Question

Prove from the following results that mercury molecules are monatomic:
(a) 10 g of mercury combines with 0.8 g of oxygen to form an oxide.
(b) 1000 mL of vapour of Hg at NTP weighs 8.923 g.
(c) The specific heat of the metal is 0.033.

Answer 1

Mercury is a metal and is liquid at standard conditions of temperature and pressure. Mercury molecules are mono-atomic in nature just like noble gases that are mono-atomic in nature.

Complete answer:
(a) We can solve this part by first calculating the empirical formula followed by the molecular formula.
Since 10 g of mercury is combining with 0.8 g of oxygen to form an oxide, therefore,
The number of atoms of mercury (in moles):
$\cfrac { mass\quad of\quad mercury }{ atomic\quad mass\quad of\quad mercury } =\cfrac { 10g }{ 200.59\quad g/mol } =0.04985\quad mole$
The number of atoms of oxygen (in moles):
$\cfrac { mass\quad of\quad oxygen }{ atomic\quad mass\quad of\quad oxygen } =\cfrac { 0.8g }{ 15.999\quad g/mol } =0.05000\quad mole$
Therefore the formula of the oxide will be ${ Hg }_{ 0.04985 }{ O }_{ 0.05000 }$. For making the subscripts in the formula in whole numbers we will divide each subscript by 0.04985.
${ \Rightarrow Hg }_{ \cfrac { 0.04985 }{ 0.04985 } }{ O }_{ \cfrac { 0.05000 }{ 0.04985 } }=HgO$
The experimental molar mass of HgO is 216.59 g/mol and the empirical formula mass of HgO is 216.589 g/mol.
Atomicity=$\cfrac { molar\quad mass }{ empirical\quad formula\quad mass } =\cfrac { 216.59\quad g/mol }{ 216.589\quad g/mol } \simeq 1$

(b) 1000 mL of Hg has a mass of=8.923 g
At NTP conditions, the volume of 1 mole of a fluid is 22.4 L or 22400 mL. Therefore,
22400 mL of Hg will have a mass of=$\cfrac { 8.923\quad g }{ 1000\quad mL } \times 22400\quad mL=199.88\quad g$
The experimental atomic mass of Hg is 200.59 g/mol which is very close to 199.88 g, hence mercury must be mono-atomic.

(c) According to Dulong-Petit’s law, the approximate atomic mass of an element in the solid state is equal to the ratio of 6.4 to the specific heat of the element.
Using the above law:
Approximate atomic mass of Hg=$\cfrac { 6.4\quad }{ 0.033 } =193.9394\quad g/mol$
The equivalent mass of Hg from part (a) will be:
$Equivalent\quad mass\quad of\quad Hg=\cfrac { mass\quad of\quad Hg }{ mass\quad of\quad oxygen } \times 8$
$\Rightarrow Equivalent\quad mass\quad of\quad Hg=\cfrac { 10\quad g }{ 0.8\quad g } \times 8=100 \quad$
Now, the valency of Hg will be equal to the ratio of the approximate atomic mass and the equivalent mass of Hg:
$Valency\quad of\quad Hg=\cfrac { 193.93 }{ 100 } \simeq 2$
The more accurate atomic mass will be=$Equivalent\quad mass\times valency$
$=100\times 2=200\quad g/mol$
Now, Atomicity=$\cfrac { approximate\quad atomic\quad mass }{ accurate\quad atomic\quad mass } =\cfrac { 193.9394\quad g/mol }{ 200\quad g/mol } \simeq 1$
Hence Mercury molecules are mono-atomic.

Note: Always remember that the volume occupied by 1 mole of a fluid at both STP (Standard Temperature and Pressure) condition and NTP (Normal Temperature and Pressure) condition is 22.4 L.