\(\color{blue} \text{Jumlah mol} = \dfrac{\text{massa zat (gram)}}{\text{Ar} \text{ atau } \text{Mr}}\)
Tentukan jumlah mol dari 28 gram besi (Ar Fe = 56).
Jawaban: 0,5 mol
\(\text{Jumlah mol besi} = \dfrac{\text{massa besi}}{\text{Ar }\ce{Fe}}\)
\(\text{Jumlah mol besi} = \dfrac{28 \text{ gram}}{56 \text{ gram/mol}}\)
\(\text{Jumlah mol besi} = \dfrac{1}{2}\text{ mol}\)
Tentukan jumlah mol dari 7 gram gas nitrogen (Ar N = 14).
Jawaban: 0,25 mol
Gas nitrogen = \(\ce{N2}\)
Massa molekul relatif (Mr) \(\ce{N2}\) = 2 × Ar N
Massa molekul relatif (Mr) \(\ce{N2}\) = 2 × 14
Massa molekul relatif (Mr) \(\ce{N2}\) = 28
\(\text{Jumlah mol gas nitrogen} = \dfrac{\text{massa nitrogen}}{\text{Mr }\ce{N2}}\)
\(\text{Jumlah mol gas nitrogen} = \dfrac{7 \text{ gram}}{28 \text{ gram/mol}}\)
\(\text{Jumlah mol gas nitrogen} = \dfrac{1}{4}\text{ mol}\)
Tentukan jumlah mol dari 200 gram kalsium karbonat.
Jawaban: 2 mol
\(\text{Mr } \ce{CaCO3} = \text{Ar } \ce{Ca} + \text{Ar } \ce{C} + 3\text{Ar } \ce{O}\)
\(\text{Mr } \ce{CaCO3} = 40 + 12 + 3(16)\)
\(\text{Mr } \ce{CaCO3} = 100\)
\(\text{Jumlah mol kalsium karbonat} = \dfrac{\text{massa kalsium karbonat}}{\text{Mr }\ce{CaCO3}}\)
\(\text{Jumlah mol kalsium karbonat} = \dfrac{200 \text{ gram}}{100 \text{ gram/mol}}\)
\(\text{Jumlah mol kalsium karbonat} = 2 \text{ mol}\)
Tentukan jumlah mol dari 29,6 gram magnesium nitrat.
Jawaban: 0,2 mol
Magnesium nitrat = \(\ce{Mg(NO3)2}\)
\(\text{Mr } \ce{Mg(NO3)2} = \text{Ar } \ce{Mg} + 2\text{Ar } \ce{N} + 6\text{Ar } \ce{O}\)
\(\text{Mr } \ce{Mg(NO3)2} = 24 + 2(14) + 6(16)\)
\(\text{Mr } \ce{Mg(NO3)2} = 24 + 28 + 96\)
\(\text{Mr } \ce{Mg(NO3)2} = 148\)
\(\text{Jumlah mol magnesium nitrat} = \dfrac{\text{massa magnesium nitrat}}{\text{Mr }\ce{Mg(NO3)2}}\)
\(\text{Jumlah mol kalsium karbonat} = \dfrac{29,6 \text{ gram}}{148 \text{ gram/mol}}\)
\(\text{Jumlah mol kalsium karbonat} = \dfrac{296 }{1480}\text{ mol}\)
\(\text{Jumlah mol kalsium karbonat} = \dfrac{1 }{5}\text{ mol}\)
Tentukan massa dari 0,30 mol \(\ce{Mg(NO3)2}\).
\(\text{Ar } \ce{Mg} = 24, \text{Ar } \ce{N} = 14, \text{Ar } \ce{O} = 16\).
Jawaban: 44,4 gram
Magnesium nitrat = \(\ce{Mg(NO3)2}\)
\(\text{Mr } \ce{Mg(NO3)2} = \text{Ar } \ce{Mg} + 2\text{Ar } \ce{N} + 6\text{Ar } \ce{O}\)
\(\text{Mr } \ce{Mg(NO3)2} = 24 + 2(14) + 6(16)\)
\(\text{Mr } \ce{Mg(NO3)2} = 24 + 28 + 96\)
\(\text{Mr } \ce{Mg(NO3)2} = 148\)
\(\text{Jumlah mol magnesium nitrat} = \dfrac{\text{massa magnesium nitrat}}{\text{Mr }\ce{Mg(NO3)2}}\)
\(0,30 \text{ mol} = \dfrac{\text{massa magnesium nitrat}}{148 \text{ gram/mol}}\)
\(0,30 \text{ mol} \times 148 \text{ gram/mol} = \text{massa magnesium nitrat}\)
\(44,4 \text{ gram} = \text{massa magnesium nitrat}\)
Jadi, massa \(\ce{Mg(NO3)2}\) adalah 44,4 gram
\(\text{Urea} = \ce{CO(NH2)2}\)
\(\text{Mr } \ce{CO(NH2)2} = \text{Ar } \ce{C} + \text{Ar } \ce{O} + 2\text{Ar } \ce{N} + 4\text{Ar } \ce{H}\)
\(\text{Mr } \ce{CO(NH2)2} = 12 + 16 + 2(14) + 4(1)\)
\(\text{Mr } \ce{CO(NH2)2} = 60\)
(A) Menghitung jumlah mol senyawa urea
\(\text{Jumlah mol urea} = \dfrac{\text{massa urea}}{\text{Mr }\ce{CO(NH2)2}}\)
\(\text{Jumlah mol urea} = \dfrac{120 \text{ gram}}{60 \text{ gram/mol}}\)
\(\text{Jumlah mol urea} = 2\text{ mol}\)
(B) Menghitung jumlah mol C
\(\ce{CO(NH2)2} \rightarrow \ce{C} + \ce{O} + 2\ce{N} + 4\ce{H}\)
\(1 \text{ molekul } \ce{CO(NH2)2}\rightarrow 1\text { atom }\ce{C}\)
\(2 \text{ mol } \ce{CO(NH2)2}\rightarrow 2\text { mol }\ce{C}\)
\(\text{Jumlah mol }\ce{C} = 2 \text{ mol}\)
(C) Menghitung jumlah mol O
\(\ce{CO(NH2)2} \rightarrow \ce{C} + \ce{O} + 2\ce{N} + 4\ce{H}\)
\(1 \text{ molekul } \ce{CO(NH2)2}\rightarrow 1\text { atom }\ce{O}\)
\(2 \text{ mol } \ce{CO(NH2)2}\rightarrow 2\text { mol }\ce{O}\)
\(\text{Jumlah mol }\ce{O} = 2 \text{ mol}\)
(D) Menghitung jumlah mol N
\(\ce{CO(NH2)2} \rightarrow \ce{C} + \ce{O} + 2\ce{N} + 4\ce{H}\)
\(1 \text{ molekul } \ce{CO(NH2)2}\rightarrow 2\text { atom }\ce{N}\)
\(2 \text{ mol } \ce{CO(NH2)2}\rightarrow 4\text { mol }\ce{N}\)
\(\text{Jumlah mol }\ce{N} = 4 \text{ mol}\)
(E) Menghitung jumlah mol H
\(\ce{CO(NH2)2} \rightarrow \ce{C} + \ce{O} + 2\ce{N} + 4\ce{H}\)
\(1 \text{ molekul } \ce{CO(NH2)2}\rightarrow 4\text { atom }\ce{H}\)
\(2 \text{ mol } \ce{CO(NH2)2}\rightarrow 8\text { mol }\ce{H}\)
\(\text{Jumlah mol }\ce{H} = 8 \text{ mol}\)
