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Element | Relative atomic mass (u) | Sample mass (g) | Number of moles in the sample |
Hydrogen | 1.01 | 1.01 | |
Magnesium | 24.31 | 24.31 | |
Carbon | 12.01 | 24.02 | |
Chlorine | 35.45 | 70.9 | |
Nitrogen | 42.08 |
Molar mass (M) is the mass of 1 mole of a chemical substance. The unit for molar mass is grams per mole or $\mathrm{g}\xb7\mathrm{mol}{}^{-1}$ .
You will remember that when the mass, in grams, of an element is equal to its relative atomic mass, the sample contains one mole of that element. This mass is called the molar mass of that element.
You may sometimes see the molar mass written as ${M}_{m}$ . We will use $M$ in this book, but you should be aware of the alternate notation.
It is worth remembering the following: On the periodic table, the relative atomic mass that is shown can be interpreted in two ways.
Element | Relative atomic mass (u) | Molar mass ( $\mathrm{g}\xb7\mathrm{mol}{}^{-1}$ ) | Mass of one mole of the element (g) |
Magnesium | 24,31 | 24,31 | 24,31 |
Lithium | 6,94 | 6,94 | 6,94 |
Oxygen | 16 | 16 | 16 |
Nitrogen | 14,01 | 14,01 | 14,01 |
Iron | 55,85 | 55,85 | 55,85 |
Calculate the number of moles of iron ( $\mathrm{Fe}$ ) in a $11,7\phantom{\rule{2pt}{0ex}}\mathrm{g}$ sample.
If we look at the periodic table, we see that the molar mass of iron is $55,85\phantom{\rule{2pt}{0ex}}\mathrm{g}\xb7\mathrm{mol}{}^{-1}$ . This means that 1 mole of iron will have a mass of $55,85\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .
If 1 mole of iron has a mass of $55,85\phantom{\rule{2pt}{0ex}}\mathrm{g}$ , then: the number of moles of iron in $111,7\phantom{\rule{2pt}{0ex}}\mathrm{g}$ must be:
There are 2 moles of iron in the sample.
You have a sample that contains 5 moles of zinc.
Molar mass of zinc is $65,38\phantom{\rule{2pt}{0ex}}\mathrm{g}\xb7\mathrm{mol}{}^{-1}$ , meaning that 1 mole of zinc has a mass of $65,38\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .
If 1 mole of zinc has a mass of $\mathrm{65,38}\phantom{\rule{2pt}{0ex}}\mathrm{g}$ , then 5 moles of zinc has a mass of: $\mathrm{65,38}\phantom{\rule{2pt}{0ex}}\mathrm{g}\times 5\phantom{\rule{2pt}{0ex}}\mathrm{mol}=\mathrm{326,9}\phantom{\rule{2pt}{0ex}}\mathrm{g}$ (answer to a)
(answer to b)
The calculations that have been used so far, can be made much simpler by using the following equation:
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