Chapter 3, Problem 3.18QP

Moles within Moles and Molar Mass

Part 1:

1. a How many hydrogen and oxygen atoms are present in 1 molecule of H₂O?
2. b How many moles of hydrogen and oxygen atoms are present in 1 mol H₂O?
3. c What are the masses of hydrogen and oxygen in 1.0 mol H₂O?
4. d What is the mass of 1.0 mol H₂O?

Part 2: Two hypothetical ionic compounds are discovered with the chemical formulas XCl₂ and YCl₂, where X and Y represent symbols of the imaginary elements. Chemical analysis of the two compounds reveals that 0.25 mol XCl₂ has a mass of 100.0 g and 0.50 mol YCl₂ has a mass of 125.0 g.

1. a What are the molar masses of XCl₂ and YCl₂?
2. b If you had 1.0-mol samples of XCl₂ and YCl₂, how would the number of chloride ions compare?
3. c If you had 1.0-mol samples of XCl₂ and YCl₂, how would the masses of elements X and Y compare?
4. d What is the mass of chloride ions present in 1.0 mol XCl₂ and 1.0 mol YCl₂?
5. e What are the molar masses of elements X and Y?
6. f How many moles of X ions and chloride ions would be present in a 200.0-g sample of XCl₂?
7. g How many grams of Y ions would be present in a 250.0-g sample of YCl₂?
8. h What would be the molar mass of the compound YBr₃?

Part 3: A minute sample of AlCl₃ is analyzed for chlorine. The analysis reveals that there are 12 chloride ions present in the sample. How many aluminum ions must be present in the sample?

1. a What is the total mass of AlCl₃ in this sample?
2. b How many moles of AlCl₂ are in this sample?

Interpretation Introduction

Interpretation:

The number of Hydrogen and oxygen atoms in 1 ${\text{H}}_{\text{2}}\text{O}$ molecule should be given.

Answer to Problem 3.18QP

One molecule of ${\text{H}}_{\text{2}}\text{O}$ contains 1 Oxygen and 2 Hydrogen atoms.

Explanation of Solution

To give the number of Hydrogen and oxygen atoms in 1 ${\text{H}}_{\text{2}}\text{O}$ molecule.

From the molecular formula of the Water molecule, it has 1 Oxygen and 2 Hydrogen atoms.

Conclusion

The number of Hydrogen and oxygen atoms in 1 ${\text{H}}_{\text{2}}\text{O}$ molecule was given.

Interpretation Introduction

Interpretation:

The number of moles of Hydrogen and oxygen atoms present in 1 mole of ${\text{H}}_{\text{2}}\text{O}$ molecule should be given.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

1 mole of ${\text{H}}_{\text{2}}\text{O}$ has 2 mole of Hydrogen and 1 mole of Oxygen.

Explanation of Solution

To give the number of moles of Hydrogen and oxygen atoms present in 1 mole of ${\text{H}}_{\text{2}}\text{O}$ molecule.

Molecule formula of water is ${\text{H}}_{\text{2}}\text{O}$, form this 1 mole of ${\text{H}}_{\text{2}}\text{O}$ has 2 mole of Hydrogen and 1 mole of Oxygen.

Conclusion

The number of moles of Hydrogen and oxygen atoms present in 1 mole of ${\text{H}}_{\text{2}}\text{O}$ molecule was given.

Interpretation Introduction

Interpretation:

The masses of Hydrogen and Oxygen in 1 mole of ${\text{H}}_{\text{2}}\text{O}$ should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

1 mole of ${\text{H}}_{\text{2}}\text{O}$ has $\text{2}\text{.0}\text{g}$ of Hydrogen and $\text{16}\text{.0}\text{g}$ of Oxygen atoms.

Explanation of Solution

To calculate the masses of Hydrogen and Oxygen in 1 mole of ${\text{H}}_{\text{2}}\text{O}$

Molar mass of Hydrogen is $\text{1}\text{.008}\text{g}$

Molar mass of Oxygen is $\text{16}\text{.0}\text{g}$

Masses of Hydrogen in 1 mole of ${\text{H}}_{\text{2}}\text{O}$

$\begin{array}{l}\text{Mass}\text{H}\text{=}\text{1}\text{.0}\text{mole}{\text{H}}_{\text{2}}\text{O}\text{×}\frac{\text{2}\text{mole}\text{of}\text{H}}{\text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}}\text{×}\frac{\text{1}\text{.008}\text{g}}{\text{1}\text{mole}\text{H}}\\ \text{=}\text{2}\text{.0}\text{g}\end{array}$

The molar mass of Hydrogen is plugged in above equation to give mass of Hydrogen in 1 mole of ${\text{H}}_{\text{2}}\text{O}$.

