Chapter U4.77, Problem 8E

Interpretation Introduction

Interpretation:

Among the given molecules, the molecule having greater number of chromium should be determined.

Concept Introduction:

Mole is SI unit which is used to measure the quantity of the substance. It is the quantity of a substance which contains same number of atoms as present in accurately 12.00 g of carbon-12 is known as mole

Number of moles is defined as the ratio of mass to the molar mass.

  $\text{Number of moles = }\frac{\text{Mass (g)}}{\text{Molar mass}}$

The ratio of mass of the given compound to the number of moles of compound is known as molar mass.

The one mole of any substance is equal to the Avogadro number i.e. $6.022\times {10}^{23}$ atoms.

Answer to Problem 8E

More number of moles of chromium is present in1.0 g chromium (IV) oxide ( ${\text{CrO}}_2$ ).

Explanation of Solution

Given information:

1.0 g chromium (II) chloride, ${\text{CrCl}}_{\text{2}}$

1.0 g chromium (III) chloride, ${\text{CrCl}}_3$

1.0 g chromium (IV) oxide, ${\text{CrO}}_{\text{2}}$

Molar mass of ${\text{CrCl}}_{\text{2}}$ = $\text{1}\times \text{51}\text{.99 g/mole +2}\times 35.45\text{ g/mole}$

= $122.89\text{ g/mole}$

Molar mass of ${\text{CrCl}}_3$ = $\text{1}\times \text{51}\text{.99 g/mole + 3}\times \text{35}\text{.45 g/mole}$

= $158.34\text{ g/mole}$

Molar mass of ${\text{CrO}}_{\text{2}}$ = $\text{1}\times 51.99\text{ g/mole + 2}\times \text{16 g/mole}$

= $83.99\text{ g/mole}$

  $\text{Number of moles = }\frac{\text{Mass (g)}}{\text{Molar mass}}$

Number of moles of ${\text{CrCl}}_{\text{2}}$ = $\frac{\text{Mass (g)}}{\text{Molar mass}}$

Put the values,

Number of moles of ${\text{CrCl}}_{\text{2}}$ = $\frac{\text{1}\text{.0 g}}{\text{122}\text{.89 g/mole}}$

Number of moles of ${\text{CrCl}}_{\text{2}}$ = $0.00813\text{ mole}$

In one mole of chromium(II) chloride, ${\text{CrCl}}_{\text{2}}$ there is one mole of chromium ion and two moles chloride ions as shown:

  ${\text{CrCl}}_{\text{2}}\to {\text{ Cr}}^{\text{2+}}{\text{ + 2Cl}}^{-}$

So, the number of moles of chromium ion in $0.00813\text{ mole}$ of chromium(II) chloride is $0.00813\text{ mole}$ .

Number of moles of ${\text{CrCl}}_3$ = $\frac{\text{Mass (g)}}{\text{Molar mass}}$

Put the values,

Number of moles of ${\text{CrCl}}_3$ = $\frac{\text{1}\text{.0 g}}{\text{158}\text{.34 g/mole}}$

Number of moles of ${\text{CrCl}}_3$ = $0.00631\text{ mole}$

In one mole of chromium(III) chloride, ${\text{CrCl}}_3$ there is one mole of chromium ion and three moles chloride ions as shown:

  ${\text{CrCl}}_3\to {\text{ Cr}}^{\text{3+}}{\text{ + 3Cl}}^{-}$

So, the number of moles of chromium ion in $0.00631\text{ mole}$ of chromium(III) chloride is $0.00631\text{ mole}$ .

Number of moles of ${\text{CrO}}_{\text{2}}$ = $\frac{\text{Mass (g)}}{\text{Molar mass}}$

Put the values,

Number of moles of ${\text{CrO}}_{\text{2}}$ = $\frac{\text{1}\text{.0 g}}{83.99\text{ g/mole}}$

Number of moles of ${\text{CrO}}_{\text{2}}$ = $0.0119\text{ mole}$

In one mole of chromium(IV)oxide, ${\text{CrO}}_{\text{2}}$ there is one mole of chromium ion and two moles oxide ions as shown:

  ${\text{CrO}}_{\text{2}}\to {\text{ Cr}}^{\text{4+}}{\text{ + 2O}}^{2-}$

So, the number of moles of chromium ion in $0.0119\text{ mole}$ of chromium(IV)oxide is $0.0119\text{ mole}$ .

Hence, on the basis of above calculations, greater number of moles of chromium is present in ${\text{CrO}}_2$ .