n(moli)=m(masa in g)/A(masa atomara in g/mol) (1)
n(moli)=4,8 g/24 g/mol=0,2 moli (Mg)
NA=6,023*1023particule...Nr. lui AVOGADRO
1mol particule substanta.........6,23*1023particule (atomi, molecule,ioni)
n moli ......... .....N particule
sau :
n(moli)=N(particule)/NA(particule/mol) (2)
N(atomi Mg)=0,2moli * 6,023*1023atomi./mol=0,2* 6,023*1023 atomi de Mg
8gH2 n H2=8g/2g/mol=4moli hidrogen (1)
1mol H2 contine 6,023 1023molecule H2.........2 * 6,23 1023atomi de H
4 moli H2 ...................... 4*2* 6,23 1023atomi de H
64g S n S= 64g/32g/mol =2 moli atomi S
N(atomi S) = 2 moli * 6,023*1023atomi./mol=2*6,023*1023 atomi de S