# How many atoms are in a gram of air?

Jun 4, 2018

Warning! Long Answer. There are 4.140 × 10^22 atoms in a gram of air.

#### Explanation:

The major components of dry air are nitrogen ( 78.08 %), oxygen (20.95 %), argon (0.93 %) and carbon dioxide (0.04 %).

The molar mass of air is the weighted average of the molar masses of its components.

That is, we multiply the molar mass of each gas by its percentage of the mixture.

$\underline{\boldsymbol{\text{Gas"color(white)(m)M_text(r)color(white)(mm)"%"color(white)(ml)"Contribution}}}$
${\text{N}}_{2} \textcolor{w h i t e}{m l l} 28.01 \textcolor{w h i t e}{m} 78.08 \textcolor{w h i t e}{m m l l} 21.870$
${\text{O}}_{2} \textcolor{w h i t e}{m l l} 32.00 \textcolor{w h i t e}{m} 20.95 \textcolor{w h i t e}{m m m} 6.704$
$\text{Ar} \textcolor{w h i t e}{m l l} 39.95 \textcolor{w h i t e}{m l l} 0.93 \textcolor{w h i t e}{m m m} 0.372$
$\underline{{\text{CO}}_{2} \textcolor{w h i t e}{m} 44.01 \textcolor{w h i t e}{m l} 0.04 \textcolor{w h i t e}{m m m l} 0.018}$
$\textcolor{w h i t e}{m m m m m m m l l} \text{Total = 28.964}$

Thus, the relative molar mass of air is 28.964.

Then, in 1 g of air,

$\text{Moles of air" = 1 color(red)(cancel(color(black)("g air"))) × "1 mol air"/(28.964 color(red)(cancel(color(black)("g air")))) = "0.034 53 mol air}$

Calculate the atoms of nitrogen

${n}_{\textrm{m o \le c \underline{e} s}} = {\text{0.034 53" color(red)(cancel(color(black)("mol air")))× "0.7808 mol N"_2/(1 color(red)(cancel(color(black)("mol air")))) = "0.026 96 mol N}}_{2}$

${n}_{\textrm{a \to m s}} = \text{0.026 96" color(red)(cancel(color(black)("molecules N"_2))) × "2 atoms N"/(1 color(red)(cancel(color(black)("molecule N"_2)))) = "0.053 92 mol N}$

"Atoms of N" = "0.053 92" color(red)(cancel(color(black)("mol N"))) × (6.022 × 10^23 "atoms N")/(1 color(red)(cancel(color(black)("mol N"))))
= 3.247 × 10^22 color(white)(l)"atoms N"

Rather than show the calculations for the other gases, I will summarize the results in a table.

ulbb("Gas"color(white)(m)"%"color(white)(mml) n_text(molecules)color(white)(ml)n_text(atoms)color(white)(m)"Atoms/"10^22
$\text{N"_2color(white)(ml)78.08color(white)(mll)"0.026 96"color(white)(mll)"0.053 92} \textcolor{w h i t e}{m m} 3.247$
$\text{O"_2color(white)(ml)20.95color(white)(mll)"0.007 234"color(white)(m)"0.014 47} \textcolor{w h i t e}{m m} 0.8713$

$\text{Ar"color(white)(mm)0.93color(white)(mll)"0.000 321"color(white)(m)"0.000 321} \textcolor{w h i t e}{m l} 0.0193$
$\underline{\text{CO"_2color(white)(m)0.04color(white)(mll)"0.000 014"color(white)(m)"0.000 041} \textcolor{w h i t e}{m l l} 0.0025}$
$\textcolor{w h i t e}{m} \text{Total =" color(white)(mm) "0.034 53"color(white)(ml)"0.068 752} \textcolor{w h i t e}{m l l} 4.140$

This 1 g of air contains 4.140 × 10^22 atoms.