Chemical Equations: Moles and Mass Calculations Revision Notes for HSC SSCE Chemistry

Chemical Equations: Moles and Mass Calculations

Reading chemical equations in terms of moles

Chemical equations provide us with important quantitative information about reactions. Consider the equation for the combustion of phosphorus:

$4\text{P(s)} + 5\text{O}_2\text{(g)} \rightarrow 2\text{P}_2\text{O}_5\text{(s)}$

This equation can be interpreted in three different ways:

Qualitatively (in words): Solid phosphorus reacts with oxygen gas to produce solid diphosphorus pentoxide.
At the particle level: Four atoms of phosphorus react with five molecules of oxygen to form two molecules of diphosphorus pentoxide.
At the mole level: Four moles of phosphorus react with five moles of oxygen to form two moles of diphosphorus pentoxide.

The third interpretation is particularly useful in laboratory work. Because individual atoms and molecules are too small to weigh practically, we work with much larger quantities measured in moles. The mole relationships in chemical equations allow us to calculate the amounts of substances involved in reactions.

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Key concept: The coefficients (numbers in front of formulas) in a balanced chemical equation represent the mole ratio of reactants and products. This ratio is the foundation for all stoichiometric calculations.

Calculating moles from chemical equations

The stoichiometric coefficients in a balanced equation tell us the exact ratio in which substances react and form. We can use these ratios to calculate how many moles of one substance are needed to react with another, or how many moles of product will form.

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Worked Example: Aluminum reacting with oxygen

Question: How many moles of oxygen are needed to react with $3.0$ mol aluminium to form $\text{Al}_2\text{O}_3$ ?

Solution steps:

Write the balanced equation: $4\text{Al(s)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{Al}_2\text{O}_3\text{(s)}$
Determine the mole ratio: From the equation, we can see that $3$ moles of $\text{O}_2$ react with $4$ moles of Al.

Therefore: $\frac{\text{moles of O}_2 \text{ required}}{\text{moles of Al given}} = \frac{3}{4}$
Rewrite as a formula: $\text{moles of O}_2 \text{ required} = 0.75 \times (\text{moles of Al given})$
Substitute and calculate: $\text{moles of O}_2 = 0.75 \times 3.0 = \mathbf{2.3 \text{ mol}}$

(Answer given to 2 significant figures to match the precision of $3.0$ )

Converting moles to masses

Once we know the mole relationships in a reaction, we can convert these to actual masses that can be measured in the laboratory. This conversion uses molar mass (the mass of one mole of a substance in grams).

Example: Mass relationships in phosphorus combustion

Using our previous equation: $4\text{P(s)} + 5\text{O}_2\text{(g)} \rightarrow 2\text{P}_2\text{O}_5\text{(s)}$

We know that four moles of phosphorus react with five moles of oxygen to form two moles of diphosphorus pentoxide.

Converting each to mass:

$4$ mol P (relative atomic mass $31.0$ ) = $4 \times 31.0 = 124.0$ g P
$5$ mol $\text{O}_2$ (relative molecular mass $32.0$ ) = $5 \times 32.0 = 160.0$ g $\text{O}_2$
$2$ mol $\text{P}_2\text{O}_5$ (relative molecular mass $142.0$ ) = $2 \times 142.0 = 284.0$ g $\text{P}_2\text{O}_5$

Therefore: $124.0$ g P reacts with $160.0$ g $\text{O}_2$ to form $284.0$ g $\text{P}_2\text{O}_5$

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Important observation: Notice that the total mass of reactants ( $124.0 + 160.0 = 284.0$ g) equals the total mass of products ( $284.0$ g). This demonstrates the law of conservation of mass - matter is neither created nor destroyed in chemical reactions.

Mass calculations from chemical equations

When solving problems involving masses of reactants and products, we need to calculate the mass of one reactant needed to react completely with a given mass of another reactant, or calculate the mass of product that can be formed from a given mass of reactant.

The five-step method

To perform mass-to-mass calculations, follow this systematic approach:

Write a balanced chemical equation for the reaction.
Calculate the number of moles of the given substance using the formula: $n = \frac{m}{MM}$ where $n$ = number of moles, $m$ = mass, and $MM$ = molar mass
Use the chemical equation to write an expression for: $\frac{\text{number of moles of required substance}}{\text{number of moles of given substance}}$ This ratio equals the ratio of stoichiometric coefficients in the balanced equation.
Calculate the number of moles of the required substance using the ratio from step 3.
Calculate the mass of required substance by rearranging the formula from step 2: $m = n \times MM$

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Visual guide to the process

The conversion pathway follows this sequence:

Start with the mass of given substance
Divide by its molar mass to get moles of given substance
Use the chemical equation to find moles of required substance
Multiply by its molar mass to get mass of required substance

This can be summarized as: Mass → Moles → Moles → Mass

Worked examples

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Worked Example: Dissolving magnesium carbonate

Question: Magnesium carbonate (magnesite ore) dissolves in hydrochloric acid to form magnesium chloride, carbon dioxide, and water. Calculate the mass of hydrochloric acid needed to dissolve $5.0$ g of magnesium carbonate.

