# Moles and molar mass  (Page 2/3)

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## Moles and mass

1. Complete the following table:
 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
2. How many atoms are there in...
1. 1 mole of a substance
2. 2 moles of calcium
3. 5 moles of phosphorus
4. $24,31\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of magnesium
5. $24,02\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of carbon

## Molar mass

Molar mass

Molar mass (M) is the mass of 1 mole of a chemical substance. The unit for molar mass is grams per mole or $\mathrm{g}·\mathrm{mol}{}^{-1}$ .

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.

1. The mass of a single, average atom of that element relative to the mass of an atom of carbon.
2. The mass of one mole of the element . This second use is the molar mass of the element.
 Element Relative atomic mass (u) Molar mass ( $\mathrm{g}·\mathrm{mol}{}^{-1}$ ) 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.

1. If we look at the periodic table, we see that the molar mass of iron is $55,85\phantom{\rule{2pt}{0ex}}\mathrm{g}·\mathrm{mol}{}^{-1}$ . This means that 1 mole of iron will have a mass of $55,85\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .

2. 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:

$\frac{111,7\mathrm{g}}{55,85\mathrm{g}·{\mathrm{mol}}^{-1}}=2\phantom{\rule{2pt}{0ex}}\mathrm{mol}$

There are 2 moles of iron in the sample.

You have a sample that contains 5 moles of zinc.

1. What is the mass of the zinc in the sample?
2. How many atoms of zinc are in the sample?
1. Molar mass of zinc is $65,38\phantom{\rule{2pt}{0ex}}\mathrm{g}·\mathrm{mol}{}^{-1}$ , meaning that 1 mole of zinc has a mass of $65,38\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .

2. If 1 mole of zinc has a mass of $65,38\phantom{\rule{2pt}{0ex}}\mathrm{g}$ , then 5 moles of zinc has a mass of: $65,38\phantom{\rule{2pt}{0ex}}\mathrm{g}×5\phantom{\rule{2pt}{0ex}}\mathrm{mol}=326,9\phantom{\rule{2pt}{0ex}}\mathrm{g}$ (answer to a)

3. $5×6,022×{10}^{23}=30,115×{10}^{23}$

## Moles and molar mass

1. Give the molar mass of each of the following elements:
1. hydrogen
2. nitrogen
3. bromine
2. Calculate the number of moles in each of the following samples:
1. $21,62\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of boron ( $\mathrm{B}$ )
2. $54,94\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of manganese ( $\mathrm{Mn}$ )
3. $100,3\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of mercury ( $\mathrm{Hg}$ )
4. $50\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of barium ( $\mathrm{Ba}$ )
5. $40\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of lead ( $\mathrm{Pb}$ )

## An equation to calculate moles and mass in chemical reactions

The calculations that have been used so far, can be made much simpler by using the following equation:

$\mathbf{n}\left(\mathrm{number of moles}\right)=\frac{\mathbf{m}\left(\mathrm{mass of substance in}\phantom{\rule{2pt}{0ex}}\mathrm{g}\right)}{\mathbf{M}\left(\mathrm{molar mass of substance in}\phantom{\rule{2pt}{0ex}}\mathrm{g}·{\mathrm{mol}}^{-1}\right)}$
Remember that when you use the equation $n=\frac{m}{M}$ , the mass is always in grams ( $\mathrm{g}$ ) and molar mass is in grams per mol ( $\mathrm{g}·\mathrm{mol}{}^{-1}$ ).

how do you translate this in Algebraic Expressions
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