# Number of I Atoms in the Iodine RDA

As we saw in How Many Atoms in the RDA of Iodine, the RDA for iodine is only 150 µg per day, or 0.000150 g per day. That seems like a miniscule amount, but it represents 7 x 1017, or 700 000 000 000 000 000 iodine atoms (about 10 million million atoms in each cm3 of our body). Our requirements for other nutrients are even smaller; for example, the RDA for vitamine B12 is only 2.4 µg/day, which represents about 1 x 1015 cobalt atoms. See the future section The Avogadro Constant for sample calculations.

Although chemists usually work with moles as units, occasionally it is helpful to refer to the actual number of atoms or molecules involved. When this is done, the symbol N is used. For example, in referring to 1 mol of iodine atoms, we could write

$$\text{n}_I = \text{1 mol} \text{ and } \text{N}_I = \text{6.022} \times \text{10}^{23}$$

Notice that NI is a pure number, rather than a quantity. To obtain such a pure number, we need a conversion factor which involves the number of particles per unit amount of substance. The appropriate factor is given the symbol NA and is called the Avogadro constant. It is defined by the equation

$\text{N}_A = \dfrac{\text{N}}{n}$

Since for any substance there are 6.022 × 1023 particles per mole, NA =6.022 × 1023/1 mol = 6.022 × 1023 mol–1.

Example $$\PageIndex{1}$$: Number of Molecules

Calculate the number of CuI molecules in 9.06 x 10-5 mol CuI.

Solution Rearranging the above equation, we obtain

$$\text{N} = \text{n} \times \text{N}_A = \text{1.18} \times \text{10}^{-6} \text{mol} \times \text{6.022} \times \text{10}^{23} \text{mol}^{-1} = \text{7.12} \times \text{10}^{17}$$

Alternatively, we might include the identity of the particles involved:

$$\text{N} = \text{1.18} \times \text{10}^{-6} \text{mol CuI} \times \dfrac{\text{6.022} \times \text{10}^{23} \text{CuI molecules}}{\text{1 mol CuI}} = \text{7.12} \times \text{10}^{17} \text{CuI molecules}$$

Notice that the above equation, which defines the Avogadro constant, has the same form as the equation which defined density. The preceding example used the Avogadro constant as a conversion factor in the same way that density was used. As in previous examples, all that is necessary is to remember that number of particles and amount of substance are related by a conversion factor, the Avogadro constant.

$\large\text{Number of particles } \large\overset{\text{Avogadro constant}}{\longleftrightarrow} \large\text{amount of substance} \\ \quad \\ \large N \large\overset{\text{N}_{\text{A}}}{\longleftrightarrow} { \space} \large n\label{2}$

As long as the units mole cancel, NA is being used correctly.

Example $$\PageIndex{2}$$: Number of Atoms

Calculate the number of I atoms in 9.06 x 10-5 mol CuI2.

Solution Rearranging the above equation, we obtain

$$\text{N} = \text{n} \times \text{N}_A = \text{1.18} \times \text{10}^{-6} \text{mol} \times \text{6.022} \times \text{10}^{23} \text{mol}^{-1} = \text{7.12} \times \text{10}^{17}$$

Including the identity of the particles involved avoids a possible mistake, because there are 1 I atoms in every CuI2, or 2 mol I atoms in every 2 mol CuI2:

$$\text{N} = \text{1.18} \times \text{10}^{-6} \text{mol CuI} \times \dfrac{\text{6.022} \times \text{10}^{23} \text{CuI molecules}}{\text{1 mol CuI}} = \text{7.12} \times \text{10}^{17} \text{CuI molecules}$$

$$\text{N}_{I_2} = \text{N}_{CuI_2} \times \dfrac{\text{2 mol I}}{\text{1 mol CuI}_2} = \text{7.12} \times \text{10}^{17} \text{CuI}_2 \text{molecules} \times \dfrac{\text{2 mol I}}{\text{1 mol CuI}_2} = \text{1.42} \times \text{10}^{18} \text{I atoms}$$

Note again, that a 100% mistake could be made by not specifying mol CuI2 or mol CuI!

From ChemPRIME: 2.9: The Avogadro Constant