Moles and Avogadro’s Number
Moles and Avogadro’s Number
It is sometimes convenient to work with a unit other than molecules when measuring the amount of substance. A mole (abbreviated mol) is defined to be the amount of a substance that contains as many atoms or molecules as there are atoms in exactly 12 grams (0.012 kg) of carbon-12. The actual number of atoms or molecules in one mole is called Avogadro’s number\(({N}_{\text{A}})\), in recognition of Italian scientist Amedeo Avogadro (1776–1856). He developed the concept of the mole, based on the hypothesis that equal volumes of gas, at the same pressure and temperature, contain equal numbers of molecules. That is, the number is independent of the type of gas. This hypothesis has been confirmed, and the value of Avogadro’s number is
\({N}_{\text{A}}=6\text{.}\text{02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}{\text{mol}}^{-1}\text{.}\)
Avogadro’s Number
One mole always contains \(6\text{.}\text{02}×{\text{10}}^{\text{23}}\) particles (atoms or molecules), independent of the element or substance. A mole of any substance has a mass in grams equal to its molecular mass, which can be calculated from the atomic masses given in the periodic table of elements.
\({N}_{\text{A}}=6\text{.}\text{02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}{\text{mol}}^{-1}\)
How big is a mole? On a macroscopic level, one mole of table tennis balls would cover the Earth to a depth of about 40 km.
Check Your Understanding
The active ingredient in a Tylenol pill is 325 mg of acetaminophen \(({\text{C}}_{8}{\text{H}}_{9}{\text{NO}}_{2})\). Find the number of active molecules of acetaminophen in a single pill.
Solution
We first need to calculate the molar mass (the mass of one mole) of acetaminophen. To do this, we need to multiply the number of atoms of each element by the element’s atomic mass.
\(\begin{array}{}(\text{8 moles of carbon})(\text{12 grams/mole})+(\text{9 moles hydrogen})(\text{1 gram/mole})\\ +(\text{1 mole nitrogen})(\text{14 grams/mole})+(\text{2 moles oxygen})(\text{16 grams/mole})=\text{151 g}\end{array}\)
Then we need to calculate the number of moles in 325 mg.
\((\cfrac{\text{325 mg}}{\text{151 grams/mole}})(\cfrac{\text{1 gram}}{\text{1000 mg}})=2.15×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{moles}\)
Then use Avogadro’s number to calculate the number of molecules.
\(N=(\text{2.15}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{moles})(\text{6.02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}\text{molecules/mole})=\text{1.30}×{\text{10}}^{\text{21}}\phantom{\rule{0.25em}{0ex}}\text{molecules}\)
Example: Calculating Moles per Cubic Meter and Liters per Mole
Calculate: (a) the number of moles in \(1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\) of gas at STP, and (b) the number of liters of gas per mole.
Strategy and Solution
(a) We are asked to find the number of moles per cubic meter, and we know from this example that the number of molecules per cubic meter at STP is \(2\text{.}\text{68}×{\text{10}}^{\text{25}}\). The number of moles can be found by dividing the number of molecules by Avogadro’s number. We let \(n\) stand for the number of moles,
\(n\phantom{\rule{0.25em}{0ex}}{\text{mol/m}}^{3}=\cfrac{N\phantom{\rule{0.25em}{0ex}}{\text{molecules/m}}^{3}}{6\text{.}\text{02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}\text{molecules/mol}}=\cfrac{2\text{.}\text{68}×{\text{10}}^{\text{25}}\phantom{\rule{0.25em}{0ex}}{\text{molecules/m}}^{3}}{6\text{.}\text{02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}\text{molecules/mol}}=\text{44}\text{.}5\phantom{\rule{0.25em}{0ex}}{\text{mol/m}}^{3}\text{.}\)
(b) Using the value obtained for the number of moles in a cubic meter, and converting cubic meters to liters, we obtain
\(\cfrac{({\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{L/m}}^{3})}{44\text{.}5\phantom{\rule{0.25em}{0ex}}{\text{mol/m}}^{3}}=\text{22}\text{.}5\phantom{\rule{0.25em}{0ex}}\text{L/mol}\text{.}\)
Discussion
This value is very close to the accepted value of 22.4 L/mol. The slight difference is due to rounding errors caused by using three-digit input. Again this number is the same for all gases. In other words, it is independent of the gas.
The (average) molar weight of air (approximately 80% \({\text{N}}_{2}\) and 20% \({\text{O}}_{2}\) is \(M=\text{28}\text{.}8\phantom{\rule{0.25em}{0ex}}\text{g}\text{.}\) Thus the mass of one cubic meter of air is 1.28 kg. If a living room has dimensions \(5\phantom{\rule{0.25em}{0ex}}\text{m}×\text{5 m}×\text{3 m,}\) the mass of air inside the room is 96 kg, which is the typical mass of a human.
Check Your Understanding
The density of air at standard conditions \((P=1\phantom{\rule{0.25em}{0ex}}\text{atm}\) and \(T=\text{20}\text{º}\text{C})\) is \(1\text{.}\text{28}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}\). At what pressure is the density \(0\text{.}{\text{64 kg/m}}^{3}\) if the temperature and number of molecules are kept constant?
Solution
The best way to approach this question is to think about what is happening. If the density drops to half its original value and no molecules are lost, then the volume must double. If we look at the equation \(\text{PV}=\text{NkT}\), we see that when the temperature is constant, the pressure is inversely proportional to volume. Therefore, if the volume doubles, the pressure must drop to half its original value, and \({P}_{\text{f}}=0\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{atm}\text{.}\)
This lesson is part of:
Temperature, Kinetic Theory, and Gas Laws