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 .4 Avogadro‘s number

Avogadro’s number 

The numbers of protons and neutrons in the nucleus of an atom of an element are related to the

atomic mass of the element, which is defined as the mass of a single atom of the element

measured in atomic mass units (u) where 1 u = 1.660 53 !" 10 #$! %g.

 &madeo &'ogadro (1!66#156) ne'er %new his own number it was named in his honour b a

*rench scientist in 1+0+ its 'alue was first estimated b osef -oschmidt, an &ustrian chemistr

teacher, in 1+5.

wing to their tin si/e, atoms and molecules cannot be counted b direct obser'ation. There

are, howe'er, a number of indirect methods that enable us to estimate the number of these

particles in a sample of an element or compound. nce this has been done, we %now the

number of formula units (to use the most general term for an combination of atoms we wish to

define) in an arbitrar weight of the substance. The number will of course depend both on the

formula of the substance and on the weight of the sample. ut if we consider a weight of

substance that is the same as its formula (molecular) weight e"pressed in grams, we ha'e onl

one number to %now &'ogadro2s number, 6.0$$115$! 4 10$3, usuall designated b  &.

hat is the special significance of this huge number, 6.0$ 4 10$37 To help ou understand this

e"tremel important point, ta%e a moment to con'ince ourself of the reasoning embodied in the

following se8uence of problems

The atomic weights of o"gen and of carbon are 16.0 and 1$.0, respecti'el. 9ow much hea'ier

is the o"gen atom in relation to carbon7

:olution &tomic weights represent the relati'e masses of different %inds of atoms. This means

that the atom of o"gen has a mass that is 16;1$ = ;3 < 1.33 as great as the mass of a carbon

atom.

The absolute mass of a carbon atom is 1$.0 unified atomic mass units (hat are these7). 9ow

man grams will a single o"gen atom weigh7

:olution The absolute mass of the carbon atom is 1$.0 u, or 1$ 4 1.6605 4 10 $! g

= 1+.+ 4 10 $! %g. The mass of the o"gen atom will be ;3 greater, or $.66 4 10 $6 %g.

 &lternati'el (1$ g;mol) > (6.0$$ 4 10$3 mol 1) 4 (;3) = $.66 4 10 $3 g.

:uppose that we ha'e carbon atoms, where is a number large enough to gi'e us a pile of

carbon atoms whose mass is 1$.0 grams. 9ow much would the same number, , of o"gen

atoms weigh7

:olution The collection of o"gen atoms would ha'e a mass of ;3 4 1$ g = 16.0 g

:o what 'alue must N  ha'e in order to ma%e the weight of a pile of N  atoms of an %ind

numericall e8ual to the atomic weight of the element7 The answer is just Avogadro’s

number N A = 6.022141527 × 102

.

http://www.woodrow.org/teachers/ci/1992/Avogadro.htmlhttp://www.woodrow.org/teachers/ci/1992/Avogadro.htmlhttp://www.chem1.com/acad/webtext/intro/AT.html#SEC6http://www.chem1.com/acad/webtext/intro/AT.html#SEC6http://www.woodrow.org/teachers/ci/1992/Avogadro.html

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