Chapter 7, Problem 7.93QA

Interpretation Introduction

To find:

a. The values of ‘$a$’ and ‘$b$’ in $U_a{O}_b$ and the charge on element $U$

b. The values of ‘$c$’ and $‘d’$ in $U_c{O}_d$ and the charge on element $U$

c. The values of ‘$x$’,’$y$’ and ‘$z$’ in $U{O}_x{\left(N{O}_3\right)}_y{\left(H_{2}O\right)}_z$

Answer to Problem 7.93QA

Solution:

a. The values of ‘$a$’ and ‘$b$’ in $U_a{O}_b$ are $1$ and $3$ respectively, and the charge on $U= 6+$

b. The values of ‘$c$’ and $‘d’$ in $U_c{O}_d$ are $3$ and $8$ respectively, and the charge on $U=5.33+$

c. The values of ‘$x$’,’$y$’ and ‘$z$’ in $U{O}_x{\left(N{O}_3\right)}_y{\left(H_{2}O\right)}_z$ are $2, 2, and 6$ respectively

Explanation of Solution

1) Concept:

Roasting of $U{O}_x{\left(N{O}_3\right)}_y{\left(H_{2}O\right)}_z$ at $400℃$ leads to loss of water molecule to form anhydrous $U{O}_x{\left(N{O}_3\right)}_y$.

This, on further heating, converts into various nitrogen oxides and uranium oxide $U_a{O}_b$ is formed as a residual.

2) Formula:

i) $Percent by mass=\frac{mass of a substance in a compound }{total mass}$

ii)

$Relative number of atoms=\frac{given mass of an element}{atomic mass}$

3) Given:

$U_a{O}_b$ contains $83.22\% U$ by mass

4) Calculation:

a. Percentage of oxygen in  $U_a{O}_b$  $= 100 - 83.22 = 16.78\%$

Ratio of the relative number of atoms of Uranium and Oxygen is

$no. of U atoms=\frac{82.33}{238.03}=0.346$

$no. of O atoms=\frac{16.78}{15.998}=1.048$

Dividing each by the least number, we get

$O atoms=\frac{0.346}{0.346 }=1$

$U atoms=\frac{1.048}{0.346}=3.03$

(Dividing the ratio by the smaller number 0.346 and taking to the nearest integer) (Atomic mass of U = 238.03g/mol)

Therefore, the value of $a=1 and b=3$

The formula for $U_a{O}_b$ is $U_1{O}_3, that is, U{O}_3.$

Calculation of charge on U:

Oxygen belonging to group 16 has common oxidation state $-2.$ Let the charge on $U$ is ‘p’

So,$p+3(-2)=0$  (since the overall charge of UO₃ is zero)

$p=6+$

b. Percentage of oxygen in $U_c{O}_d$ $= 100- 84.8 = 15.20\%$

Ratio of the relative number of atoms of uranium is:

$no. of U atoms=\frac{84.8}{238.03}=0.356$

$no. of O atoms=\frac{15.20}{15.998}=0.95011$

Dividing each by the least number, we get

$O atoms=\frac{0.356}{0.356 }=1$

$U atoms=\frac{0.95011}{0.356}=2.66$

Multiplying each of the numbers by 3 to get the whole numbers for the number of each atoms, we get:

$O atoms=\frac{0.356}{0.356 }=1 \times 3=3$

$U atoms=\frac{0.95011}{0.356}=2.66 \times 3=8.00$

Therefore, the value of $c=3 and b=8,$ and so, the molecular formula for $U_c{O}_d$ is $U_3{O}_{8.}$

Calculation of charge on U:

Oxygen belonging to group 16 has common oxidation state $-2$. Let the charge on U be ‘q.’

So, $3q+8(-2)=0$(since the overall charge of $U_3{O}_8$ is zero)

$q=5.33+$

c. The decomposition reaction for the compound $U{O}_x\left(N{O}_3{)}_y\right(H_2{O)}_z$ is given as

$U{O}_x\left(N{O}_3{)}_y\right(H_2{O)}_z\to U{O}_x(N{O}_3{)}_y+z{H}_{2}O\to U_n{O}_m$

$1.328g 1.042g 0.742g$

Therefore, the mass of ‘z’ moles of water = $1.328g-1.042g=0.286g$

Number of moles of water is calculated as

$mol H_{2}O=\frac{0.286g}{18g/mol}=0.016$

Mass of nitrogen is calculated as

$mass of nitrogen= 1.042g-0.742g=0.3g$

Number of atoms of nitrogen is

$mol N= \frac{0.3g}{14g/mol}=0.021$

By the principle of atom conservation,

$n=1$ and from part a.) $m=3$

We assume mass of $U{O}_x(N{O}_3{)}_y=M g$

So, equating moles, the following equation can be written:

$\frac{0.742}{286.03}=\frac{1.042}{M}$

286.03 = molar mass of $U{O}_3$

$M=401.6$

Hence,

$\frac{1.328}{401.6+18z}=\frac{0.742}{286.03}$

$z=6$(approx to nearest integer)

By the principle of atom conservation,

$x+3y=8$

Since ‘x’ and ‘y’ have to be integers by observation,

$x=2 a nd y=2$

So, the molecular formula for $U{O}_x(N{O}_3{)}_y$ is $U{O}_2(N{O}_3{)}_2$ while that for $U{O}_x\left(N{O}_3{)}_y\right(H_2{O)}_z$ is $U{O}_2(N{O}_3{)}_2{\left(H_{2}O\right)}_6$

Conclusion:

From the percent composition for elements given for a compound, it is possible to get the molecular formulas for the compounds.