Question

3. The Earth has a mass of 5.98 x 10²⁴ kg. Theaverage mass of the atoms that make up the Earth is 40 u. Howmany atoms are in the Earth? Show all work!

( u = atomic mass unit = 1.661 x10^-27 kg )

Answer 1

Concepts and reason

The concept of atomic mass and number of atoms are used to solve this question. The mass of an atom is given in the unit of ‘atomic mass unit’ so, first convert the mass of an atom into kilograms. Then find the number of atoms in the Earth.

Fundamentals

Atomic Mass:

The atomic mass is the mass of an atom which is expressed in terms of atomic mass unit. Generally, the atomic mass is expressed in ‘u’ or ‘amu’ (atomic mass unit).

Conversion of ‘u’ into ‘kg’:

$1\;{\rm{u}} = 1.661 \times {10^{ - 27}}\;{\rm{kg}}$

Number of atoms:

The total number of atoms in a sample is equal to the total mass of that sample divided by mass of one atom.

$N = \frac{M}{m}$

Here, $M$ is the mass of the sample and $m$ is the mass of an atom.

Conversion of ‘u’ into ‘kg’:

$1\;{\rm{u}} = 1.661 \times {10^{ - 27}}\;{\rm{kg}}$

Mass of an atom in the Earth:

$\begin{array}{c}\\m = 40\;{\rm{u}}\\\\ = 40{\rm{ u}}\left( {\frac{{1.661 \times {{10}^{ - 27}}\;{\rm{kg}}}}{{1{\rm{ u}}}}} \right)\\\\ = 66.44 \times {10^{ - 27}}\;{\rm{kg}}\\\end{array}$

Expression to find the number of atoms in the Earth:

${N_E} = \frac{{{M_E}}}{m}$

Here, ${M_E}$ is the mass of the earth and m is the mass of the atom in the earth.

Substitute $5.98 \times {10^{24}}\;{\rm{kg}}$ for ${M_E}$ and $66.44 \times {10^{ - 27}}\;{\rm{kg}}$ for $m$ .

$\begin{array}{c}\\{N_E} = \frac{{5.98 \times {{10}^{24}}\;{\rm{kg}}}}{{66.44 \times {{10}^{ - 27}}\;{\rm{kg}}}}\\\\ = 8.94 \times {10^{49}}\;{\rm{atoms}}\\\\ \approx {\bf{9}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{49}}}}\;{\bf{atoms}}\\\end{array}$

Ans:

The number of atoms in the Earth are ${\bf{9}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{49}}}}$ .