Question

3.4: the mole

1. consider the definition of an atom.

a. why do scientists use the mole?
b. how many particles are in a single mole?
c. this is called _______________ number and is abbreviated as na.

1. how many particles are in 1.25 moles?
2. how many atoms are in a 3.2 mol sample of chlorine?
3. what is the mass of 2.18 mol of cu?
4. how many moles are in 1.0 × 10⁸ atoms?

Explanation:

1. a.

Atoms are extremely small. Counting individual atoms is impractical. The mole provides a convenient way to deal with large numbers of atoms (or other particles). It is a unit that allows scientists to relate the microscopic world of atoms to the macroscopic world of measurable quantities in experiments.

1. b.

By definition, there are \(6.022\times 10^{23}\) particles in a single mole. This number is known as Avogadro's number.

1. c.

This is called Avogadro’s number and is abbreviated as \(N_A\).

2.

We use the formula \(N = n\times N_A\), where \(n = 1.25\space mol\) and \(N_A=6.022\times 10^{23}\space mol^{-1}\).

$$N=(1.25\space mol)\times(6.022\times 10^{23}\space mol^{-1}) = 7.5275\times 10^{23}$$

3.

Chlorine (\(Cl_2\)) is a di - atomic molecule. But if we assume the question is about chlorine atoms (if it was about \(Cl_2\) molecules, we would need to adjust). Using \(N = n\times N_A\), with \(n = 3.2\space mol\) and \(N_A = 6.022\times 10^{23}\space mol^{-1}\)

$$N=(3.2\space mol)\times(6.022\times 10^{23}\space mol^{-1})=1.92704\times 10^{24}$$

4.

The molar mass of \(Cu\) (\(M\)) is \(63.55\space g/mol\). Using the formula \(m=n\times M\), where \(n = 2.18\space mol\) and \(M = 63.55\space g/mol\)

$$m=(2.18\space mol)\times(63.55\space g/mol)=138.539\space g$$

5.

Using the formula \(n=\frac{N}{N_A}\), where \(N = 1.0\times 10^{8}\) and \(N_A=6.022\times 10^{23}\space mol^{-1}\)

$$n=\frac{1.0\times 10^{8}}{6.022\times 10^{23}\space mol^{-1}}\approx1.66\times 10^{-16}\space mol$$

Answer:

1. a. Atoms are too small to count individually; the mole helps deal with large numbers.
2. b. \(6.022\times 10^{23}\)
3. c. Avogadro’s
4. \(7.53\times 10^{23}\) particles
5. \(1.93\times 10^{24}\) atoms (assuming atomic chlorine)
6. \(138.54\space g\)
7. \(1.66\times 10^{-16}\space mol\)