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% \(\mathrm{Z}\).

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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.

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Most popular questions from this chapter

Aspirin \(\left(\mathrm{C}, \mathrm{H}_{8} \mathrm{O}_{4}\right)\) is synthesized by reacting salicylic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) with acetic anhydride \(\left(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\right)\). The balanced equa- tion is $$ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3} \longrightarrow \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} $$ a. What mass of acetic anhydride is needed to completely consume \(1.00 \times 10^{2} \mathrm{~g}\) salicylic acid? b. What is the maximum mass of aspirin (the theoretical yield) that could be produced in this reaction?

A compound contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\). Combustion of \(35.0 \mathrm{mg}\) of the compound produces \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\). What is the empirical formula of the compound?

In 1987 the first substance to act as a superconductor at a temperature above that of liquid nitrogen \((77 \mathrm{~K})\) was discovered. The approximate formula of this substance is \(\mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{7} .\) Calculate the percent composition by mass of this material.

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{~kg} \mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction and assume that the NO produced in the third step is not recycled.

Consider the reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ Identify the limiting reagent in each of the reaction mixtures given below: a. 50 molecules of \(\mathrm{H}_{2}\) and 25 molecules of \(\mathrm{O}_{2}\) b. 100 molecules of \(\mathrm{H}_{2}\) and 40 molecules of \(\mathrm{O}_{2}\) c. 100 molecules of \(\mathrm{H}_{2}\) and 100 molecules of \(\mathrm{O}_{2}\) d. \(0.50 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}\). e. \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}_{2}\) f. \(1.0 \mathrm{~g} \mathrm{H}_{2}\) and \(0.25 \mathrm{~mol} \mathrm{O}_{2}\) g. \(5.00 \mathrm{~g} \mathrm{H}_{2}\) and \(56.00 \mathrm{~g} \mathrm{O}_{2}\)

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