As we experienced in The lot of Substance: Moles, there is no relationship between the mass or volume that a substance and the number of molecules. However we then characterized the quantity of substance, n, to represent the number of particles. The quantity is as such useful in determining exactly how much of every substance will react. When 1 g of Hg, or 1 cm3 the Hg, reacts v a fixed or volume of Br2 the is not concerned the coefficients that the centregalilee.comical equation, 1 mol the Hg constantly reacts through 1 mol the Br2, since one atom of Hg reacts through one molecule the Br2:

What we require is a convenient way to transform masses come amounts, and also the crucial conversion factor is referred to as the molar mass. A molar amount is one which has been divided by the quantity of substance. For example, an extremely useful molar quantity is the molar massive M:

< extMolar mass=frac extmass extamount the substance\ extitM ext(g/mol)=frac extitm ext (g) extitn ext (mol)>

It is often convenient to express physical amounts per unit amount of substance (per mole), since in this method equal numbers of atoms or molecules room being compared. Together molar quantities often tell united state something around the atom or molecule themselves. Because that example, if the molar volume the one heavy is larger than that of another, that is reasonable come assume that the molecule of the first substance are bigger than those that the second. (Comparing the molar volumes of liquids, and also especially gases, would not necessarily offer the same information due to the fact that the molecules would certainly not be as tightly packed.)

It is almost trivial to acquire the molar mass, since atomic and also molecular weights express in grams provide us the masses the 1 mol of substance.

You are watching: Mercury has an atomic mass of 200.59 amu

Example (PageIndex1): Molar Mass

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

Solution

a) The atomic weight of mercury is 200.59, and also so 1 mol Hg weighs 200.59 g.

( M_ extHg=fracm_ extHgn_ extHg = frac ext200 ext.59 g ext1 mol= ext200.59 g mol^-1)

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

< extM_ extHg_2 extBr_2 =frac extm_ extHg_2 extBr_2 extn_ extHg_2 extBr_2 = ext560.98 g mol^-1>

Example (PageIndex2): Moles

Calculate the lot of octane (C8H18) in 500 g the this liquid.

Solution

Any difficulty involving interconversion the mass and also amount of substance calls for molar mass

< extitM=8cdot12.01 ext + 18cdot1.008 ext g mol^ ext-1= 114.2 ext g mol^ ext-1>

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

< extitn= extitmcdot extconversion factor= extmcdot frac 1 extitM= 500 ext gcdot frac ext1 mol ext114.2 g= ext4.38 mol>

In this case the mutual of the molar mass to be the proper conversion factor.

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

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Example (PageIndex3): Molecules

How countless molecules would be existing in 25.0 ml that pure carbon tetrachloride (CCl4)?

Solution

In previous examples, we confirmed that the variety of molecules might be derived from the lot of problem by utilizing the Avogadro constant. The amount of substance may be derived from mass by making use of the molar mass, and also mass native volume by way of density. A road map to the solution of this trouble is

< extVolumexrightarrow extdensity extmass overset extMolar masslongleftrightarrow extamountoverset extAvogadro constantlongleftrightarrow extnumber of molecules>

or in shorthand notation

< extVxrightarrow ho extitmxrightarrow extitM extitnxrightarrow extitN_ extitA extitN>

The road map tells united state that we should look increase the density of CCl4:

< ho=1.595 ext g ext cm^-3>

The molar mass should be calculated native the Table of atomic Weights.

< extitM=left(12.01 + 4cdot35.45 ight) ext g mol^-1=153.81 ext g mol^-1>

and us recall the the Avogadro constant is

< extitN_ extitA=6.022cdot10^23 ext mol^-1>

The last amount (N) in the road map can then be acquired by beginning with the first (V) and applying succeeding conversion factors:

eginalign extitN&=& 25 ext cm^3cdot frac ext1.595 g 1 ext cm^3cdot frac 1 ext mol ext153.81 gcdot frac 6.022cdot10^23 ext molecules1 ext mol \&=&1.56cdot10^23 ext molecules endalign