bonded together. These compounds feature specific chemical bonds that join atoms together, resulting in a stable arrangement where the properties of the compound can differ vastly from those of the individual elements.
  • The binding of iodine and fluorine in different proportions leads to various compounds, as was the case in our exercise.
  • Such compounds exhibit unique physical and chemical properties that depend on the elements and the ratios in which they are combined.
  • For example, changes in the mass ratio of the fluorine and iodine contribute to the diverse characteristics and reactivity of the compounds.
By connecting the mass ratios to atomic theory principles, our exercise provides a practical example of how chemical compounds are constructed. The ability of elements to form multiple compounds by varying their combining ratios underscores a key aspect of chemical behavior and underscores the applicability of the law of multiple proportions in explaining the existence of diverse chemical substances.

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Most popular questions from this chapter

The element chromium (Cr) consists of four naturally occurring isotopes with atomic masses 49.9460,51.9405,52.9407 , and \(53.9389 \mathrm{u}\). The relative abundances of these four isotopes are \(4.3,83.8,9.5,\) and \(2.4 \%\), respectively. From these data, calculate the atomic weight of chromium.

Give the chemical names of each of the following familiar compounds: (a) \(\mathrm{NaCl}\) (table salt), (b) \(\mathrm{NaHCO}_{3}\) (baking soda), (c) NaOCl (in many bleaches),(d) \(\mathrm{NaOH}\) (caustic soda), (e) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) (smelling salts), (f) \(\mathrm{CaSO}_{4}\) (plaster of Paris).

Each of the following isotopes is used in medicine. Indicate the number of protons and neutrons in each isotope: (a) samarium-153, (b) lutetium-177, (c) bismuth-213, (d) molybdenum-99, (e) lead-212, (f) caesium-131.

Locate each of the following elements in the periodic table; give its name and atomic number, and indicate whether it is a metal, metalloid, or nonmetal: (a) \(\mathrm{Hg},(\mathbf{b}) \mathrm{At},(\mathbf{c}) \mathrm{Mo},\) (d) \(\mathrm{W},(\mathbf{e}) \mathrm{Sn},(\mathbf{f}) \mathrm{V},(\mathbf{g}) \mathrm{K}\).

From the following list of elements \(-\mathrm{Mg}, \mathrm{Li}, \mathrm{T}, \mathrm{Pb}, \mathrm{Se}, \mathrm{Cl}, \mathrm{Xe}, \mathrm{Si},\) C-pick the one that best fits each description. Use each element only once: (a) an alkali metal, \((\mathbf{b})\) an alkaline earth metal, \((\mathbf{c})\) a noble gas, (d) a halogen, \((\mathbf{e})\) a metalloidin group \(14,(\mathbf{f})\) a nonmetal listed in group \(14,(\mathbf{g})\) a metal that forms a \(3+\) ion, \((\mathbf{h})\) a nonmetal that forms a 2 - ion, (i) an element that is used as radiation shielding.

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