Relating Moles with Other Quantities

This is chemistry, by the way.

Moles and number of particles

\(moles=\frac{number\ of\ particles}{6.02\times10^{23}}\)

Moles and mass

\(moles=\frac{mass}{A_r\ or\ M_r}\)

Moles and the volume of a gas (at R.T.P. only!)

Assuming \(dm^3\) as a unit,

\(moles=\frac{volume\ of\ gas}{24}\)

If you're dealing with \(cm^3\),

\(moles=\frac{volume\ of\ gas}{24000}\)

The difference between cubic centimetres and decimetres is 1000. Decimetres > centimetres, so multiply and divide accordingly.

Moles and concentration (solutions)

If you're dealing with moles,

\(concentration=\frac{moles}{volume\ of\ solution}\). The unit is \(mol\ dm^{-3}\).

If you're dealing with mass,

\(concentration=\frac{mass}{volume\ of\ solution}\). The unit is \(g\ dm^{-3}\).

Adjust the \(dm^3\) or \(cm^3\) as needed.

Percentage yield

\(percentage\ yield=\frac{actual\ yield}{experimental/theoretical\ yield}\times100\)

Percentage purity

\(percentage\ purity=\frac{mass/moles\ of\ pure\ substance}{mass/moles\ of\ impure\ substance}\times100\)

Note: Room temperature and pressure is 25 degrees celcius and 1 atm. 1 mol of gas will occupy \(24dm^3\) in such conditions.

This article was written on 13/09/2023. If you have any thoughts, feel free to send me an email with them. Have a nice day!