The Molar Mass

Submitted by ChemPRIME Staff on Wed, 12/08/2010 - 23:44


As we saw in The Amount of Substance: Moles, there is no relationship between the mass or volume of a substance and the number of molecules. But we then defined the amount of substance, n, to represent the number of particles. The amount is therefore useful in determining how much of each substance will react. While 1 g of Hg, or 1 cm3 of Hg, reacts with a mass or volume of Br2 that is not related to the coefficients of the chemical equationA representation of a chemical reaction in which chemical symbols represent reactants on the left side and products on the right side., 1 mol of Hg always reacts with 1 mol of Br2, since one atom of Hg reacts with one molecule of Br2:

1 Hg (l) + 1 Br2 (l) → 1 HgBr2 (s)
1 atom 1 molecule 1 molecule
1.00 g 0.797 g 1.797 g
1.00 cm3 3.47 cm3 0.30 cm3
1.00 mol Hg 1.00 mol Br2 1.00 mol HgBr2

What we need is a convenient way to convert masses to amounts, and the necessary conversion factorA relationship between two units of measure that is derived from the proportionality of one quantity to another; for example, the mass of a substances is proportional to its volume and the conversion factor from volume to mass is density. is called the molar massThe mass of a mole of substance; the same as molecular weight for molecular substances.. A molar quantity is one which has been divided by the amount of substance. For example, an extremely useful molar quantity is the molar mass M:

\text{Molar mass}=\frac{\text{mass}}{\text{amount of substance}}      

M~(g/mol) ~=~\frac{m~(g)}{n~(mol)}      (1)

It is often convenient to express physical quantities per unitA particular measure of a physical quantity that is used to express the magnitude of the physical quantity; for example, the meter is the unit of the physical quantity, length. amount of substance (per mole), because in this way equal numbers of atoms or molecules are being compared. Such molar quantities often tell us something about the atoms or molecules themselves. For example, if the molar volume of one solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. is larger than that of another, it is reasonable to assume that the molecules of the first substance are larger than those of the second. (Comparing the molar volumes of liquids, and especially gases, would not necessarily give the same information since the molecules would not be as tightly packed.)

It is almost trivial to obtain the molar mass, since atomic and molecular weights expressed in grams give us the masses of 1 mol of substance.



EXAMPLE 1 Obtain the molar mass of (a) Hg and (b) Hg2Br2.


SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture.

a) The atomic weightThe average mass of the naturally occurring isotopes of an element, taking into account the different natural abundances of the isotopes. Expressed relative to the value of exactly 12 for carbon-12; also called atomic mass. of mercury is 200.59, and so 1 mol Hg weighs 200.59 g.


M_{\text{Hg}}=\frac{\text{m}_{\text{Hg}}}{n_{\text{Hg}}}=\frac{\text{200}\text{.59 g}}{\text{1 mol}}=\text{200}\text{.59 g mol}^{-1}


b) Similarly, for Hg2Br2 the molecular weight is 560.98, and so


M_{\text{Hg}_{\text{2}}\text{Br}_{\text{2}}}=\frac{\text{m}_{\text{Hg}_{\text{2}}\text{Br}_{\text{2}}}}{n_{\text{Hg}_{\text{2}}\text{Br}_{\text{2}}}}=\text{560}\text{.98 g mol}^{-1}


The molar mass is numerically the same as the atomic or molecular weight, but it has units of grams per mole. The equation, which defines the molar mass, has the same form as those defining density, and the Avogadro constant. As in the case of density or the Avogadro constant, it is not necessary to memorize or manipulate a formula. Simply remember that mass and amount of substance are related via molar mass.


\text{Mass }\overset{\text{Molar mass}}{\longleftrightarrow}\text{amount of substance    }m\overset{M}{\longleftrightarrow}n


The molar mass is easily obtained from atomic weights and may be used as a conversion factor, provided the units cancel.



EXAMPLE 2 Calculate the amount of octane (C8H18) in 500 g of this liquid.


Solution Any problem involving interconversion of mass and amount of substance requires molar mass


M = (8 × 12.01 + 18 × 1.008) g mol–1 = 114.2 g mol–1


The amount of substance will be the mass times a conversion factor which permits cancellation of units:


n = m × conversion factor = m × \frac{\text{1}}{M} = 500 g × \frac{\text{1 mol}}{\text{114}\text{.2 g}} = 4.38 mol


In this case the reciprocal of the molar mass was the appropriate conversion factor.


The Avogadro constant, molar mass, and density may be used in combination to solve more complicated problems.



EXAMPLE 3 How many molecules would be present in 25.0 ml of pure carbon tetrachloride (CCl4)?


Solution In previous examples, we showed that the number of molecules may be obtained from the amount of substance by using the Avogadro constant. The amount of substance may be obtained from mass by using the molar mass, and mass from volume by means of density. A road map to the solution of this problem is


\text{Volume}\xrightarrow{\text{density}}\text{mass }\overset{\text{Molar mass}}{\longleftrightarrow}\text{amount}\overset{\text{Avogadro constant}}{\longleftrightarrow}\text{number of molecules}


or in shorthand notation


V\xrightarrow{\rho }m\xrightarrow{M}n\xrightarrow{\text{N}_{\text{A}}}N


The road map tells us that we must look up the density of CCl4:


ρ = 1.595 g cm–3


The molar mass must be calculated from the Table of Atomic Weights.


M = (12.01 + 4 × 35.45) g mol–1 = 153.81 g mol–1


and we recall that the Avogadro constant is


NA = 6.022 × 1023 mol–1


The last quantity (N) in the road map can then be obtained by starting with the first (V) and applying successive conversion factors:


\begin{align}
  & \text{N}=\text{25}\text{.0 cm}^{\text{3}}\times \frac{\text{1}\text{.595 g}}{\text{1 cm}^{\text{3}}}\times \frac{\text{1 mol}}{\text{153}\text{.81 g}}\times \frac{\text{6}\text{.022}\times \text{10}^{\text{23}}\text{ molecules}}{\text{1 mol}} \ 
 & \text{   }=\text{1}\text{.56}\times \text{10}^{\text{23}}\text{ molecules} \ 
\end{align}


Notice that in this problem we had to combine techniques from previous examples. To do this you must remember relationships among quantities. For example, a volume was given, and we knew it could be converted to the corresponding mass by means of density, and so we looked up the density in a table. By writing a road map, or at least seeing it in your mind’s eye, you can keep track of such relationships, determine what conversion factors are needed, and then use them to solve the problem.