Moles and molar mass  (Page 3/3)

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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 $114,95\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .

Calculate the number of atoms there are in a sample of aluminium that weighs $80,94\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .

1. $n=\frac{m}{M}=\frac{80,94}{26,98}=3\phantom{\rule{2pt}{0ex}}\mathrm{moles}$
2. Number of atoms in 3 mol aluminium $=3×6,022×10{}^{23}$

There are $18,069×10{}^{23}$ aluminium atoms in a sample of $80,94\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .

Some simple calculations

1. Calculate the number of moles in each of the following samples:
1. $5,6\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of calcium
2. $0,02\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of manganese
3. $40\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of aluminium
2. A lead sinker has a mass of $5\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .
1. Calculate the number of moles of lead the sinker contains.
2. How many lead atoms are in the sinker?
3. Calculate the mass of each of the following samples:
1. $2,5\phantom{\rule{2pt}{0ex}}\mathrm{mol}$ magnesium
2. $12\phantom{\rule{2pt}{0ex}}\mathrm{mol}$ lithium
3. $4,5×10{}^{25}$ atoms of silicon

Molecules and compounds

So far, we have only discussed moles, mass and molar mass in relation to elements . But what happens if we are dealing with a molecule or some other chemical compound? Do the same concepts and rules apply? The answer is 'yes'. However, you need to remember that all your calculations will apply to the whole molecule . So, when you calculate the molar mass of a molecule, you will need to add the molar mass of each atom in that compound. Also, the number of moles will also apply to the whole molecule. For example, if you have one mole of nitric acid ( $\mathrm{HNO}{}_{3}$ ), it means you have $6,022×{10}^{23}$ molecules of nitric acid in the sample. This also means that there are $6,022×{10}^{23}$ atoms of hydrogen, $6,022×{10}^{23}$ atoms of nitrogen and ( $3×6,022×{10}^{23}$ ) atoms of oxygen in the sample.

In a balanced chemical equation, the number that is written in front of the element or compound, shows the mole ratio in which the reactants combine to form a product. If there are no numbers in front of the element symbol, this means the number is '1'.

e.g. ${\mathrm{N}}_{2}+3{\mathrm{H}}_{2}\to 2\mathrm{N}{\mathrm{H}}_{3}$

In this reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

Calculate the molar mass of $\mathrm{H}{}_{2}\mathrm{SO}{}_{4}$ .

1. Hydrogen $=1,008\phantom{\rule{2pt}{0ex}}\mathrm{g}·\mathrm{mol}{}^{-1}$ ; Sulphur = $=32,07\phantom{\rule{2pt}{0ex}}\mathrm{g}·\mathrm{mol}{}^{-1}$ ; Oxygen $=16\phantom{\rule{2pt}{0ex}}\mathrm{g}·\mathrm{mol}{}^{-1}$

2. ${M}_{\left({H}_{2}S{O}_{4}\right)}=\left(2×1,008\right)+\left(32,07\right)+\left(4×16\right)=98,09\phantom{\rule{2pt}{0ex}}\mathrm{g}·{\mathrm{mol}}^{-1}$

Calculate the number of moles there are in $1\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ of $\mathrm{MgCl}{}_{2}$ .

1. $n=\frac{m}{M}$
1. Convert mass into grams
$m=1\mathrm{kg}×1 000=1 000\phantom{\rule{2pt}{0ex}}\mathrm{g}$
2. Calculate the molar mass of $\mathrm{MgCl}{}_{2}$ .
${M}_{\left({\mathrm{MgCl}}_{2}\right)}=24,31+\left(2×35,45\right)=95,21\phantom{\rule{2pt}{0ex}}\mathrm{g}·{\mathrm{mol}}^{-1}$
2. $n=\frac{1000}{95,21}=10,5\phantom{\rule{2pt}{0ex}}\mathrm{mol}$

There are $10,5\phantom{\rule{2pt}{0ex}}\mathrm{moles}$ of magnesium chloride in a $1\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ sample.

Barium chloride and sulphuric acid react according to the following equation to produce barium sulphate and hydrochloric acid.

${\mathrm{BaCl}}_{2}+{\mathrm{H}}_{2}{\mathrm{SO}}_{4}\to {\mathrm{BaSO}}_{4}+2\mathrm{HCl}$

If you have $2\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of $\mathrm{BaCl}{}_{2}$ ...