Masses of Oxygen in 1 mole of ${\text{H}}_{\text{2}}\text{O}$

$\begin{array}{l}\text{Mass}\text{H}\text{=}\text{1}\text{.0}\text{mole}{\text{H}}_{\text{2}}\text{O}\text{×}\frac{\text{1}\text{mole}\text{of}\text{O}}{\text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}}\text{×}\frac{\text{16}\text{.0}\text{g}}{\text{1}\text{mole}\text{O}}\\ \text{=}\text{16}\text{.0}\text{g}\end{array}$

The molar mass of Oxygen is plugged in above equation to give mass of Oxygen in 1 mole of ${\text{H}}_{\text{2}}\text{O}$.

1 mole of ${\text{H}}_{\text{2}}\text{O}$ has $\text{2}\text{.0}\text{g}$ of Hydrogen and $\text{16}\text{.0}\text{g}$ of Oxygen atoms.

Conclusion

The masses of Hydrogen and Oxygen in 1 mole of ${\text{H}}_{\text{2}}\text{O}$ was calculated.

Interpretation Introduction

Interpretation:

Mass of 1 mole ${\text{H}}_{\text{2}}\text{O}$ should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

Mass of 1 mole ${\text{H}}_{\text{2}}\text{O}$ is $\text{18}\text{.0}\text{g/mole}$.

Explanation of Solution

To calculate the mass of 1.0 mole of ${\text{H}}_{\text{2}}\text{O}$.

Molar mass of Hydrogen is $\text{1}\text{.008}\text{g}$

Molar mass of Oxygen is $\text{16}\text{.0}\text{g}$

Masses of Hydrogen in 1 mole of ${\text{H}}_{\text{2}}\text{O}$

$\begin{array}{l}\text{Mass}\text{H}\text{=}\text{1}\text{.0}\text{mole}{\text{H}}_{\text{2}}\text{O}\text{×}\frac{\text{2}\text{mole}\text{of}\text{H}}{\text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}}\text{×}\frac{\text{1}\text{.008}\text{g}}{\text{1}\text{mole}\text{H}}\\ \text{=}\text{2}\text{.0}\text{g}\end{array}$

The molar mass of Hydrogen is plugged in above equation to give mass of Hydrogen in 1 mole of ${\text{H}}_{\text{2}}\text{O}$.

Masses of Oxygen in 1 mole of ${\text{H}}_{\text{2}}\text{O}$

$\begin{array}{l}\text{Mass}\text{H}\text{=}\text{1}\text{.0}\text{mole}{\text{H}}_{\text{2}}\text{O}\text{×}\frac{\text{1}\text{mole}\text{of}\text{O}}{\text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}}\text{×}\frac{\text{16}\text{.0}\text{g}}{\text{1}\text{mole}\text{O}}\\ \text{=}\text{16}\text{.0}\text{g}\end{array}$

The molar mass of Oxygen is plugged in above equation to give mass of Oxygen in 1 mole of ${\text{H}}_{\text{2}}\text{O}$.

1 mole of ${\text{H}}_{\text{2}}\text{O}$ has $\text{2}\text{.0}\text{g}$ of Hydrogen and $\text{16}\text{.0}\text{g}$ of Oxygen atoms.

Sum the above calculated masses of Hydrogen and Oxygen to gives mass of 1 mole ${\text{H}}_{\text{2}}\text{O}$

Conclusion

Mass of 1 mole ${\text{H}}_{\text{2}}\text{O}$ was calculated.

Interpretation Introduction

Interpretation:

The molar masses of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

The molar mass of ${\text{XCl}}_{\text{2}}$ is $\text{400}\text{g/mole}$

The molar mass of ${\text{YCl}}_{\text{2}}$ is $\text{250}\text{g/mole}$

Explanation of Solution

To calculate the molar masses of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$

Given,

$\text{0}\text{.25}$ mole of ${\text{XCl}}_{\text{2}}$ has mass of $\text{100}\text{.0}\text{g}$

$\text{0}\text{.50}$ mole of ${\text{YCl}}_{\text{2}}$ has mass of $\text{125}\text{.0 g}$

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Rearrange the above equation to give a result of plugged the values.

