Soal dan Pembahasan Hubungan Mol dengan Massa Molar

Soal 1

 Diketahui $Ar$ C = 12, O = 16, dan Fe = 56. Berapakah massa molar Fe dan molekul CO$_2$?

Jawab :
Perbandingan massa 1 atom C-12 : massa 1 atom Fe : massa 1 molekul CO$_2$ = 12 : 56 : 44.
Massa 1 mol C-12 = 12 gram, maka
massa 1 mol Fe = $ \frac{56}{12} \times 12 = 56 \, $ gram
massa 1 mol CO$_2 = \frac{44}{12} \times 12 = 44 \, $ gram
Jadi, massa 1 mol Fe = $Ar$ Fe, dan massa 1 mol CO$_2$ = $Mr $ CO$_2$.

Soal 2

Hitunglah massa dari :
a). 5 mol besi ($Ar$ Fe = 56)
b). 0,75 mol urea CO(NH$_2$)$_2$ ($Ar$ C = 12, O = 16, N = 14, H = 1)
c). 0,5 mol O$_2$ ($Ar$ O = 16)

Jawab :
a). massa besi $ = n \times A_r \, Fe = 5 \times 56 = 280 \, $ gram.
b). massa urea $ = n \times M_r \, CO[NH_2]_2 = 0,75 \times 60 = 45 \, $ gram.
c). massa urea $ = n \times M_r \, O_2 = 0,5 \times 32 = 16 \, $ gram.

Soal 3

 Hitunglah banyaknya mol dari:
a). 2,3 gram natrium (Ar Na = 23)
b). 45 gram C$_6$H$_{12}$O$_6$ ( Ar C = 12, H = 1, dan O = 16)
c). 35,1 gram NaCl (Ar Na = 23, dan Cl = 35,5)
d). 196,5 gram seng (Ar Zn = 65,5)

Jawab :
a). mol Na $ = \frac{m}{A_r \, Na} = \frac{2,3}{23} = 0,1 \, $ mol.
b). mol C$_6$H$_{12}$O$_6 = \frac{m}{M_r \, C_6H_{12}O_6} = \frac{45}{180} = 0,25 \, $ mol.
c). mol NaCl $ = \frac{m}{M_r \, NaCl} = \frac{35,1}{58,5} = 0,6 \, $ mol.
d). mol Zn $ = \frac{m}{A_r \, Zn} = \frac{196,5}{65,5} = 3 \, $ mol.