What is the total no. Of atoms present in 25.0 mg of camphor
mols camphor = grams/molar mass = 25.0/152.24 = approx 0.15 but that's an estimate you should recalculate it.
The number of molecules is 6.02E23 molecules in 1 mol so in 0.15 mols there will be approx 0.15 x 6.02E23 = approx 1E21. The formula is C10H16O. There are 27 atoms in 1 molecule so 27 x 1E21 = approx ?.
Again, you should recalculate each step.
To find the total number of atoms present in 25.0 mg of camphor, we can follow these steps:
1. Determine the molar mass of camphor (C10H16O), which is the sum of the atomic masses of each element in the compound.
- Atomic mass of carbon (C) = 12.01 g/mol
- Atomic mass of hydrogen (H) = 1.01 g/mol
- Atomic mass of oxygen (O) = 16.00 g/mol
Multiply the atomic mass of each element by the number of atoms in one molecule of camphor:
- Carbon: 12.01 g/mol x 10 = 120.10 g/mol
- Hydrogen: 1.01 g/mol x 16 = 16.16 g/mol
- Oxygen: 16.00 g/mol x 1 = 16.00 g/mol
Add up the masses of all the elements to find the molar mass of camphor:
- Molar mass of camphor = 120.10 g/mol + 16.16 g/mol + 16.00 g/mol = 152.26 g/mol
2. Convert the given mass of camphor to moles using the molar mass.
- Moles of camphor = Mass of camphor / Molar mass of camphor
- Moles of camphor = 0.025 g / 152.26 g/mol = 0.000164 moles
3. Use Avogadro's number, which is 6.022 x 10^23 atoms/mol, to calculate the total number of atoms.
- Number of atoms = Moles of camphor x Avogadro's number
- Number of atoms = 0.000164 moles x 6.022 x 10^23 atoms/mol
Calculating this equation will give you the total number of atoms present in 25.0 mg of camphor.