Question

Which of the following would weigh the highest?

• A. $0.2$ mole of sucrose $({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11})$
• B. $2$ moles of ${\mathrm{CO}}_2$
• C. $2$ moles of ${\mathrm{CaCO}}_3$
• D. $10$ moles of ${\mathrm{H}}_2\mathrm{O}$

Answer 1

The correct option is C

$2$ moles of ${\mathrm{CaCO}}_3$

Explanation for correct option:

C. $2$ moles of ${\mathrm{CaCO}}_3$

Step 1: Molar mass:

• “The molar mass of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance”

Step 2: Calculation of molar mass of ${\mathrm{CaCO}}_3$:

$$\begin{gathered} \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{CaCO}}_3=\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{Ca}+\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{C}+\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{O} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{CaCO}}_3=40\mathrm{g}+12\mathrm{g}+(3\times 16)\mathrm{g} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{CaCO}}_3=100\mathrm{g} \end{gathered}$$

Step 3: Calculation of $2$ moles of ${\mathrm{CaCO}}_3$:

$2\mathrm{moles}\mathrm{of}{\mathrm{CaCO}}_3=2\times 100\mathrm{g}=200\mathrm{g}$

Explanation for incorrect options:

A. $0.2$ mole of sucrose $({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11})$

Step 1: Calculation of molar mass of $({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11})$:

$$\begin{gathered} \mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}=\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{C}+\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{H}+\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{O} \\ \mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}=(12\times 12)\mathrm{g}+(1\times 22)\mathrm{g}+(16\times 11)\mathrm{g} \\ \mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}=144+22+176\mathrm{g} \\ \mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}=342\mathrm{g} \end{gathered}$$:

Step 2: Calculation of $0.2$ moles of $({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11})$:

$0.2\mathrm{moles}\mathrm{of}{\mathrm{C}}_{11}{\mathrm{H}}_{22}{\mathrm{O}}_{11}=0.2\times 342\mathrm{g}=68.4\mathrm{g}$

B. $2$ moles of ${\mathrm{CO}}_2$

Step 1: Calculation of molar mass of ${\mathrm{CO}}_2$:

$$\begin{gathered} \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{CO}}_2=\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{C}+\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{O} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{CO}}_2=12\mathrm{g}+(2\times 16)\mathrm{g}=12+32\mathrm{g} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{CO}}_2=44\mathrm{g} \end{gathered}$$

Step 2: Calculation of $2$ moles of ${\mathrm{CO}}_2$:

$2\mathrm{moles}\mathrm{of}{\mathrm{CO}}_2=2\times 44\mathrm{g}=88\mathrm{g}$

D. $10$ moles of ${\mathrm{H}}_2\mathrm{O}$

Step 1: Calculation of molar mass of ${\mathrm{H}}_2\mathrm{O}$:

$$\begin{gathered} \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{H}}_2\mathrm{O}=\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{H}+\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{O} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{H}}_2\mathrm{O}=\left(1\times 2\right)\mathrm{g}+\left(16\right)\mathrm{g}=2+16\mathrm{g} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{H}}_2\mathrm{O}=18\mathrm{g} \end{gathered}$$

Step 2: Calculation of $10$ moles of ${\mathrm{H}}_2\mathrm{O}$:

$10\mathrm{moles}\mathrm{of}{\mathrm{H}}_2\mathrm{O}=10\times 18\mathrm{g}=180\mathrm{g}$

Hence, the correct option is C i.e. $2$ moles of ${\mathrm{CaCO}}_3$would weigh the highest.