Question

Which of the following has the greatest number of molecules?
A. 44g $C{{O}_{2}}$
B. 48g ${{O}_{3}}$
C. 8g ${{H}_{2}}$
D. 64g $S{{O}_{2}}$

Answer 1

Think about how to calculate the number of moles of any substance and how many molecules one mole of any substance contains. That will give you an idea about how many molecules are present in the given substance.

Complete answer:
- Before we solve this question, let us first remember what the molar mass of a substance really is. The molar mass of any chemical substance is its mass that holds an Avogadro number of its molecules. For example, molar Mass for ${{O}_{2}}$ is 32 grams so 32 grams of ${{O}_{2}}$ will contain $6.022\times {{10}^{23}}$ i.e. the Avogadro’s number of molecules
- Now, let us analyse all the given molecules and evaluate their molar masses from the atomic masses of the atoms that make the molecule:
$\begin{aligned} & \text{Molar mass of }C{{O}_{2}}=(12\times 1)+(16\times 2) \\\ & \text{Molar mass of }C{{O}_{2}}=44g/mol \\\ \end{aligned}$
$\begin{aligned} & \text{Molar mass of }{{O}_{3}}=(16\times 3) \\\ & \text{Molar mass of }{{O}_{3}}=48g/mol \\\ \end{aligned}$
$\begin{aligned} & \text{Molar mass of }{{H}_{2}}=(2\times 1) \\\ & \text{Molar mass of }{{H}_{2}}=2g/mol \\\ \end{aligned}$
$\begin{aligned} & \text{Molar mass of }S{{O}_{2}}=(32\times 1)+(16\times 2) \\\ & \text{Molar mass of }S{{O}_{2}}=64g/mol \\\ \end{aligned}$
We know that one mole of any substance contains $6.022\times {{10}^{23}}$ molecules irrespective of the nature of the substance. Thus, the substance with the greatest number of moles will have the most molecules. The formula used will be:
$\text{Number of moles = }\frac{\text{given mass}}{\text{molar mass}}$
- For $C{{O}_{2}}$

& \text{Number of moles = }\frac{44g}{44g/mol} \\\ & \text{Number of moles = 1}mol \\\ \end{aligned}$$ \- For ${{O}_{2}}$ $$\begin{aligned} & \text{Number of moles}=\frac{48g}{48g/mol} \\\ & \text{Number of moles}=1mol \\\ \end{aligned}$$ \- For ${{H}_{2}}$ $$\begin{aligned} & \text{Number of moles}=\frac{8g}{2g/mol} \\\ & \text{Number of moles}=4mol \\\ \end{aligned}$$ \- For $S{{O}_{2}}$ $$\begin{aligned} & \text{Number of moles}=\frac{64g}{64g/mol} \\\ & \text{Number of moles}=1mol \\\ \end{aligned}$$ \- We see that only ${{H}_{2}}$ has 4mol of gas present and it will have the highest number of molecules. **Therefore, the answer to this question is ‘C. 8g ${{H}_{2}}$’** **Note:** Be careful while calculating the molar mass and do not confuse it with the atomic mass. Do not get confused between the number of moles that are present and the number of molecules that will be present. You have to multiply the number of moles with the Avogadro’s number to get the number of molecules.