What is the total number of atoms present in 25.0 mg of camphor, `C_(10)H_(16)O`?

Options

A. `9.89xx10^(19)`
B. `9.89xx10^(20)`
C. `6.02xx10^(20)`
D. `2.67xx10^(21)`

Answer

To find the total number of atoms present in 25.0 mg of camphor (C₁₀H₁₆O), we can follow these steps:

### Step 1: Calculate the molar mass of camphor (C₁₀H₁₆O)
- Carbon (C): 12 g/mol
- Hydrogen (H): 1 g/mol
- Oxygen (O): 16 g/mol

The molar mass of camphor is calculated as follows:
\[ \text{Molar mass} = (10 \times 12) + (16 \times 1) + (1 \times 16) = 120 + 16 + 16 = 152 \text{ g/mol} \]

### Step 2: Convert the mass of camphor from mg to g
Given mass = 25.0 mg
\[ \text{Mass in grams} = 25.0 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}} = 0.025 \text{ g} \]

### Step 3: Calculate the number of moles of camphor
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
\[ \text{Number of moles} = \frac{0.025 \text{ g}}{152 \text{ g/mol}} \approx 0.0001645 \text{ moles} \]

### Step 4: Calculate the total number of molecules of camphor
Using Avogadro's number (6.022 × 10²³ molecules/mol):
\[ \text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number} \]
\[ \text{Number of molecules} = 0.0001645 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 9.89 \times 10^{19} \text{ molecules} \]

### Step 5: Calculate the total number of atoms in camphor
Each molecule of camphor (C₁₀H₁₆O) contains:
- 10 Carbon atoms
- 16 Hydrogen atoms
- 1 Oxygen atom

Total atoms per molecule = 10 + 16 + 1 = 27 atoms

Now, we calculate the total number of atoms:
\[ \text{Total number of atoms} = \text{Number of molecules} \times \text{Number of atoms per molecule} \]
\[ \text{Total number of atoms} = 9.89 \times 10^{19} \text{ molecules} \times 27 \text{ atoms/molecule} \approx 2.67 \times 10^{21} \text{ atoms} \]

### Final Answer
The total number of atoms present in 25.0 mg of camphor is approximately \(2.67 \times 10^{21}\) atoms.