$\begin{array}{l}\text{Molar}\text{mass}{\text{XCl}}_{\text{2}}\text{=}\frac{\text{100}\text{.0}\text{g}}{\text{0}\text{.25}\text{mole}}\\ \text{=}\text{400}\text{g/mole}\end{array}$

$\begin{array}{l}\text{Molar}\text{mass}{\text{YCl}}_{\text{2}}\text{=}\frac{\text{125}\text{.0}\text{g}}{\text{0}\text{.50}\text{mole}}\\ \text{=}\text{250}\text{g/mole}\end{array}$

Plugged the given moles and mass of the compounds to give the molar masses of the compound.

Conclusion

The molar masses of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ was calculated.

Interpretation Introduction

Interpretation:

The number of Chlorine atoms in 1 moles of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ should be explained.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

1 mole of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ compounds are have same number of Chlorine atoms.

Explanation of Solution

To explain the he number of Chlorine atoms in 1 moles of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$

Formula of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ are having same number of Chlorine atoms hence, 1 mole of each compounds are have same number of Chlorine atoms.

Conclusion

The number of Chlorine atoms in 1 moles of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ was explained.

Interpretation Introduction

Interpretation:

The mass of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ should be comparatively explained.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

The mass of $\text{X}$ is higher than $\text{Y}$.

Explanation of Solution

To comparatively explain the mass of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$

Given,

$\text{0}\text{.25}$ mole of ${\text{XCl}}_{\text{2}}$ has mass of $\text{100}\text{.0}\text{g}$

$\text{0}\text{.50}$ mole of ${\text{YCl}}_{\text{2}}$ has mass of $\text{125}\text{.0 g}$

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Rearrange the above equation to give a result of plugged the values.

$\begin{array}{l}\text{Molar}\text{mass}{\text{XCl}}_{\text{2}}\text{=}\frac{\text{100}\text{.0}\text{g}}{\text{0}\text{.25}\text{mole}}\\ \text{=}\text{400}\text{g/mole}\end{array}$

$\begin{array}{l}\text{Molar}\text{mass}{\text{YCl}}_{\text{2}}\text{=}\frac{\text{125}\text{.0}\text{g}}{\text{0}\text{.50}\text{mole}}\\ \text{=}\text{250}\text{g/mole}\end{array}$

Plugged the given moles and mass of the compounds to give the molar masses of the compound.

From the above mole calculation, the mole of ${\text{XCl}}_{\text{2}}$ is higher than the mole of ${\text{YCl}}_{\text{2}}$ because of mass of $\text{X}$ is higher than $\text{Y}$.

Hence, the mass of $\text{X}$ is higher than $\text{Y}$.

Conclusion

The mass of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ was comparatively explained.

Interpretation Introduction

Interpretation:

The mass of Chlorine ion in 1 moles of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

The mass of Chlorine in 1 moles of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ are same and it is $\text{71}\text{g}$.

Explanation of Solution

To calculate the mass of Chlorine ion in 1 moles of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$

Given,

Molar mass of Chlorine is $\text{35}\text{.45 g}$

Molecular formula of given compound is ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$

From this, both compounds are having same number of atoms so the mass of chlorine atoms present in 1 moles of both ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ are same.

The mass of Chlorine in 1 moles of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ is,

$\begin{array}{l}\text{=}\text{1}\text{.0}\text{mole}\text{×}\frac{\text{2}\text{mole}\text{of}{\text{Cl}}^{\text{-}}\text{ion}}{\text{1}\text{mole}{\text{XCl}}_{\text{2}}}\text{×}\frac{\text{35}\text{.45}\text{g}}{\text{1}\text{mole}\text{Cl}}\\ \text{=}\text{71}\text{g}\end{array}$

Molar mass of Chlorine and is plugged in above equation to give the mass of Chlorine 1 moles of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$.

Conclusion

The mass of Chlorine ion in 1 moles of ${\text{XCl}}_{\text{2}}$ and ${\text{YCl}}_{\text{2}}$ should be calculated.

