/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A 1.0 -g sample of carbon dioxid... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 1.0 -g sample of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is fully decomposed into its elements, yielding 0.273 g of carbon and 0.727 \(\mathrm{g}\) of oxygen. (a) What is the ratio of the mass of \(\mathrm{O}\) to \(\mathrm{C} ?\) (b) If a sample of a different compound decomposes into 0.429 \(\mathrm{g}\) of carbon and 0.571 \(\mathrm{g}\) of oxygen, what is its ratio of the mass of \(\mathrm{O}\) to \(\mathrm{C} ?(\mathbf{c})\) According to Dalton's atomic theory, what is the empirical formula of the second compound?

Short Answer

Expert verified
The mass ratio of O to C for the first compound is \( \frac{0.727}{0.273} \approx 2.66 \) and for the second compound is \( \frac{0.571}{0.429} \approx 1.33 \). The empirical formula of the second compound can be found using the mole ratio, resulting in the empirical formula as \( C_{1}O_{1} \) or simply CO.

Step by step solution

01

Calculation of mass ratio for the first compound

To calculate the mass ratio of O to C for the first compound, we have the following masses: Mass of Carbon (C) = 0.273 g Mass of Oxygen (O) = 0.727 g Now we can find the ratio of the mass of O to C: \( Ratio_{O:C} = \frac{Mass_{O}}{Mass_{C}} \) \( Ratio_{O:C} = \frac{0.727}{0.273} \)
02

Calculation of mass ratio for the second compound

For the second compound, we have the following masses: Mass of Carbon (C) = 0.429 g Mass of Oxygen (O) = 0.571 g Now we can find the ratio of the mass of O to C: \( Ratio_{O:C} = \frac{Mass_{O}}{Mass_{C}} \) \( Ratio_{O:C} = \frac{0.571}{0.429} \)
03

Calculating the empirical formula of the second compound

To find the empirical formula for the second compound, we will look at the ratio of O to C and compare it with the atomic masses of Oxygen and Carbon. The atomic mass of Oxygen (O) is 16 g/mol, and Carbon (C) is 12 g/mol. Now, we can calculate the mole ratio of O to C: \( Mole \ Ratio_{O:C} = \frac{\frac{0.571}{16}}{\frac{0.429}{12}} \) Simplify the ratio by finding the simplest whole number ratio: Now that we have the mole ratio in simplest whole number ratio, we can write the empirical formula for the second compound. Let the empirical formula is CXOY, where X and Y are the simplest whole-number quantities of C and O, respectively. Now we have the empirical formula for the second compound.

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 91Ó°ÊÓ!

Key Concepts

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

Mass Ratio
In chemistry, understanding the mass ratio of elements in a compound is key to determining its composition. To find the mass ratio of oxygen to carbon in a compound, we divide the mass of oxygen by the mass of carbon. For example, in the decomposition of carbon dioxide
  • Mass of Carbon ( \( C \)) = 0.273 g
  • Mass of Oxygen ( \( O \)) = 0.727 g
Thus, the mass ratio of O to C is calculated as:\[Ratio_{O:C} = \frac{Mass_{O}}{Mass_{C}} = \frac{0.727}{0.273}\]This ratio helps to understand the proportion of each element in the compound. Similarly, for another compound decomposing into 0.429 g of carbon and 0.571 g of oxygen, the mass ratio of O to C would be:\[Ratio_{O:C} = \frac{0.571}{0.429}\]Knowing these ratios is crucial when comparing different compounds and understanding their chemical properties.
Dalton's Atomic Theory
Dalton's Atomic Theory is a fundamental concept in chemistry that provides insights into the nature of matter. According to this theory, elements are composed of tiny, indivisible particles called atoms. Here are some key points:
  • Each atom of an element is identical in mass and properties.
  • Atoms of different elements have distinct masses and properties.
  • Atoms combine in fixed, simple ratios to form compounds.
In the context of calculating empirical formulas, Dalton's theory helps us understand how atoms combine in specific ratios. Using the mass ratios calculated, these values can be converted to mole ratios, helping us deterministically create the empirical formula of a compound. For example, from the mass ratio of O to C in the second compound, we can convert this to a simple whole number atom ratio, using atomic masses: - Atomic mass of Oxygen (O) = 16 g/mol - Atomic mass of Carbon (C) = 12 g/mol This reflects the fixed and simple way atoms combine, aligning with Dalton's idea about chemical compounds.
Chemical Decomposition
Chemical decomposition is the process where a single compound breaks down into two or more elements or new compounds. This can be thought of as "reverse synthesis." Understanding it is essential in chemistry because it helps us study the composition of compounds. When a compound like carbon dioxide decomposes, it can form elemental carbon and oxygen. - Example: CO₂ decomposing into carbon (C) and oxygen (O). The decomposition reaction provides a way to verify the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction. When analyzing decomposition reactions, recognizing the resulting mass ratios of elements aids in determining the compound’s original composition. This is instrumental when deciding the empirical formula. By observing the decomposition processes and calculating mass ratios, like in the given problem, students can apply knowledge of chemical bonding and composition to broader chemical analyses.

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

What fraction of the \(\alpha\) particles in Rutherford's gold foil experiment are scattered at large angles? Assume the gold foil is two layers thick, as shown in Figure \(2.9,\) and that the approximate diameters of a gold atom and its nucleus are 2.7 A and \(1.0 \times 10^{-4} \hat{\mathrm{A}}\) , respectively. Hint: Calculate the cross sectional area occupied by the nucleus as a fraction of that occupied by the atom. Assume that the gold nuclei in each layer are offset from each other.

A charged particle moves between two electrically charged plates, as shown here.\(\begin{array}{l}{\text { (a) What is the sign of the electrical charge on the particle? }} \\ {\text { (b) As the charge on the plates is increased, would you ex- }} \\ {\text { pect the bending to increase, decrease, or stay the same? }}\end{array}$$\begin{array}{l}{\text { (c) As the mass of the particle is increased while the speed of }} \\ {\text { the particles remains the same, would you expect the bending }} \\ {\text { to increase, decrease, or stay the same? [Section } 2.2 ]}\end{array}\)

Which of the three subatomic particles was discovered first- the proton, neutron, or electron? Which one was discovered last?

Consider an atom of 10 \(\mathrm{B}\) . (a) How many protons, neutrons, and electrons does this atom contain? (b) What is the symbol of the atom obtained by adding one proton to \(^{10} \mathrm{B} ?\) (c) What is the symbol of the atom obtained by adding one neutron to 10 \(\mathrm{B}\) ? (c) What is the symbol of the atom obtained by adding one neutron to 10 \(\mathrm{B} ?(\mathbf{d})\) Are either of the atoms obtained in parts (b) and (c) isotopes of 10 \(\mathrm{B} ?\) If so which one?

Answer the following questions without referring to Table \(2.1 :\) (a) What are the main subatomic particles that make up the atom? (b) What is the relative charge (in multiples of the electronic charge ) of each of the particles? (c) Which of the particles is the most massive? (d) Which is the least massive?

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.