Question

What is the mass in grams of \[3.75 \times {10^{15}}\]atoms of gold?

Answer 1

Hint: We should have knowledge of the formula to calculate number of moles and we should also know that \[1{\text{mol}}\] of substance contains avogadro's number of atoms. Avogadro number of atoms is denoted by \[{{\text{N}}_a}\].
Formula used for calculation of number of moles is shown below:
\[{\text{n = }}\dfrac{{\text{m}}}{{\text{M}}}\]
Where n is number of mole
m is given mass
M is molar mass
\[6.022 \times {10^{23}}{\text{atoms}} = 1{\text{mole}}\]

Complete answer:
First step is to find the number of moles present in \[3.75 \times {10^{15}}\] atoms of gold.
We know that \[1{\text{ mol}}\]of substance contains avogadro's number of atoms. Now, let’s calculate the number of atoms in \[3.75 \times {10^{15}}\] atoms of gold.
\[6.022 \times {10^{23}}{\text{atoms}} = 1{\text{mole}}\]
\[{\text{3}}{\text{.75}} \times {\text{1}}{{\text{0}}^{15}}{\text{atoms = ?}}\]
\[{\text{ = }}\dfrac{{{\text{3}}{\text{.75}} \times {\text{1}}{{\text{0}}^{15}}{\text{atoms}} \times 1{\text{mole}}}}{{6.022 \times {{10}^{23}}{\text{atoms}}}}\]
On simplification we get,
\[ = 6.227 \times {10^{ - 9}}{\text{mol}}\]
Second step is to calculate the mass of the atom by using the formula of the number of moles.
We know that, atomic mass of gold is \[197{\text{g/mol}}\]
\[{\text{n = }}\dfrac{{\text{m}}}{{\text{M}}}\]
Where n is number of mole
m is given mass
M is molar mass
Now, calculate the mass of gold atom, by substituting the values in the above formula:
\[6.227 \times {10^{ - 9}}{\text{mol}} = \dfrac{{\text{m}}}{{197{\text{g/mol}}}}\]
\[{\text{m}} = 6.227 \times {10^{ - 9}}{\text{mol}} \times 197{\text{g/mol}}\]
On multiplication we get,
\[{\text{m}} = 1.2267 \times {10^{ - 6}}{\text{g}}\]
Hence, the mass of the gold atom is \[1.2267 \times {10^{ - 6}}g\].

Additional information:
We have to remember that the gold is a metal atom which belongs to the periodic table. Its atomic number is \[79\] and mass number is \[197\]. Its symbol is \[{\text{Au}}\] which is derived from its latin name that is aurum.

Note:
We should be kept in mind that by using dimensional analysis we can easily get the units. By using the dimensional analysis there is w/very less chance of getting wrong units. Never forget to write the units.