Calculations Using Moles

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Calculations Using Moles Revision

Calculations Using Moles

The mole is one of the most, if not the most, important quantity in chemistry. One mole is equal to $6.022 \times 10^{23}$ of whatever it is that is being measured. To give a sense of just how massive a number this is, if you were to line up one mole of world cup standard footballs, the line would be around a thousand times longer than the width of the milky way.

What is a Mole?

The value of a mole is fixed, it does not change with the substance being discussed, i.e. one mole of iron, one mole of electrons, and one methane molecules both contain $6.022 \times 10^{23}$ particles. This number is known as the Avogadro constant and is typically give the symbols L or N_A.

For any given substance, the mass of one mole ( $6.022 \times 10^{23}$ particles) of a substance will be equal to the relative mass of said substance. This means that one mole of carbon, with a relative atomic mass of $12$ weighs exactly $12\text{ g}$ . One mole of methane molecules, with a relative formula mass of $16$ weighs exactly $16\text{ g}$ . This is particularly useful for chemists, as it means we can easily convert between the moles of substances and its mass in grams.

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Calculating the Number of Moles

Calculating how many moles of a substance we have is fairly straight forward. The mass in grams of a substance and the number of moles present are related by a simple formula:

$\text{Moles of Substance}=\frac{\text{Mass of Substance in grams}}{\text{Relative Mass of Substance (M}_r\text{)}}$

It is very common for chemists to talk about the number of moles involved in a reaction, instead of the masses or volumes taking place. This is because the ratios of moles involved in reactions will always remain constant. For example, in the reaction:

$\text{Mg(OH)}_2 + 2 \text{HCl} \rarr \text{MgCL}_2 + \text{H}_2\text{O}$

The ratio of moles of magnesium hydroxide (Mg(OH)₂) to moles of hydrochloric acid (HCl) will always be $1:2$ . Therefore, if we are told that $0.5$ mol of magnesium hydroxide is used in a reaction, we can deduce that $1$ mol of hydrochloric acid must have been used. This rule applies across the arrow as well. From the equation we know that the ratio of the moles magnesium hydroxide reacted to moles of magnesium chloride formed is $1:1$ .

If we know the number of moles of a substbstance then we can also calculate its mass. The above formula can be rearranged to give:

$\text{Mass of Substance} = \text{Moles of Substance} \times \text{Relative Mass of Substance}$

By combining this with the ratios rule above, we can use the number of moles at the start of a reaction to predict the mass of any individual products formed.

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Using Moles to Balance Equations

One of the useful things about moles is that we can use them to help us balance an equation. If we know how many grams of two substances react, and their molecular masses, we can calculate the ratio of moles in the reaction. Often, this ratio will not be simple whole numbers. However the numbers used in balancing equations must be whole numbers. As such, once we have calculated the ratio of moles in a reaction, we then either round up or down or multiply or divide the ratio to find the simplest whole number ratio.

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Example 1: Calculating the Number of Moles

A student weighs out a $\textcolor{#00bfa8}{3.24\text{ g}}$ sample of iron (III) oxide $\left(\text{M}_r = \textcolor{#f21cc2}{160}\right)$ for a reaction. Calculate the number of moles present in the sample:

[1 mark]

$\text{Number of}$ $\text{Moles Iron (III)}$ $\begin{aligned} \text{Oxide} &= \frac{\text{Mass of Iron (III) Oxide}}{\text{M}_r\text{ Iron(III) Oxide}}\\&= \frac{\textcolor{#00bfa8}{3.24}}{\textcolor{#f21cc2}{160}}\\ &= \textcolor{#008d65}{0.02 \text{ mol}}\end{aligned}$

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Example 2: Calculating Mass from Moles

A student is given a sample of propane $\left(\text{M}_r = \textcolor{#00bfa8}{44}\right)$ . The sample is known to contain $\textcolor{#f21cc2}{2.5 \text{ moles}}$ . Calculate the mass of the sample:

[1 mark]

$\text{Mass of Propane}$ $=$ $\text{Moles of Propane}$ $\times \text{M}_r\text{ of Propane}$ $\\ = \textcolor{#00bfa8}{2.5} \times \textcolor{#f21cc2}{44}\\ = \textcolor{#008d65}{110 \text{ g}}$

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Example 3: Calculating Masses from Equations

A reaction is carried out between potassium hydroxide $\left(\text{KOH, M}_r =\textcolor{#00bfa8}{56}\right)$ and sulfuric acid $\left(\text{H}_2\text{SO}_4\text{, M}_r =\textcolor{#f21cc2}{98}\right)$ . Below is the balanced equation for this reaction:

$2\text{KOH} + \text{H}_2\text{SO}_4 \rarr \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}$

$\textcolor{#327399}{5.98\text{ g}}$ of potassium hydroxide were used in the reaction. Predict the mass of potassium sulfate $(\text{K}_2\text{SO}_4\text{, M}_r =\textcolor{#a233ff}{174})$ produced:

[3 marks]

This is more involved than the above calculations. The thing to do with questions like these is to break them down into steps.

Step 1: Calculate the moles of $\text{KOH}$ used:

$\text{Moles KOH Used} = \frac{\text{Mass KOH Used}}{\text{M}_r \text{ KOH}}\\ =\frac{\textcolor{#00bfa8}{5.98}}{\textcolor{#10a6f3}{56}}\\ =\textcolor{#008d65}{0.107 \text{ mol}}$

Step 2: Calculate the moles of $\text{K}_2\text{SO}_4$ produced:

From the equation we know that $2$ moles of $\text{KOH}$ form $1$ mole of $\text{K}_2\text{SO}_4$ . Therefore, we can state that the ratio between $\text{KOH}$ and $\text{K}_2\text{SO}_4$ is $\text{2:1}$ .

