Hubungan Mol dengan Massa Zat

\(\text{Jumlah mol besi} = \dfrac{\text{massa besi}}{\text{Ar }\ce{Fe}}\) \(\text{Jumlah mol besi} = \dfrac{28 \text{ gram}}{56 \text{ gram/mol}}\) \(\text{Jumlah mol besi} = \dfrac{1}{2}\text{ mol}\)

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 \(\text{Jumlah mol gas nitrogen} = \dfrac{\text{massa nitrogen}}{\text{Mr }\ce{N2}}\) \(\text{Jumlah mol gas nitrogen} = \dfrac{7 \text{ gram}}{28 \text{ gram/mol}}\) \(\text{Jumlah mol gas nitrogen} = \dfrac{1}{4}\text{ mol}\)

\(\text{Mr } \ce{CaCO3} = \text{Ar } \ce{Ca} +  \text{Ar } \ce{C} + 3\text{Ar } \ce{O}\) \(\text{Mr } \ce{CaCO3} = 40 + 12 + 3(16)\) \(\text{Mr } \ce{CaCO3} = 100\) \(\text{Jumlah mol kalsium karbonat} = \dfrac{\text{massa kalsium karbonat}}{\text{Mr }\ce{CaCO3}}\) \(\text{Jumlah mol kalsium karbonat} = \dfrac{200 \text{ gram}}{100 \text{ gram/mol}}\) \(\text{Jumlah mol kalsium karbonat} = 2 \text{ mol}\)

\(\text{Mr } \ce{Mg(NO3)2} = \text{Ar } \ce{Mg} +  2\text{Ar } \ce{N} + 6\text{Ar } \ce{O}\) \(\text{Mr } \ce{Mg(NO3)2} = 24 + 2(14) + 6(16)\) \(\text{Mr } \ce{Mg(NO3)2} = 24 + 28 + 96\) \(\text{Mr } \ce{Mg(NO3)2} = 148\) \(\text{Jumlah mol magnesium nitrat} = \dfrac{\text{massa magnesium nitrat}}{\text{Mr }\ce{Mg(NO3)2}}\) \(\text{Jumlah mol kalsium karbonat} = \dfrac{29,6 \text{ gram}}{148 \text{ gram/mol}}\) \(\text{Jumlah mol kalsium karbonat} = \dfrac{296 }{1480}\text{ mol}\) \(\text{Jumlah mol kalsium karbonat} = \dfrac{1 }{5}\text{ mol}\)

\(\text{Mr } \ce{Mg(NO3)2} = \text{Ar } \ce{Mg} +  2\text{Ar } \ce{N} + 6\text{Ar } \ce{O}\) \(\text{Mr } \ce{Mg(NO3)2} = 24 + 2(14) + 6(16)\) \(\text{Mr } \ce{Mg(NO3)2} = 24 + 28 + 96\) \(\text{Mr } \ce{Mg(NO3)2} = 148\) \(\text{Jumlah mol magnesium nitrat} = \dfrac{\text{massa magnesium nitrat}}{\text{Mr }\ce{Mg(NO3)2}}\) \(0,30 \text{ mol} = \dfrac{\text{massa magnesium nitrat}}{148 \text{ gram/mol}}\) \(0,30 \text{ mol} \times 148 \text{ gram/mol} = \text{massa magnesium nitrat}\) \(44,4 \text{ gram} = \text{massa magnesium nitrat}\) Jadi, massa \(\ce{Mg(NO3)2}\) adalah 44,4 gram

\(\text{Mr } \ce{CO(NH2)2} = \text{Ar } \ce{C} +  \text{Ar } \ce{O} + 2\text{Ar } \ce{N} + 4\text{Ar } \ce{H}\) \(\text{Mr } \ce{CO(NH2)2} = 12 +  16 + 2(14) + 4(1)\) \(\text{Mr } \ce{CO(NH2)2} = 60\) (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 \(\ce{CO(NH2)2} \rightarrow \ce{C} + \ce{O} + 2\ce{N} + 4\ce{H}\) \(1 \text{ molekul } \ce{CO(NH2)2}\rightarrow 4\text { atom }\ce{H}\) \(2 \text{ mol } \ce{CO(NH2)2}\rightarrow 8\text { mol }\ce{H}\) \(\text{Jumlah mol }\ce{H} = 8 \text{ mol}\)