1. What quantity (in g) of $\mathrm{H}{}_{2}\mathrm{SO}{}_{4}$ will you need for the reaction so that all the barium chloride is used up?
2. What mass of $\mathrm{HCl}$ is produced during the reaction?
1. $n=\frac{m}{M}=\frac{2}{208,24}=0,0096\phantom{\rule{2pt}{0ex}}\mathrm{mol}$
2. According to the balanced equation, 1 mole of $\mathrm{BaCl}{}_{2}$ will react with 1 mole of $\mathrm{H}{}_{2}\mathrm{SO}{}_{4}$ . Therefore, if $0,0096\phantom{\rule{2pt}{0ex}}\mathrm{mol}$ of $\mathrm{BaCl}{}_{2}$ react, then there must be the same number of moles of $\mathrm{H}{}_{2}\mathrm{SO}{}_{4}$ that react because their mole ratio is 1:1.

3. $m=n×M=0,0096×98,086=0,94\phantom{\rule{2pt}{0ex}}\mathrm{g}$

4. According to the balanced equation, 2 moles of $\mathrm{HCl}$ are produced for every 1 mole of the two reactants. Therefore the number of moles of $\mathrm{HCl}$ produced is ( $2×0,0096$ ), which equals $0,0096\phantom{\rule{2pt}{0ex}}\mathrm{moles}$ .

5. $m=n×M=0,0192×35,73=0,69\phantom{\rule{2pt}{0ex}}\mathrm{g}$

Group work : understanding moles, molecules and avogadro's number

Divide into groups of three and spend about 20 minutes answering the following questions together:

1. What are the units of the mole? Hint: Check the definition of the mole.
2. You have a $56\phantom{\rule{2pt}{0ex}}\mathrm{g}$ sample of iron sulphide ( $\mathrm{FeS}$ )
1. How many moles of $\mathrm{FeS}$ are there in the sample?
2. How many molecules of $\mathrm{FeS}$ are there in the sample?
3. What is the difference between a mole and a molecule?
3. The exact size of Avogadro's number is sometimes difficult to imagine.
1. Write down Avogadro's number without using scientific notation.
2. How long would it take to count to Avogadro's number? You can assume that you can count two numbers in each second.

1. Calculate the molar mass of the following chemical compounds:
1. $\mathrm{KOH}$
2. $\mathrm{FeCl}{}_{3}$
3. ${\mathrm{Mg\left(OH\right)}}_{2}$
2. How many moles are present in:
1. $10\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of $\mathrm{Na}{}_{2}$ SO ${}_{4}$
2. $34\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of $\mathrm{Ca\left(OH\right)}{}_{2}$
3. $2,45×10{}^{23}$ molecules of $\mathrm{CH}{}_{4}$ ?
3. For a sample of $0,2\phantom{\rule{2pt}{0ex}}\mathrm{moles}$ of potassium bromide ( $\mathrm{KBr}$ ), calculate...
1. the number of moles of ${\mathrm{K}}^{+}$ ions
2. the number of moles of ${\mathrm{Br}}^{-}$ ions
4. You have a sample containing $3\phantom{\rule{2pt}{0ex}}\mathrm{moles}$ of calcium chloride.
1. What is the chemical formula of calcium chloride?
2. How many calcium atoms are in the sample?
5. Calculate the mass of:
1. $3\phantom{\rule{2pt}{0ex}}\mathrm{moles}$ of $\mathrm{NH}{}_{4}\mathrm{OH}$
2. $4,2\phantom{\rule{2pt}{0ex}}\mathrm{moles}$ of $\mathrm{Ca}\left(\mathrm{NO}{}_{3}\right){}_{2}$
6. $96,2\phantom{\rule{2pt}{0ex}}\mathrm{g}$ sulphur reacts with an unknown quantity of zinc according to the following equation: $\mathrm{Zn}+\mathrm{S}\to \mathrm{ZnS}$
1. What mass of zinc will you need for the reaction, if all the sulphur is to be used up?
2. What mass of zinc sulphide will this reaction produce?
7. Calcium chloride reacts with carbonic acid to produce calcium carbonate and hydrochloric acid according to the following equation: ${\mathrm{CaCl}}_{2}+{\mathrm{H}}_{2}{\mathrm{CO}}_{3}\to {\mathrm{CaCO}}_{3}+2\mathrm{HCl}$ If you want to produce $10\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of calcium carbonate through this chemical reaction, what quantity (in g) of calcium chloride will you need at the start of the reaction?

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