Količina tvari ili množina tvari označava se sa n , a njezina mjerna jedinica je mol atoma u 12 g (0.012 kg) ugljikovog izotopa 12 C.
Formula za množinu tvari glasi:
n
=
m
M
=
N
L
=
V
0
V
m
{\displaystyle n={\frac {m}{M}}={\frac {N}{L}}={\frac {V^{0}}{V_{m}}}}
Pri čemu je:
Napomena : ? označava vrijednost koja ovisi o zadatku.
Izračunajte množinu tvari klora u 200 g.
m
(
C
l
2
)
=
200
g
{\displaystyle m(Cl_{2})=200g}
n
(
C
l
2
)
=
?
{\displaystyle n(Cl_{2})=?}
n
(
C
l
2
)
=
m
(
C
l
2
)
M
(
C
l
2
)
=
200
g
(
2
⋅
35
,
45
)
⋅
g
m
o
l
−
1
=
200
g
70
,
9
g
m
o
l
−
1
=
2
,
82
m
o
l
{\displaystyle n(Cl_{2})={\frac {m(Cl_{2})}{M(Cl_{2})}}={\frac {200~g}{(2\cdot 35,45)\cdot gmol^{-1}}}={\frac {200~g}{70,9~gmol^{-1}}}=2,82~mol}
Izračunaj množinu iona natrija u 0,5 g kuhinjske soli.
m
(
N
a
C
l
)
=
0
,
5
g
{\displaystyle m(NaCl)=0,5g}
n
(
N
a
+
)
=
?
{\displaystyle n(Na^{+})=?}
n
(
N
a
C
l
)
=
m
(
N
a
C
l
)
M
(
N
a
C
l
)
=
0
,
5
g
(
22
,
99
+
35
,
45
)
⋅
g
m
o
l
−
1
=
0
,
5
g
58
,
44
g
m
o
l
−
1
=
8
,
56
⋅
10
−
3
m
o
l
{\displaystyle n(NaCl)={\frac {m(NaCl)}{M(NaCl)}}={\frac {0,5~g}{(22,99+35,45)\cdot gmol^{-1}}}={\frac {0,5~g}{58,44~gmol^{-1}}}=8,56\cdot 10^{-3}~mol}
N
a
C
l
⟶
N
a
+
+
C
l
−
⇒
n
(
N
a
C
l
)
n
(
N
a
+
)
=
1
1
⇒
n
(
N
a
+
)
=
n
(
N
a
C
l
)
{\displaystyle NaCl\longrightarrow Na^{+}+Cl^{-}\qquad \Rightarrow \qquad {\frac {n(NaCl)}{n(Na^{+})}}={\frac {1}{1}}\qquad \Rightarrow \qquad n(Na^{+})=n(NaCl)}
n
(
N
a
+
)
=
n
(
N
a
C
l
)
=
8
,
56
⋅
10
−
3
m
o
l
{\displaystyle {\begin{alignedat}{2}n(Na^{+})&=n(NaCl)\\&=8,56\cdot 10^{-3}~mol\\\end{alignedat}}}
U uzorku kromova(III) klorida mase 0,08 g izračunaj množinu kloridnih iona.
m
(
C
r
C
l
3
)
=
0
,
08
g
{\displaystyle m(CrCl_{3})=0,08g}
n
(
C
l
−
)
=
?
{\displaystyle n(Cl^{-})=?}
n
(
C
r
C
l
3
)
=
m
(
C
r
C
l
3
)
M
(
C
r
C
l
3
)
=
0
,
08
g
(
52
,
00
+
3
⋅
35
,
45
)
⋅
g
m
o
l
−
1
=
0
,
08
g
158
,
35
g
m
o
l
−
1
=
5
,
05
⋅
10
−
4
m
o
l
{\displaystyle n(CrCl_{3})={\frac {m(CrCl_{3})}{M(CrCl_{3})}}={\frac {0,08~g}{(52,00+3\cdot 35,45)\cdot gmol^{-1}}}={\frac {0,08~g}{158,35~gmol^{-1}}}=5,05\cdot 10^{-4}~mol}
C
r
C
l
3
⟶
C
r
3
+
+
3
C
l
−
⇒
n
(
C
r
C
l
3
)
n
(
C
l
−
)
=
1
3
⇒
n
(
C
l
−
)
=
3
⋅
n
(
C
r
C
l
3
)
{\displaystyle CrCl_{3}\longrightarrow Cr^{3+}+3~Cl^{-}\qquad \Rightarrow \qquad {\frac {n(CrCl_{3})}{n(Cl^{-})}}={\frac {1}{3}}\qquad \Rightarrow \qquad n(Cl^{-})=3\cdot n(CrCl_{3})}
n
(
C
l
−
)
=
3
⋅
n
(
C
r
C
l
3
)
=
3
⋅
5
,
05
⋅
10
−
4
m
o
l
=
1
,
52
⋅
10
−
3
m
o
l
{\displaystyle {\begin{alignedat}{2}n(Cl^{-})&=3\cdot n(CrCl_{3})\\&=3\cdot 5,05\cdot 10^{-4}~mol\\&=1,52\cdot 10^{-3}~mol\\\end{alignedat}}}
Primjeri 2 i 3 djelomično su preuzeti iz:A. Petreski - B. Sever, Zbirka riješenih primjera i zadataka iz opće kemije, PROFIL international, 4. izdanje, Zagreb, 1997. 
