From the list of compounds below identify the molecules (find out the no. of molecules):
1) Propanone C3H6O Mr:58
2) Chromium Trioxide Cr2O3 Mr: 152
3) Rubidium Carbonate Rb2CO3 Mr: 231
4) Neon Ne Mr:20.2
5) Nitrophenol C6H5NO3 Mr: 139
6) Dinitrogentetraoxide N2O4 Mr:92
Please show every single bit of working out including full equations that aren’t abbreviated. Please help.
To find the number of molecules in each compound, we need to use Avogadro's number and the molar mass of each compound.
Avogadro's number (NA) is approximately 6.022 × 10^23. It represents the number of atoms or molecules in one mole of a substance.
The relationship between moles (n), number of particles (N), and Avogadro's number is given by the equation:
N = n * NA
To find the number of molecules (N), we can rearrange the equation to solve for n:
n = N / NA
Let's calculate the number of molecules for each compound:
1) Propanone (C3H6O)
Molar mass of propanone (Mr) = 58 g/mol
To find the number of moles (n) of propanone, we divide the molar mass by the mass of the compound:
n = 58 g / 58 g/mol = 1 mole
Using Avogadro's number (NA), we can find the number of molecules:
N = n * NA = 1 mole * 6.022 × 10^23 molecules/mole = 6.022 × 10^23 molecules
Therefore, there are 6.022 × 10^23 molecules in one mole of propanone.
2) Chromium Trioxide (Cr2O3)
Molar mass of chromium trioxide (Mr) = 152 g/mol
n = 152 g / 152 g/mol = 1 mole
N = n * NA = 1 mole * 6.022 × 10^23 molecules/mole = 6.022 × 10^23 molecules
There are 6.022 × 10^23 molecules in one mole of chromium trioxide.
3) Rubidium Carbonate (Rb2CO3)
Molar mass of rubidium carbonate (Mr) = 231 g/mol
n = 231 g / 231 g/mol = 1 mole
N = n * NA = 1 mole * 6.022 × 10^23 molecules/mole = 6.022 × 10^23 molecules
There are 6.022 × 10^23 molecules in one mole of rubidium carbonate.
4) Neon (Ne)
Molar mass of neon (Mr) = 20.2 g/mol
n = 20.2 g / 20.2 g/mol = 1 mole
N = n * NA = 1 mole * 6.022 × 10^23 molecules/mole = 6.022 × 10^23 molecules
There are 6.022 × 10^23 molecules in one mole of neon.
5) Nitrophenol (C6H5NO3)
Molar mass of nitrophenol (Mr) = 139 g/mol
n = 139 g / 139 g/mol = 1 mole
N = n * NA = 1 mole * 6.022 × 10^23 molecules/mole = 6.022 × 10^23 molecules
There are 6.022 × 10^23 molecules in one mole of nitrophenol.
6) Dinitrogen Tetraoxide (N2O4)
Molar mass of dinitrogen tetraoxide (Mr) = 92 g/mol
n = 92 g / 92 g/mol = 1 mole
N = n * NA = 1 mole * 6.022 × 10^23 molecules/mole = 6.022 × 10^23 molecules
There are 6.022 × 10^23 molecules in one mole of dinitrogen tetraoxide.