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.

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!

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

There are two different isotopes of bromine atoms. Under normal conditions, elemental bromine consists of \(\mathrm{Br}_{2}\) molecules, and the mass of a \(\mathrm{Br}_{2}\) molecule is the sum of the masses of the two atoms in the molecule. The mass spectrum of \(\mathrm{Br}_{2}\) consists of three peaks: $$ \begin{array}{lc} \hline \text { Mass (u) } & \text { Relative Size } \\ \hline 157.836 & 0.2569 \\ 159.834 & 0.4999 \\ 161.832 & 0.2431 \\ \hline \end{array} $$ (a) What is the origin of each peak (of what isotopes does each consist)? (b) What is the mass of each isotope? (c) Determine the average molecular mass of a \(\mathrm{Br}_{2}\) molecule. (d) Determine the average atomic mass of a bromine atom. (e) Calculate the abundances of the two isotopes.

Naturally occurring lead has the following isotopic abundances: $$ \begin{array}{lcc} \hline \text { Isotope } & \text { Abundance (\%) } & \text { Atomic mass (u) } \\ \hline{ }^{204} \mathrm{~Pb} & 1.4 & 203.9730 \\ { }^{206} \mathrm{~Pb} & 24.1 & 205.9744 \\ { }^{207} \mathrm{~Pb} & 22.1 & 206.9759 \\ { }^{208} \mathrm{~Pb} & 52.4 & 207.9766 \\ \hline \end{array} $$ (a) What is the average atomic mass of \(\mathrm{Pb}\) ? (b) Sketch the mass spectrum of \(\mathrm{Pb}\).

Bromine has two naturally occurring isotopes, bromine- 79 (atomic mass \(=78.9183 \mathrm{u} ; \quad\) abundance \(=50.69 \%)\) a \(\mathrm{nd}\) bromine-81 \(\quad\) (atomic mass \(=80.9163 \mathrm{u} ;\) abundance \(=\) $49.31 \%) .$ Calculate the atomic weight of bromine.

Name the following ionic compounds: (a) \(\mathrm{Li}_{2} \mathrm{O},\) (b) \(\mathrm{FeCl}_{3}\), (c) \(\mathrm{NaClO}\) (d) \(\mathrm{CaSO}_{3}\), (e) \(\mathrm{Cu}(\mathrm{OH})_{2}\) (f) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}\), (g) \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\) (h) \(\mathrm{Cr}_{2}\left(\mathrm{CO}_{3}\right)_{3}\) (i) $\mathrm{K}_{2} \mathrm{CrO}_{4},(\mathbf{j})\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$.

Fill in the blanks in the following table: $$ \begin{array}{l|l|l|l|} \hline \text { Cation } & \text { Anion } & \text { Formula } & \text { Name } \\\ \hline \mathrm{Ni}^{2+} & \mathrm{CH}_{3} \mathrm{COO} & & \text { Sodium carbonate } \\ & & & \\ \mathrm{Ca}^{2+} & \mathrm{F}^{-} & & \\ & & \mathrm{Cu}\left(\mathrm{ClO}_{4}\right)_{2} & \\ & & & \\ & & & \\ \hline \end{array} $$

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