Question

In 1811, Avogadro calculated the formula of camphor by means of elemental chemical analysis and by measuring the density of its vapour. Avogadro found the density to be 3.84 g/L when he made the measurement at 210°C at 1 atm pressure. Which of the following is the correct formula of camphor?
A) ${{\text{C}}_{\text{10}}}{{\text{H}}_{14}}\text{O}$
B) ${{\text{C}}_{\text{10}}}{{\text{H}}_{16}}\text{O}$
C) ${{\text{C}}_{\text{10}}}{{\text{H}}_{18}}{{\text{O}}_{2}}$
D) ${{\text{C}}_{\text{10}}}{{\text{H}}_{18}}\text{O}$
E) None of these

Answer 1

Hint: To answer this question, we need to know the basics of Avogadro’s law. This law states that the equal volume of all gases have the same number of molecules at the same temperature and pressure. Keeping this in mind, we can proceed with the given problem.

Complete step by step answer:
At first we have to apply the formula,
$\text{PV = }\dfrac{\text{d}}{\text{M}}\text{ RT}$
So, from the formula we can get the equation for M.
So, $M =$$\dfrac{\text{dRT}}{\text{PV}}$.
In the above equation,
d represents the density
P represents the Pressure
V represents the volume
M represents the molar mass
R represents the constant value, which depends on the unit of pressure
T represents the temperature taken in Kelvin
So now putting the values we get,
$M =$ $\dfrac{\text{3}\text{.84 }\!\!\times\!\!\text{ 0}\text{.0821 }\!\!\times\!\!\text{ 483}}{\text{1}}$=152
The molecular weight of ${{\text{C}}_{\text{10}}}{{\text{H}}_{14}}\text{O}$is $\text{(12}\text{.0107 }\!\!\times\!\!\text{ 10) + (14 }\!\!\times\!\!\text{ 1) + 16 = 203}\text{.0825}$. This does not match with the molecular weight derived above. So, this is not the answer.
The molecular weight of ${{\text{C}}_{\text{10}}}{{\text{H}}_{16}}\text{O}$is $\text{(12}\text{.0107 }\!\!\times\!\!\text{ 10) + (1 }\!\!\times\!\!\text{ 16) + 16 = 152}\text{.107}$. This matches with the molecular weight derived above. So, this is the answer.
The molecular weight of ${{\text{C}}_{\text{10}}}{{\text{H}}_{18}}{{\text{O}}_{2}}$is $\text{(12}\text{.0107 }\!\!\times\!\!\text{ 10) + (1 }\!\!\times\!\!\text{ 18) + (16 }\!\!\times\!\!\text{ 2) = 170}\text{.107}$. This does not match with the molecular weight derived above. So, this is not the answer.
The molecular weight of ${{\text{C}}_{\text{10}}}{{\text{H}}_{18}}\text{O}$is $\text{(12}\text{.0107 }\!\!\times\!\!\text{ 10) + (1 }\!\!\times\!\!\text{ 18) + 16 = 154}\text{.107}$. This does not match with the molecular weight derived above. So, this is not the answer.
Hence, the correct answer is Option B.

Note: When no information is given in the question, we need to assume the gas to be in standard temperature and pressure conditions. At STP, the standard temperature is 273 K and standard pressure is 1 atm.