0.2 Atomic masses and molecular formulas  (Page 9/10)

Since this seems like a useful number of atoms, we will give it a name. N is called a mole of atoms. We don’t need to know what N is to know that we can find a mole of atoms simply by finding the mass of a sample: 12.01 g of carbon, 1.008 g of hydrogen, 15.99 g of oxygen, and so on. (For historical reasons, the value of N which is a mole of atoms is called “Avogadro’s number,” in his honor but not because he discovered the number. Avogadro’s number is given the symbol N _A . The number of particles in a mole is approximately 6.022×10 ²³ , although we will almost never need this number when doing chemical calculations.)

Since we know the mass of one mole of a substance, we can find the number of moles in a sample of that substance just by finding the mass. Consider a sample of carbon with mass 24.02 g. This is twice the mass of one mole, so it must contain twice the number of particles as one mole. This must be two moles of particles. That example was easy, but what if we have 30.02 g of carbon? Since one mole has mass 12.01 g, then 30.02 g must contain 30.02/12.01 moles = 2.5 moles. Even more generally, then, if we have a sample of an element has mass m and the atomic mass of the element is M, the number of moles of atoms, n , is

$n\mathrm{=}\frac{m}{M}$

Since one mole contains a fixed number of particles, regardless of the type of particle, calculating the number of moles n is a way of counting the number of particles in a sample with mass m . For example, in the 100.0 g sample of the compound above, we have 40.0 g of carbon, 53.3 g of oxygen, and 6.7 g of hydrogen. We can calculate the number of moles of atoms of each element using the equation above:

$n_C\mathrm{=}\frac{\text{40}\text{.}\mathrm{0g}}{\text{12}\text{.}\mathrm{0g}\mathrm{/}\text{mol}}\mathrm{=}3\text{.}\text{33}\text{moles}$

$n_O\mathrm{=}\frac{\text{53}\text{.}\mathrm{3g}}{\text{16}\text{.}\mathrm{0g}\mathrm{/}\text{mol}}\mathrm{=}3\text{.}\text{33}\text{moles}$

$n_H\mathrm{=}\frac{6\text{.}\mathrm{7g}}{1\text{.}\mathrm{0g}\mathrm{/}\text{mol}}\mathrm{=}6\text{.}\text{67}\text{moles}$

A mole is a fixed number of particles. Therefore, the ratio of the numbers of moles is also the same as the ratio of the numbers of atoms. In the data above, this means that the ratio of the number of moles of carbon, oxygen, and hydrogen is 1:1:2, and therefore the ratio of the three types of atoms in the compound is also 1:1:2. This suggests that the compound has molecular formula COH ₂ .

However, this is just the ratio of the atoms of each type, and does not give the number of atoms of each type. Thus the molecular formula could just as easily be C ₂ O ₂ H ₄ or C ₃ O ₃ H ₆ . Since the formula COH ₂ is based on empirical mass ratio data, we refer to this as the empirical formula of the compound. To determine the molecular formula, we need to determine the relative mass of a molecule of the compound, i.e. the molecular mass. One way to do so is based on the Law of Combining Volumes, Avogadro’s Hypothesis, and the Ideal Gas Law. To illustrate, however, if we were to find that the relative mass of one molecule of the compound is 60.0, we could conclude that the molecular formula is C ₂ O ₂ H ₄ .

Counting the relative number of particles in a sample of a substance by measuring the mass and calculating the number of moles allows us to do “chemical algebra,” calculations of the masses of materials that react and are produced during chemical reactions.