To get the number of moles of $\text{K}_2\text{SO}_4$ we need to divide the number of moles of $\text{KOH}$ by $2$ :

$\text{Moles of K}_2\text{SO}_4 = \frac{\text{Moles of KOH}}{2}\\ =\frac{0.107}{2}\\ =\textcolor{#008d65}{0.054 \text{ mol}}$

Step 3: Use the moles of $\text{K}_2\text{SO}_4$ to calculate the mass produced:

$\text{Mass of K}_2\text{SO}_4 = \text{Moles of K}_2\text{SO}_4 \times \text{M}_r \text{ K}_2\text{SO}_4\\ = 0.054 \times \textcolor{#a233ff}{174}\\ = \textcolor{#008d65}{9.4 \text{ g}}$

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Example 4: Using Moles to Balance Equations

In a reaction between lithium chloride ( $\text{LiCl}$ , $\text{M}_r =\textcolor{#00bfa8}{42.5}$ ) and sulfuric acid ( $\text{H}_2\text{SO}_4$ , $\text{M}_r =\textcolor{#f21cc2}{98}$ ), $\textcolor{#327399}{7.8\text{ g}}$ of $\text{LiCl}$ was found to react with $\textcolor{#a233ff}{7.7\text{ g}}$ of $\text{H}_2\text{SO}_4$ . The equation of the reaction is given below:

$x\text{LiCl} + \text{H}_2\text{SO}_4 \rarr \text{Li}_2\text{SO}_4 + 2\text{HCl}$

Using the information given, deduce the value of $x$ to balance the equation:

[4 marks]

This may seem like a daunting task at first. The best way to tackle a question like this, like above, is to split it up into steps.

Step 1: Calculate the moles of $\text{LiCl}$ that have reacted.

$\text{Moles LiCl} = \frac{\textcolor{#327399}{7.8}}{\textcolor{#00bfa8}{42.5}} =\textcolor{#008d65}{0.18 \text{ mol}}$

Step 2: Calculate the moles of $\text{H}_2\text{SO}_4$ that have reacted.

$\text{Moles H}_2{SO}_4 = \frac{\textcolor{#a233ff}{7.7}}{\textcolor{#f21cc2}{98}} =\textcolor{#008d65}{0.079 \text{ mol}}$

Step 3: Divide the moles of $\text{LiCl}$ by the moles of $\text{H}_2\text{SO}_4$ to find the ratio between them.

$\frac{0.18}{0.079} = \textcolor{#008d65}{2.3}$

$\text{Moles LiCl} = 2.3 \times \text{Moles H}_2\text{SO}_4$

$\text{Ratio}=1:2.3$

Step 4: Round up or down to get determine the value of $x$

$1:2.3 \approx 1:2$

$\textcolor{#008d65}{x = 2}$

Substituting for $x$ in the equation above gives:

$\textcolor{#008d65}{2}\text{LiCl} + \text{H}_2\text{SO}_4 \rarr \text{Li}_2\text{SO}_4 + 2\text{HCl}$

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Calculations Using Moles Example Questions

Question 1: A student weighs out $6.5\text{ g}$ of a sample of phosphoric acid (H₃PO₄, $\text{M}_r =98$ ). Calculate the moles present in the sample.

[1 mark]

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$\begin{aligned}\text{Moles H}_3\text{PO}_4 &= \frac{\text{Mass H}_3\text{PO}_4}{\text{M}_r \text{ H}_3\text{PO}_4}\\ &= \frac{6.5}{98}\\ & = \underline{0.067 \text{ mol}}\end{aligned}$

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Question 2: Nitrogen and Hydrogen react in the following reaction:

$\text{N}_2 + 3\text{H}_2 \rarr 2\text{NH}_3$

Calculate the mass of Hydrogen needed for reaction if $13.5 \text{ g}$ of nitrogen is used.

[3 marks]

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Calculation should be broken down into steps (one mark per step):

1. $\text{Moles N}_2 = \frac{13.5}{28} = \underline{0.5 \text{ mol}}$

2. $\text{Moles H}_2 = 3 \times \text{ Moles N}_2 = \underline{1.5 \text{ mol}}$

3. $\text{Mass H}_2 = 1.5 \times 2 = \underline{2.9 \text{ g}}$

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Question 3: Glowing hot iron (Fe) will react with gaseous chlorine (Cl₂) to form iron (III) chloride (FeCl₃) in the following reaction:

$2\text{Fe} + X\text{Cl}_2 \rarr 2\text{FeCl}_3$

An experiment is carried out to find the value of X. In the experiment it was found that $11.00\text{ g}$ of iron reacted with $20.58\text{ g}$ of chlorine. Using this information, determine the value of X.

[4 marks]

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Calculation should be broken down into steps (one mark per step):

Step 1: Calculate the moles of Fe

$\text{Moles of Fe} = \frac{11}{56} = \underline{0.196 \text{ mol}}$

Step 2: Calculate the moles of Cl₂.

$\text{Moles of Cl}_2 = \frac{20.58}{70} = \underline{0.294\text{ mol}}$

Step 3: Calculate the ratio.

$\frac{0.294}{0.196} = \underline{1.5}$

$\text{Ratio} = \underline{1:1.5}$

Step 4: Deduce the simplest whole number ratio.

$\text{Simplest Whole Number Ratio} = 1:1.5 \times 2 = \underline{2:3}$
$\text{X} = \underline{3}$