Solution:

Write the skeletal equation, then balance it:
- Skeletal: $\text{MgCO}_3 + \text{HCl} \rightarrow \text{MgCl}_2 + \text{CO}_2 + \text{H}_2\text{O}$
- Balanced: $\text{MgCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}$
Express moles required in terms of moles given: $\text{Moles of HCl required} = 2 \times (\text{moles of MgCO}_3 \text{ given})$
Calculate molar mass of $\text{MgCO}_3$ : $MM = 24.31 + 12.01 + (3 \times 16.00) = 84.32 \text{ g mol}^{-1}$
Calculate moles of $\text{MgCO}_3$ given: $n = \frac{5.0}{84.32} = 0.0593 \text{ mol}$ (Keep extra digits for now; round at the end)
Calculate moles of HCl needed: $n_{\text{HCl}} = 2 \times 0.0593 = 0.1186 \text{ mol}$
Calculate molar mass of HCl: $MM = 1.008 + 35.45 = 36.46 \text{ g mol}^{-1}$
Calculate mass of HCl: $m = 0.1186 \times 36.46 = \mathbf{4.3 \text{ g}}$ (Rounded to 2 significant figures to match $5.0$ )

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Worked Example: Producing iron(III) oxide

Question: Calculate the amount of iron needed to react with oxygen to produce $3.5$ g iron(III) oxide, $\text{Fe}_2\text{O}_3$ .

Given equation: $4\text{Fe(s)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{Fe}_2\text{O}_3\text{(s)}$

Solution:

The equation is already balanced ✓
Calculate molar mass of $\text{Fe}_2\text{O}_3$ : $MM = (2 \times 55.85) + (3 \times 16.00) = 159.7 \text{ g mol}^{-1}$
Calculate moles of $\text{Fe}_2\text{O}_3$ to be produced: $n = \frac{3.5}{159.7} = 0.0219 \text{ mol}$
Use the equation to find moles of Fe needed: From the balanced equation, $4$ moles of Fe produce $2$ moles of $\text{Fe}_2\text{O}_3$ . $\text{Moles Fe needed} = \frac{4}{2} \times (\text{moles of Fe}_2\text{O}_3) = 2 \times 0.0219 = 0.0418 \text{ mol}$
Calculate mass of Fe required:
- Molar mass of Fe = $55.85$ g mol $^{-1}$
- Mass = $0.0418 \times 55.85 = \mathbf{2.5 \text{ g}}$ (Rounded to 2 significant figures)

Exam tips

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Tips for Success in Calculations

Always balance your equation first - all calculations depend on correct stoichiometric ratios.
Show your working clearly - write out the mole ratio and each calculation step.
Keep extra digits during calculations - only round your final answer to match the significant figures of the given data.
Check your answer makes sense - if you're asked for a mass of product, it should be less than the total mass of reactants (unless you're calculating one reactant from another).
Include units throughout your calculations - this helps avoid errors and makes your work easier to follow.
State symbols are important - they can affect how you interpret the question, though they don't affect the mole calculations.

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Common Mistake to Avoid

Never begin calculations without first ensuring your chemical equation is properly balanced. Using an unbalanced equation will give incorrect mole ratios and lead to wrong answers throughout your calculation.

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Key Points to Remember

Chemical equations can be read in terms of moles, not just individual particles. The coefficients tell us the mole ratio of substances in the reaction.
To convert between moles and mass, use the formula: $n = \frac{m}{MM}$ where $n$ is moles, $m$ is mass in grams, and $MM$ is molar mass in g mol $^{-1}$ .
For mass-to-mass calculations, follow the pathway: mass of given substance → moles of given substance → moles of required substance → mass of required substance.
The mole ratio comes directly from the stoichiometric coefficients in the balanced equation. For example, in $4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$ , the ratio of Al to $\text{O}_2$ is $4:3$ .
Always balance your equation first, keep extra significant figures during calculations, and round only your final answer to match the precision of the given data.

Chemical Equations: Moles and Mass Calculations (HSC SSCE Chemistry): Revision Notes