1.7 Atomic and Molecular Masses

After having some idea about the terms atoms and molecules, it is appropriate here to understand what do we mean by atomic and molecular masses.

Atomic Mass

One atomic mass unit is defined as a mass exactly equal to one-twelfth of the mass of one carbon-12 atom.

\( \text { Atomic weight of an element }=\frac{\text { Weight of } 1 \text { atom of element }}{1 / 12 \times \text { weight of } 1 \text { atom of } \mathrm{C}-12}\)

Here, Carbon-12 is one of the isotopes of carbon and can be represented as \({ }^{12} \mathrm{C}\). In this system, \({ }^{12} \mathrm{C}\) is assigned a mass of exactly 12 atomic mass unit (amu) and masses of all other atoms are given relative to this standard.

And \(1 \mathrm{~amu}=1.66056 \times 10^{-24} \mathrm{~g}\)
Mass of an atom of hydrogen \( =1.6736 \times 10^{-24} \mathrm{~g} \)
Thus, in terms of amu, the mass of hydrogen atom \(=\frac{1.6736 \times 10^{-24} \mathrm{~g}}{1.66056 \times 10^{-24} \mathrm{~g}}\)
\(=1.0078 \mathrm{~amu}\)
\(=1.0080 \mathrm{~amu}\)
Similarly, the mass of oxygen – \(16\left({ }^{16} \mathrm{O}\right)\) atom would be \(15.995 \mathrm{~amu}\).

At present, ‘amu‘ has been replaced by ‘ \(\mathbf{u}\) ‘, which is known as unified mass. When we use atomic masses of elements in calculations, we actually use average atomic masses of elements.

Average Atomic Mass

Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance (percent occurrence), the average atomic mass of that element can be computed. For example, carbon has the following three isotopes with relative abundances and masses as shown against each of them.
\(
\begin{array}{|lcc|}
\hline \text { Isotope } & \begin{array}{c}
\text { Relative } \\
\text { Abundance } \\
\text { (%) }
\end{array} & \begin{array}{c}
\text { Atomic } \\
\text { Mass (amu) }
\end{array} \\
\hline{ }^{12} \mathrm{C} & 98.892 & 12 \\
{ }^{13} \mathrm{C} & 1.108 & 13.00335 \\
{ }^{14} \mathrm{C} & 2 \times 10^{-10} & 14.00317 \\
\hline
\end{array}
\)

From the above data, the average atomic mass of carbon will come out to be:
\((0.98892)(12 \mathrm{~u})+(0.01108)(13.00335 \mathrm{~u})+\) \(\left(2 \times 10^{-12}\right)(14.00317 \mathrm{~u})=12.011 \mathrm{~u}\)

Similarly, average atomic masses for other elements can be calculated. In the periodic table of elements, the atomic masses mentioned for different elements actually represent their average atomic masses.

Determination of atomic weight

Atomic weight \(\times\) specific heat \(=6.4(\mathrm{app})\).

Molecular Mass

Molecular mass is the sum of the atomic masses of the elements present in a molecule. It is obtained by multiplying the atomic mass of each element by the number of its atoms and adding them together. For example, the molecular mass of methane, which contains one carbon atom and four hydrogen atoms, can be obtained as follows:
The molecular mass of methane,
\(
\begin{aligned}
&\left(\mathrm{CH}_{4}\right)=(12.011 \mathrm{~u})+4(1.008 \mathrm{~u}) \\
&=16.043 \mathrm{~u}
\end{aligned}
\)
Similarly, molecular mass of water \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) \(=2 \times\) atomic mass of hydrogen \(+1 \times\) atomic mass of oxygen
\(
\begin{aligned}
&=2(1.008 \mathrm{~u})+16.00 \mathrm{~u} \\
&=18.02 \mathrm{~u}
\end{aligned}
\)

Formula Mass

Some substances, such as sodium chloride, do not contain discrete molecules as their constituent units. In such compounds, positive (sodium ion) and negative (chloride ion) entities are arranged in a three-dimensional structure, as shown in Fig. 1.10.

It may be noted that in sodium chloride, one \(\mathrm{Na}^{+}\)ion is surrounded by six \(\mathrm{Cl}^{-}\)ion and vice-versa.

The formula, such as \(\mathrm{NaCl}\), is used to calculate the formula mass instead of molecular mass as in the solid state sodium chloride does not exist as a single entity.
Thus, the formula mass of sodium chloride is the atomic mass of sodium + atomic mass of chlorine
\(
=23.0 \mathrm{u}+35.5 \mathrm{u}=58.5 \mathrm{u}
\)

Example:
Calculate the molecular mass of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) molecule.
Answer:
Molecular mass of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) \(=6(12.011 \mathrm{~u})+12(1.008 \mathrm{~u})+6(16.00 \mathrm{~u})\)
\(=(72.066 \mathrm{~u})+(12.096 \mathrm{~u})+(96.00 \mathrm{~u})=180.162 \mathrm{~u}\)

Molecular weight:
It is the number of times a molecule of any compound is heavier than \(1 / 12\) th of an atom of \(\mathrm{C}-12\)
\(
\text { Molecular weight }=\frac{\text { Weight of one molecule }}{1 / 12 \times \text { weight of one } \mathrm{C}-12 \text { atom }}
\)

Determination of molecular weight:
(i) Vapour density method:
\(
\text { Vapour density }=\frac{\text { Wt. of a certain vol. of a gas or vapour under certain temperature and pressure }}{\text { Wt. of the same volume of } \mathrm{H}_2 \text { under same temperature and pressure}}
\)
Molecular weight \(=2 \times\) vapour density

(ii) Diffusion method:
(a) It is based on Graham’s law of diffusion.
(b) Graham’s law states that: The rate of diffusion of different gases, under similar conditions of temperature and pressure are inversely proportional to the square roots of their density (or molecular weights).
\(
\frac{r_1}{r_2}=\sqrt{\frac{d_2}{d_1}}=\sqrt{\frac{M_2}{M_1}}
\)