/*! 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 148 A substance \(\mathrm{X}_{2} Z\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A substance \(\mathrm{X}_{2} Z\) has the composition (by mass) of \(40.0 \% \mathrm{X}\) and \(60.0 \% \mathrm{Z}\). What is the composition (by mass) of the compound \(\mathrm{XZ}_{2}\) ?

Short Answer

Expert verified
The composition (by mass) of the compound \(\mathrm{XZ}_{2}\) is 25% \(\mathrm{X}\) and 75% \(\mathrm{Z}\).

Step by step solution

01

Find the Mass Ratio of X and Z in X2Z

Given the substance \(\mathrm{X}_{2}\mathrm{Z}\) with a 40% mass of X and 60% mass of Z, we can find the mass ratio of \(\mathrm{X}\) and \(\mathrm{Z}\) in the compound. As the percentages already add up to 100%, we can simply use these values as the ratio. Mass Ratio of X and Z in \(\mathrm{X}_{2}\mathrm{Z}\) = 40/60
02

Simplify the Mass Ratio

To make it easy to work with, let's simplify the mass ratio from step 1. Mass Ratio of X and Z in \(\mathrm{X}_{2}\mathrm{Z}\) = 40/60 = 2/3 That means, in the compound \(\mathrm{X}_{2}\mathrm{Z}\), there are 2 parts of X for every 3 parts of Z.
03

Set up the Mass Ratio for XZ2

Now, let's find the mass ratio of \(\mathrm{X}\) and \(\mathrm{Z}\) in the compound \(\mathrm{XZ}_{2}\). As there are 2 parts of Z for every part of X in \(\mathrm{XZ}_{2}\), we will multiply the mass of Z in \(\mathrm{X}_{2}\mathrm{Z}\) by 2 to find the mass ratio of \(\mathrm{X}\) and \(\mathrm{Z}\) in \(\mathrm{XZ}_{2}\). Mass Ratio of X and Z in \(\mathrm{XZ}_{2}\) = 2/6
04

Simplify the Mass Ratio for XZ2

As we did in step 2, let's simplify the mass ratio from step 3. Mass Ratio of X and Z in \(\mathrm{XZ}_{2}\) = 2/6 = 1/3 That means there are 1 part of X for every 3 parts of Z in the compound \(\mathrm{XZ}_{2}\).
05

Calculate the Percentage Composition of XZ2

In step 4, we found the mass ratio of \(\mathrm{X}\) and \(\mathrm{Z}\) in \(\mathrm{XZ}_{2}\) as 1/3. Now, we can find the percentage composition (by mass) of \(\mathrm{X}\) and \(\mathrm{Z}\) in the compound. Percentage of X in \(\mathrm{XZ}_{2}\) = (1 part of X / 4 total parts) * 100% = 25% Percentage of Z in \(\mathrm{XZ}_{2}\) = (3 parts of Z / 4 total parts) * 100% = 75% So, the composition (by mass) of the compound \(\mathrm{XZ}_{2}\) is 25% \(\mathrm{X}\) and 75% \(\mathrm{Z}\).

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
Understanding the mass ratio in a compound is crucial for various calculations in chemistry. The mass ratio represents how much of one substance is present compared to another within a compound. It's especially useful when we want to understand the proportions of elements in a molecule.

In the exercise, the mass ratio is determined directly from the given percentage composition. With X constituting 40% and Z 60% of the compound X2Z, the initial mass ratio comes out as 40:60. Simplified, it turns into a ratio of 2:3, meaning for every two parts of X, there are three parts of Z. This ratio forms the basis for understanding the composition of other compounds involving the same elements.

When dealing with mass ratios, remember to simplify the ratios whenever possible. This simplification can make subsequent calculations more straightforward, as seen when this ratio is used to find the composition of XZ2.
Percent Composition by Mass
The percent composition by mass refers to the percentage by mass of each element in a compound. It's an expression of the mass fraction of each element presented as a percentage. This concept is fundamental in chemistry, as it helps us understand the makeup of substances and allows for the conversion between mass and moles of elements and compounds.

To calculate the percent composition, one simply divides the mass of the element in a mole of the compound by the mass of a mole of the compound and multiplies by 100%. In our exercise, after establishing the mass ratio of X to Z in XZ2 as 1:3, we sum the parts, leading to four total parts, and assign percentages accordingly. We find that X makes up 25% (1 out of 4 parts) and Z makes up 75% (3 out of 4 parts) of the compound's mass.

This percent composition is a critical step that links the conceptual mass ratio to practical applications in stoichiometry and chemical analysis.
Stoichiometry
Stoichiometry is the section of chemistry that involves the calculation of reactants and products in chemical reactions. It builds upon the concepts of mass ratio and percent composition by mass to predict the outcomes of reactions. Stoichiometry is governed by the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.

In practice, stoichiometry allows us to calculate things like how much reactant is needed to produce a desired amount of product, or what the yield of a reaction might be. In the context of the exercise, if we were to react X and Z to form XZ2, we'd use the percent compositions and mass ratios we've calculated to ensure that reactants are mixed in the correct proportions.

Effective stoichiometry depends on a solid understanding of the relationships between the different elements and compounds involved in a reaction, highlighting the significance of the previous concepts in real-world applications.

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

A compound that contains only carbon, hydrogen, and oxygen is \(48.64 \% \mathrm{C}\) and \(8.16 \% \mathrm{H}\) by mass. What is the empirical formula of this substance?

The space shuttle environmental control system handles excess \(\mathrm{CO}_{2}\) (which the astronauts breathe out; it is \(4.0 \%\) by mass of exhaled air) by reacting it with lithium hydroxide, LiOH, pellets to form lithium carbonate, \(\mathrm{Li}_{2} \mathrm{CO}_{3}\), and water. If there are 7 astronauts on board the shuttle, and each exhales \(20 . \mathrm{L}\) of air per minute, how long could clean air be generated if there were \(25,000 \mathrm{~g}\) of LiOH pellets available for each shuttle mission? Assume the density of air is \(0.0010 \mathrm{~g} / \mathrm{mL}\).

Consider a mixture of potassium chloride and potassium nitrate that is \(43.2 \%\) potassium by mass. What is the percent \(\mathrm{KCl}\) by mass of the original mixture?

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(\mathrm{g})\) or \(\mathrm{NO}_{2}(\mathrm{~g})\) according to these unbalanced equations: $$ \begin{array}{l} \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ In a certain experiment \(2.00 \mathrm{~mol} \mathrm{NH}_{3}(g)\) and \(10.00 \mathrm{~mol}\) \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, \(6.75 \mathrm{~mol} \mathrm{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations, because you cannot assume that the two reactions will occur with equal probability.)

When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) In a certain experiment, \(20.00 \mathrm{~g}\) iron metal was reacted with \(11.20 \mathrm{~g}\) oxygen gas. After the experiment, the iron was totally consumed, and \(3.24 \mathrm{~g}\) oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.

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.