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Chapter 3: Problem 151

A substance \(\mathrm{X}_{2} \mathrm{Z}\) 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 of the compound XZ2 in terms of the atomic masses of X and Z is: X = \( \frac{0.4 * (2 * mX + mZ)}{mX + 2 * mZ} \) * 100 % Z = \( \frac{0.6 * (2 * mX + mZ)}{mX + 2 * mZ} \) * 100 %

Step by step solution

01

Determine the ratio of X and Z in X2Z

For the compound X2Z, we have 2 moles of X and 1 mole of Z. Therefore, the ratio of moles of X to moles of Z is 2:1.
02

Define variables for the atomic masses

In order to simplify the calculations, let's assign variable names to the atomic masses of X and Z. Let mX be the atomic mass of X and mZ be the atomic mass of Z.
03

Calculate the mass of one mole of X2Z

Given the ratio of moles and the atomic masses, we can calculate the mass of one mole of X2Z. The mass of one mole of X2Z is: 2 * mX + mZ
04

Determine the composition of X2Z in terms of mX and mZ

Using the given mass percentages of X and Z in X2Z, we can write the following equations: Mass of X in one mole of X2Z = 0.4 * (2 * mX + mZ) Mass of Z in one mole of X2Z = 0.6 * (2 * mX + mZ)
05

Find the composition of XZ2 in terms of mX and mZ

XZ2 has a mole ratio of 1 mole of X to 2 moles of Z. So the mass of one mole of XZ2 is: mX + 2 * mZ
06

Find the mass percentage of X in XZ2

To find the mass percentage of X in XZ2, divide the mass of X in one mole of XZ2 by the mass of one mole of XZ2, and multiply by 100: Percentage of X in XZ2 = \( \frac{mX}{mX + 2 * mZ} * 100 \) Using the equation for the mass of X in one mole of X2Z, divide both sides by (2 * mX + mZ): Percentage of X in XZ2 = \( \frac{0.4 * (2 * mX + mZ)}{mX + 2 * mZ} \)
07

Find the mass percentage of Z in XZ2

Similar to the previous step, divide the mass of Z in one mole of XZ2 by the mass of one mole of XZ2, and multiply by 100: Percentage of Z in XZ2 = \( \frac{2 * mZ}{mX + 2 * mZ} * 100 \) Using the equation for the mass of Z in one mole of X2Z, divide both sides by (2 * mX + mZ): Percentage of Z in XZ2 = \( \frac{0.6 * (2 * mX + mZ)}{mX + 2 * mZ} \) Now we have the composition of XZ2 in terms of mX and mZ: X = \( \frac{0.4 * (2 * mX + mZ)}{mX + 2 * mZ} \) * 100 % Z = \( \frac{0.6 * (2 * mX + mZ)}{mX + 2 * mZ} \) * 100 %

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Most popular questions from this chapter

The aspirin substitute, acetaminophen \(\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{~N}\right)\), is produced by the following three-step synthesis: I. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{~N}(s)+3 \mathrm{H}_{2}(g)+\mathrm{HCl}(a q) \longrightarrow\) $$ \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{ONCl}(s)+2 \mathrm{H}_{2} \mathrm{O}(t) $$ II. \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+\mathrm{NaOH}(a q) \longrightarrow\) $$ \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{H}_{2} \mathrm{O}(t)+\mathrm{NaCl}(a q) $$ III. \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}(l) \longrightarrow\) $$ \mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{~N}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(l) $$ The first two reactions have percent yields of \(87 \%\) and \(98 \%\) by mass, respectively. The overall reaction yields 3 moles of acetaminophen product for every 4 moles of \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{O}_{3} \mathrm{~N}\) reacted. a. What is the percent yield by mass for the overall process? b. What is the percent yield by mass of Step III?

A compound contains \(47.08 \%\) carbon, \(6.59 \%\) hydrogen, and \(46.33 \%\) chlorine by mass; the molar mass of the compound is \(153 \mathrm{~g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?

Bornite \(\left(\mathrm{Cu}_{3} \mathrm{FeS}_{3}\right)\) is a copper ore used in the production of copper. When heated, the following reaction occurs: \(2 \mathrm{Cu}_{3} \mathrm{FeS}_{3}(s)+7 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{Cu}(s)+2 \mathrm{FeO}(s)+6 \mathrm{SO}_{2}(g)\) If \(2.50\) metric tons of bornite is reacted with excess \(\mathrm{O}_{2}\) and the process has an \(86.3 \%\) yield of copper, what mass of copper is produced?

What number of atoms of nitrogen are present in \(5.00 \mathrm{~g}\) of each of the following? a. glycine, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{~N}\) b. magnesium nitride c. calcium nitrate d. dinitrogen tetroxide

What amount (moles) is represented by each of these samples? a. \(20.0 \mathrm{mg}\) caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\) b. \(2.72 \times 10^{21}\) molecules of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) c. \(1.50 \mathrm{~g}\) of dry ice, \(\mathrm{CO}_{2}\)
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