Interpretation Introduction

Interpretation:

The molar masses of X and Y elements should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

The molar mass of $\text{X}$ is $\text{329}\text{.1}\text{g/mole}$

The molar mass of $\text{Y}$ is $\text{179}\text{.1}\text{g/mole}$

Explanation of Solution

To calculate the molar masses X and Y elements.

Given,

$\text{0}\text{.25}$ mole of ${\text{XCl}}_{\text{2}}$ has mass of $\text{100}\text{.0}\text{g}$

$\text{0}\text{.50}$ mole of ${\text{YCl}}_{\text{2}}$ has mass of $\text{125}\text{.0 g}$

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Rearrange the above equation to give a result of plugged the values.

$\begin{array}{l}\text{Molar}\text{mass}{\text{XCl}}_{\text{2}}\text{=}\frac{\text{100}\text{.0}\text{g}}{\text{0}\text{.25}\text{mole}}\\ \text{=}\text{400}\text{g/mole}\end{array}$

$\begin{array}{l}\text{Molar}\text{mass}{\text{YCl}}_{\text{2}}\text{=}\frac{\text{125}\text{.0}\text{g}}{\text{0}\text{.50}\text{mole}}\\ \text{=}\text{250}\text{g/mole}\end{array}$

Plugged the given moles and mass of the compounds to give the molar masses of the compound.

We know the molar mass of Chlorine is $\text{70}\text{.90}\text{g}$

$\begin{array}{l}\text{Molar}\text{mass}\text{of}\text{X}\text{=}\text{400-70}\text{.90}\\ \text{=}\text{329}\text{.1}\text{g/mole}\\ \\ \text{Molar}\text{mass}\text{of}\text{Y}\text{=}\text{250-70}\text{.90}\\ \text{=}\text{179}\text{.1}\text{g/mole}\end{array}$

Subtracts the mass of Chlorine from The calculated molar mass of each elements to gives the molar mass of X and Y.

Conclusion

The molar masses of X and Y elements were calculated.

Interpretation Introduction

Interpretation:

The number of moles of chlorine and X ions present in $\text{200}\text{.0}\text{g}$ of ${\text{XCl}}_{\text{2}}$ should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

The number of moles of Chlorine ions present in $\text{200}\text{.0}\text{g}$ of ${\text{XCl}}_{\text{2}}$ is $\text{1}\text{.0}\text{mole}\text{Cl}$

The number of moles of X ions present in $\text{200}\text{.0}\text{g}$ of ${\text{XCl}}_{\text{2}}$ is $\text{0}\text{.50}\text{mole}\text{X}$

Explanation of Solution

To calculate the number of moles of chlorine and X ions present in $\text{200}\text{.0}\text{g}$ of ${\text{XCl}}_{\text{2}}$.

Given,

$\text{0}\text{.25}$ mole of ${\text{XCl}}_{\text{2}}$ has mass of $\text{100}\text{.0}\text{g}$

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Rearrange the above equation to give a result of plugged the values.

$\begin{array}{l}\text{Molar}\text{mass}{\text{XCl}}_{\text{2}}\text{=}\frac{\text{100}\text{.0}\text{g}}{\text{0}\text{.25}\text{mole}}\\ \text{=}\text{400}\text{g/mole}\end{array}$

The number of moles of chlorine and X ions present in $\text{200}\text{.0}\text{g}$ of ${\text{XCl}}_{\text{2}}$ is,

$\begin{array}{l}\text{Mole of }\text{X}\text{ion}\text{=}200.0g\times \frac{\text{1}\text{mole}{\text{XCl}}_{\text{2}}}{\text{400}\text{g}}\text{×}\frac{\text{1}\text{mole}\text{X}\text{ions}}{\text{1}\text{mole}{\text{XCl}}_{\text{2}}}\\ \text{=}\text{0}\text{.50}\text{mole}\text{X}\end{array}$

The number of moles of chlorine and Cl ions present in $\text{200}\text{.0}\text{g}$ of ${\text{XCl}}_{\text{2}}$ is,

$\begin{array}{l}\text{Mole of }\text{X}\text{ion}\text{=}\text{200}\text{.0}\text{g}\frac{\text{1}\text{mole}{\text{XCl}}_{\text{2}}}{\text{400}\text{g}}\text{×}\frac{\text{2}\text{mole}{\text{Cl}}^{\text{-}}\text{ions}}{\text{1}\text{mole}{\text{XCl}}_{\text{2}}}\\ \text{=}\text{1}\text{.0}\text{mole}\text{Cl}\end{array}$

The number of Chlorine and X ions present in 1 molecule and mass of given sample are plugged in above equation to give moles of chlorine and X ions present in $\text{200}\text{.0}\text{g}$ of ${\text{XCl}}_{\text{2}}$.

