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

How many moles of \(\mathrm{O}\) atoms are in \(4.25 \mathrm{~mol}\) of calcium carbonate, \(\mathrm{CaCO}_{3}\), the chief constituent of seashells?

Short Answer

Expert verified
There are 12.75 moles of oxygen atoms in 4.25 moles of calcium carbonate.

Step by step solution

01

Determine the molecular formula of calcium carbonate

Identify calcium carbonate by its chemical formula, which is \( \text{CaCO}_3 \). This indicates that each molecule has one calcium atom (Ca), one carbon atom (C), and three oxygen atoms (O).
02

Calculate the total number of oxygen atoms per molecule of calcium carbonate

Since there are three oxygen atoms in each molecule of calcium carbonate, to find the total number of moles of oxygen atoms, multiply the moles of calcium carbonate by 3.
03

Calculate the number of moles of oxygen atoms

Multiply the given amount of calcium carbonate in moles (4.25 mol) by the number of oxygen atoms per molecule (3) to find the total moles of oxygen atoms: \( 4.25 \text{ mol CaCO}_3 \times 3 \text{ mol O atoms/mol CaCO}_3 = 12.75 \text{ mol O atoms}\).

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.

Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. It is a fundamental concept in chemistry that allows us to predict the amounts of substances consumed and produced in a given reaction. When dealing with stoichiometry, it is vital to understand the mole ratio, which is the proportion of reactants to products based on the balanced chemical equation.

To solve stoichiometric problems, such as determining the amount of reactants needed or products formed, we use the coefficients in the balanced chemical equation. In our example involving calcium carbonate, \( \text{CaCO}_3 \), there is no chemical reaction per se, but the concept still applies when breaking down the compound into its constituent elements. Since the molecular formula indicates a 1:3 ratio of calcium to oxygen atoms, we know that for every mole of calcium carbonate, there are three moles of oxygen atoms. Understanding this ratio is fundamental to solving the given problem.
Molecular Formula
The molecular formula represents the actual number of atoms of each element in a molecule. In the case of calcium carbonate, the molecular formula is \( \text{CaCO}_3 \), denoting it contains one atom of calcium (Ca), one atom of carbon (C), and three atoms of oxygen (O). Knowing this is crucial because it gives us the specifics needed to calculate the amount of each element in a given sample.

The formula showcases that these elements are in a fixed ratio, which is constant and does not change regardless of the quantity of the substance we have. So, if we need to know the moles of oxygen in a sample, we look at the ratio indicated by the formula. For every single molecule of calcium carbonate, there will always be precisely three oxygen atoms, a fact that is essential to the resolution of stoichiometric problems.
Mole Concept
The mole concept is a bridge between the microscopic world of atoms and the macroscopic realm that we can measure in the laboratory. One mole is defined as exactly \(6.022 \times 10^{23}\) entities, be it atoms, molecules, ions, or other particles. This number is known as Avogadro's number.

When working with moles, we can translate between the count of particles and the mass using molar mass, but in the context of our problem, we are concerned with the mole ratio. As we know, the molecular formula of calcium carbonate informs us that for every mole of this compound, we have three moles of oxygen atoms. This direct relationship allows us to scale our calculations up from the molecular to the molar level, as demonstrated in the exercise, where \(4.25 \text{ mol}\) of calcium carbonate equate to \(4.25 \times 3 = 12.75 \text{ mol}\) of oxygen atoms, making complex quantitative analysis of compounds accessible and manageable.

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

Cinnamic acid, a compound related to the flavor component of cinnamon, is \(72.96 \%\) carbon, \(5.40 \%\) hydrogen, and the rest is oxygen. What is the empirical formula of this acid?

The following are empirical formulas and the masses per mole for three compounds. What are their molecular formulas? (a) \(\mathrm{NaS}_{2} \mathrm{O}_{3} ; 270.4 \mathrm{~g} / \mathrm{mol}\) (c) \(\mathrm{C}_{2} \mathrm{HCl} ; 181.4 \mathrm{~g} / \mathrm{mol}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{2} \mathrm{Cl} ; 147.0 \mathrm{~g} / \mathrm{mol}\)

How many moles of \(\mathrm{Bi}\) atoms are needed to combine with \(1.58 \mathrm{~mol}\) of \(\mathrm{O}\) atoms to make \(\mathrm{Bi}_{2} \mathrm{O}_{3}\) ?

Chlorine atoms cause chain reactions in the stratosphere that destroy ozone that protects the earth's surface from ultraviolet radiation. The chlorine atoms come from chlorofluorocarbons, compounds that contain carbon, fluorine, and chlorine, which were used for many years as refrigerants. One of these compounds is Freon-12, \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\). If a sample contains \(1.0 \times 10^{-9} \mathrm{~g}\) of \(\mathrm{Cl}\), how many grams of \(\mathrm{F}\) should be present if all of the \(\mathrm{F}\) and \(\mathrm{Cl}\) atoms in the sample came from \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) molecules?

Oxygen gas can be produced in the laboratory by decomposition of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) $$ \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) $$ How many \(\mathrm{kg}\) of \(\mathrm{O}_{2}\) can be produced from \(1.0 \mathrm{~kg}\) of \(\mathrm{KClO}_{3} ?\)

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