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

Determine the mass percent of each of the elements in the fungicide copper(II) oleate, \(\mathrm{Cu}\left(\mathrm{C}_{18} \mathrm{H}_{33} \mathrm{O}_{2}\right)_{2}\)

Short Answer

Expert verified
The mass percent of each element in Copper(II) oleate can be determined by the method of calculating mass percent using atomic masses and mole concept. The mass percent of an element = (mass of the element in one mole of the compound / molar mass of the compound) x 100. It's necessary to compute the molar mass of Copper(II) oleate as the first step, and then use it to calculate the mass percent of each element.

Step by step solution

01

Determine the Molecular Mass of Copper(II) Oleate

Molecular weight of Copper(II) oleate, Cu(C18H33O2)2 can be calculated by adding the atomic masses of its constituent atoms. It is important to use the correct atomic weights in atomic mass units (amu). The atomic weights of copper, carbon, hydrogen, and oxygen are 63.55 amu, 12.01 amu, 1.01 amu, and 16.00 amu respectively. Then multiply each of these by the number of each atom in the molecule and sum up the results. So, the molecular weight of Copper(II) oleate = 63.55 + 2*[(18*12.01) + (33*1.01) + (2*16.00)] amu.
02

Calculate the Mass Percent

The mass percent of an element is calculated by dividing the total mass of each element in one molecule by the molar mass of copper(II) oleate. Then, multiply the result by 100 to get a percentage. The formulas are as follows: \%Cu = [mass of Cu / molar mass of copper(II) oleate] x100. \%C = [mass of C / molar mass of copper(II) oleate] x100. \%H = [mass of H / molar mass of copper(II) oleate] x100. \%O = [mass of O / molar mass of copper(II) oleate] x100.
03

Execute the Calculations

Now substitute the earlier determined values into the respective formulas to find out the mass percentages of copper, carbon, hydrogen, and oxygen in copper(II) oleate.

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.

Molecular Weight Determination
Determining the molecular weight of a compound like copper(II) oleate is a fundamental step in understanding its chemical composition. To do this accurately, you need to sum the atomic masses of all atoms in the molecule. This requires knowledge of both the chemical formula and the atomic weights of the elements involved.
Copper(II) oleate is denoted by \({\text{Cu}(\text{C}_{18}\text{H}_{33}\text{O}_{2})_{2}}\). The molecular weight can be calculated as follows:
  • Copper (Cu): One atom at 63.55 amu.
  • Carbon (C): 18 carbon atoms in each oleate group. With two oleate groups, it's \(18 \times 2 = 36\) atoms total, each with an atomic mass of 12.01 amu.
  • Hydrogen (H): 33 hydrogen atoms per oleate contributes \(33 \times 2 = 66\) atoms, each with an atomic mass of 1.01 amu.
  • Oxygen (O): 2 oxygen atoms per oleate total to \(2 \times 2 = 4\), each with an atomic weight of 16.00 amu.
By adding these up, the molecular weight of copper(II) oleate can be derived and used in further chemical analysis.
Chemical Composition Analysis
Once the molecular weight is determined, you move on to calculating the mass percent of each element. Mass percent indicates how much of the mass of a molecule is contributed by a particular element. This is crucial for understanding the role of each element in a compound's properties and reactions.
To find the mass percent of an element:
  • Divide the mass of the element in one molecule by the total molar mass of the compound.
  • Multiply the result by 100 to convert it into a percentage.
For example, the mass percent of copper \(%Cu\) is calculated as \[ \frac{\text{mass of Cu}}{\text{molar mass of copper(II) oleate}} \times 100 \].Follow similar steps for carbon, hydrogen, and oxygen. This analysis helps chemists understand the proportions of each element in the compound, facilitating insights into its chemical behavior.
Copper(II) Oleate
Copper(II) oleate is an interesting compound commonly used in agriculture as a fungicide. Its structural complexity means understanding its chemical composition is necessary for effective application and handling.
The chemical formula of copper(II) oleate, \(\text{Cu}(\text{C}_{18}\text{H}_{33}\text{O}_{2})_{2}\), reveals that twice the oleate ion is associated with a single copper ion. This setup not only affects the molecular weight but also the physical and chemical properties of the compound.
As with many copper compounds, it plays a crucial role in pest management. To maximize its efficiency, precise knowledge of its chemical makeup—down to each element's mass percent—is vital. By understanding these components, users can optimize use and environmental impact, highlighting the significance of molecular weight determination and composition analysis.

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

To deposit exactly one mole of \(\mathrm{Ag}\) from an aqueous solution containing \(\mathrm{Ag}^{+}\) requires a quantity of electricity known as one faraday (F). The electrodeposition requires that each \(\mathrm{Ag}^{+}\) ion gain one electron to become an Ag atom. Use appropriate physical constants listed on the inside back cover to obtain a precise value of the Avogadro constant, \(N_{A}\).

Appendix E describes a useful study aid known as concept mapping. Using the method presented in Appendix \(\mathrm{E},\) construct a concept map illustrating the different concepts in Sections \(3-2\) and \(3-3\).

The insecticide dieldrin contains carbon, hydrogen, oxygen, and chlorine. When burned in an excess of oxygen, a 1.510 g sample yields \(2.094 \mathrm{g} \mathrm{CO}_{2}\) and \(0.286 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) The compound has a molecular mass of 381 u and has half as many chlorine atoms as carbon atoms. What is the molecular formula of dieldrin?

Dry air is essentially a mixture of the following entities: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{Ar},\) and \(\mathrm{CO}_{2} .\) The composition of dry air, in mole percent, is \(78.08 \% \mathrm{N}_{2}, 20.95 \% \mathrm{O}_{2}, 0.93 \% \mathrm{Ar}\) and \(0.04 \% \mathrm{CO}_{2}\). (a) What is the mass, in grams, of a sample of air that contains exactly one mole of the entities? (b) Dry air also contains other entities in much smaller amounts. For example, the mole percent of krypton (Kr) is about \(1.14 \times 10^{-4} \% .\) Given that the density of dry air is about \(1.2 \mathrm{g} / \mathrm{L}\) at room temperature, what mass of krypton could be obtained from exactly one cubic meter of dry air?

Determine the mass percent of each of the elements in the antimalarial drug quinine, \(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{N}_{2} \mathrm{O}_{2}\)

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