The number of moles of Chlorine ions present in $\text{200}\text{.0}\text{g}$ of ${\text{XCl}}_{\text{2}}$ is $\text{1}\text{.0}\text{mole}\text{Cl}$

The number of moles of X ions present in $\text{200}\text{.0}\text{g}$ of ${\text{XCl}}_{\text{2}}$ is $\text{0}\text{.50}\text{mole}\text{X}$

Conclusion

The number of moles of chlorine and X ions present in $\text{200}\text{.0}\text{g}$ of ${\text{XCl}}_{\text{2}}$ was calculated.

Interpretation Introduction

Interpretation:

The gram mass of Y ions present in $\text{250}\text{.0}\text{g}$ of ${\text{YCl}}_{\text{2}}$ should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

The gram mass of Y ions present in $\text{250}\text{.0}\text{g}$ of ${\text{YCl}}_{\text{2}}$ is $\text{179}\text{.10}\text{g}$.

Explanation of Solution

To calculate the gram mass of Y ions present in $\text{250}\text{.0}\text{g}$ of ${\text{YCl}}_{\text{2}}$

Given,

$\text{0}\text{.50}$ mole of ${\text{YCl}}_{\text{2}}$ has mass of $\text{125}\text{.0 g}$

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Rearrange the above equation to give a result of plugged the values.

$\begin{array}{l}\text{Molar}\text{mass}{\text{YCl}}_{\text{2}}\text{=}\frac{\text{125}\text{.0}\text{g}}{\text{0}\text{.50}\text{mole}}\\ \text{=}\text{250}\text{g/mole}\end{array}$

The number of moles of chlorine and Y ions present in $\text{250}\text{.0}\text{g}$ of ${\text{YCl}}_{\text{2}}$ is,

$\begin{array}{l}\text{Mole of }\text{X}\text{ion}\text{=}\text{250}\text{.0}\text{g×}\frac{\text{1}\text{mole}{\text{YCl}}_{\text{2}}}{\text{250}\text{g}}\text{×}\frac{\text{1}\text{mole}\text{Y}\text{ions}}{\text{1}\text{mole}{\text{YCl}}_{\text{2}}}\text{×}\frac{\text{179}\text{.10}\text{g}}{\text{1}\text{mole}\text{Y}}\\ \text{=}\text{179}\text{.10}\text{g}\end{array}$

The molar mass of Y and mass of sample are plugged in above equation to gives the gram mass of Y ions present in $\text{250}\text{.0}\text{g}$ of ${\text{YCl}}_{\text{2}}$.

The gram mass of Y ions present in $\text{250}\text{.0}\text{g}$ of ${\text{YCl}}_{\text{2}}$ is $\text{179}\text{.10}\text{g}$.

Conclusion

The gram mass of Y ions present in $\text{250}\text{.0}\text{g}$ of ${\text{YCl}}_{\text{2}}$ was calculated.

Interpretation Introduction

Interpretation:

The molar mass of ${\text{YBr}}_3$ should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

The molar mass of ${\text{YBr}}_3$ is $\text{418}\text{.8}\text{g}$

Explanation of Solution

To calculate the molar mass of ${\text{YBr}}_3$.

Given,

$\text{0}\text{.50}$ mole of ${\text{YCl}}_{\text{2}}$ has mass of $\text{125}\text{.0 g}$

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Rearrange the above equation to give a result of plugged the values.

$\begin{array}{l}\text{Molar}\text{mass}{\text{YCl}}_{\text{2}}\text{=}\frac{\text{125}\text{.0}\text{g}}{\text{0}\text{.50}\text{mole}}\\ \text{=}\text{250}\text{g/mole}\end{array}$

Plugged the given moles and mass of the compounds to give the molar masses of the compound.

We know the molar mass of Chlorine is $\text{70}\text{.90}\text{g}$

$\begin{array}{l}\text{Molar}\text{mass}\text{of}\text{Y}\text{=}\text{250-70}\text{.90}\\ \text{=}\text{179}\text{.1}\text{g/mole}\end{array}$

Subtracts the mass of Chlorine from The calculated molar mass of Y to gives the molar mass of Y.

