# 0.9 Quantitative aspects of chemical change  (Page 3/10)

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 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 ( $\mathrm{g}·\mathrm{mol}{}^{-1}$ ).

The equation can also be used to calculate mass and molar mass, using the following equations:

$m=n×M$

and

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

The following diagram may help to remember the relationship between these three variables. You need to imagine that the horizontal line is like a 'division' sign and that the vertical line is like a 'multiplication' sign. So, for example, if you want to calculate 'M', then the remaining two letters in the triangle are 'm' and 'n' and 'm' is above 'n' with a division sign between them. In your calculation then, 'm' will be the numerator and 'n' will be the denominator.

Calculate the number of moles of copper there are in a sample that weighs $127\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .

1. $n=\frac{m}{M}$
2. $n=\frac{127}{63,55}=2$

There are 2 moles of copper in the sample.

You are given a 5 mol sample of sodium. What mass of sodium is in the sample?

1. $m=n×M$
2. M ${}_{\mathrm{Na}}=22,99\phantom{\rule{2pt}{0ex}}\mathrm{g}·\mathrm{mol}{}^{-1}$

Therefore,

$m=5×22,99=114,95\phantom{\rule{2pt}{0ex}}\mathrm{g}$

The sample of sodium has a mass of $114,95\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .