Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 14

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:(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 of fluorine per gram of iodine for each compound can be represented as follows: Compound 1: ratio = \(\frac{F1}{I1}\) Compound 2: ratio = \(\frac{F2}{I2}\) Compound 3: ratio = \(\frac{F3}{I3}\) These ratios indicate the proportions of fluorine and iodine atoms in each compound. If these values are close to small whole numbers, it supports the atomic theory, which proposes that atoms combine in fixed proportions to form chemical compounds.

Step by step solution

01

Calculate the mass of fluorine per gram of iodine

To calculate the mass of fluorine per gram of iodine in each compound, we must first find the ratios for each compound: Ratio in Compound 1: F1/I1 Ratio in Compound 2: F2/I2 Ratio in Compound 3: F3/I3
02

Input the given values

Since we have not received any numerical values for the experiment, we will use the given variables to showcase this example.
03

Analyze ratios and the atomic theory

In the second part of the problem, we have to discuss how the ratios (calculated in step 1) support the atomic theory. The atomic theory states that matter is composed of atoms, and when atoms combine to form chemical compounds, they do so in fixed proportions. The ratios calculated from step 1 represent the atomic proportions involved in each of the compounds being analyzed. If these values are close to a small whole number, it can be inferred that the data supports 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.

Start your free trial

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

Two substances have the same molecular and empirical formulas. Does this mean that they must be the same compound?

Each of the following elements is capable of forming an ion in chemical reactions. By referring to the periodic table, predict the charge of the most stable ion of each: (a) Mg, (b) Al, \((\mathbf{c}) \mathrm{K},(\mathbf{d}) \mathrm{S},(\mathbf{e}) \mathrm{F}\)

Selenium, an element required nutritionally in trace quantities, forms compounds analogous to sulfur. Name the following ions: (a) \(\operatorname{SeO}_{4}^{2-},(\mathbf{b}) \mathrm{Se}^{2-},(\mathbf{c}) \mathrm{HSe}^{-},(\mathbf{d}) \mathrm{HSeO}_{3}^{-}\)

A cube of gold that is 1.00 \(\mathrm{cm}\) on a side has a mass of 19.3 \(\mathrm{g}\) A single gold atom has a mass of 197.0 amu. (a) How many gold atoms are in the cube? (b) From the information given,estimate the diameter in \(A\) of a single gold atom. (c) What assumptions did you make in arriving at your answer for part (b)?

For each of the following elements, write its chemical symbol, determine the name of the group to which it belongs (Table \(2.3 ),\) and indicate whether it is a metal, metalloid, or nonmetal: (a) potassium, (b) iodine, (c) magnesium, (d) argon, (e) sulfur.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation
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