The molar mass of Bromine is $\text{79}\text{.90}\text{g}$

The given compound ${\text{YBr}}_3$ has 3 bromine atom hence molar mass of ${\text{YBr}}_3$ is,

$\begin{array}{l}\text{Molar}\text{mass}\text{of}{\text{YBr}}_{\text{3}}\text{=}\text{179}\text{.10}\text{+}\text{3×79}\text{.90}\text{g}\\ \text{=}\text{418}\text{.8}\text{g}\end{array}$

To sum the masses of Y with 3 bromine to give molar mass of ${\text{YBr}}_3$

The molar mass of ${\text{YBr}}_3$ is $\text{418}\text{.8}\text{g}$

Conclusion

The molar mass of ${\text{YBr}}_3$ was calculated.

Interpretation Introduction

Interpretation:

The total mass of given sample should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

The total mass of given sample is $\text{8}\text{.86}\text{×10}{}^{\text{-22}}\text{g}$

Explanation of Solution

To calculate the total mass of given sample.

Given,

12 Chloride ions present in given sample.

In 1 molecule of ${\text{AlCl}}_{\text{3}}$ has 3 Chloride ions and 1 aluminum ion.

So the 4 unit of ${\text{AlCl}}_{\text{3}}$ has12 Chloride ions and 4 aluminum ions.

The sample has 4 ${\text{AlCl}}_{\text{3}}$ units.

The mass of sample is,

Molar mass of ${\text{AlCl}}_{\text{3}}$ is $\text{133}\text{.33 g}$

$\begin{array}{l}\text{=}\text{4}{\text{AlCl}}_{\text{3}}\text{units×}\frac{\text{1}\text{mole}{\text{AlCl}}_{\text{3}}}{\text{6}{\text{.23×10}}^{\text{23}}}\text{×}\frac{\text{133}\text{.33}\text{g}{\text{AlCl}}_{\text{3}}}{\text{1}\text{mole}{\text{AlCl}}_{\text{3}}}\\ \text{=}\text{8}\text{.86}\text{×10}{}^{\text{-22}}\text{g}\end{array}$

The number of ${\text{AlCl}}_{\text{3}}$ units and molar mass of ${\text{AlCl}}_{\text{3}}$ are plugged in above equation to give mass of total sample.

The total mass of given sample is $\text{8}\text{.86}\text{×10}{}^{\text{-22}}\text{g}$

Conclusion

The total mass of given sample was calculated.

Interpretation Introduction

Interpretation:

The mole of ${\text{AlCl}}_{\text{3}}$ in given sample should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.18QP

The mole of ${\text{AlCl}}_{\text{3}}$ in given sample is $\text{6}\text{.64}\text{×10}{}^{\text{-24}}\text{mole}$

Explanation of Solution

To calculate the mole of ${\text{AlCl}}_{\text{3}}$ in given sample.

Given,

12 Chloride ions present in given sample.

In 1 molecule of ${\text{AlCl}}_{\text{3}}$ has 3 Chloride ions and 1 aluminum ion.

So the 4 unit of ${\text{AlCl}}_{\text{3}}$ has12 Chloride ions and 4 aluminum ions.

The sample has 4 ${\text{AlCl}}_{\text{3}}$ units.

The mole of ${\text{AlCl}}_{\text{3}}$ in given sample is,

$\begin{array}{l}\text{=}\text{4}{\text{AlCl}}_{\text{3}}\text{units×}\frac{\text{1}\text{mole}{\text{AlCl}}_{\text{3}}}{\text{6}{\text{.23×10}}^{\text{23}}}\\ \text{=}\text{6}\text{.64}\text{×10}{}^{\text{-24}}\text{mole}\end{array}$

The number of ${\text{AlCl}}_{\text{3}}$ units is plugged in above equation to give the mole of ${\text{AlCl}}_{\text{3}}$ in given sample.

The mole of ${\text{AlCl}}_{\text{3}}$ in given sample is $\text{6}\text{.64}\text{×10}{}^{\text{-24}}\text{mole}$

Conclusion

The mole of ${\text{AlCl}}_{\text{3}}$ in given sample was calculated.