Collections:Texts:Chemistry:2.8 The Molar Mass

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2.8 THE MOLAR MASS


It is often convenient to express physical quantities per unit 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 solid 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.)

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=\frac{m}{n}      (2.5)


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

Solution

a) The atomic weight 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{-1}}}=\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. Equation (2.5), which defines the molar mass, has the same form as Eq. (1.1), which defined density, and Eq. (2.4), which defined 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 2.6 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 2.7 How many molecules would be present in 25.0 ml of pure carbon tetrachloride (CCl4)?

Solution In Example 2.4 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 (Example 2.6), and mass from volume by means of density (Example 1.12). 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 in Table 1.4:


ρ = 1.595 g cm–3


The molar mass must be calculated from 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 three 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.


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