Reacting Masses and Volumes of Solutions and Gases

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1. Calculating Reacting Masses from Equations

What this means: When you have a balanced chemical equation, you can work out how much of each substance reacts and how much product forms.

The basic steps:

1. Write the balanced chemical equation
2. Calculate the number of moles of the substance you know about
3. Use the mole ratio from the equation to find moles of the substance you want
4. Convert moles back to mass if needed

Key formula:

Where:

• n = number of moles
• mass = mass in grams (g)
• M_r = relative molecular mass (add up all the atomic masses in the formula)

Example 1: Simple mass calculation

Consider the reaction: $4Na(s) + O_2(g) \rightarrow 2Na_2O(s)$

Question: How much sodium reacts exactly with 3.20 g of oxygen?

Solution:

• Step 1: Calculate moles of $O_2 = \frac{3.2}{32} = 0.1 \text{ mol}$
• Step 2: From the equation, 4 moles of Na react with 1 mole of $O_2$
• Step 3: So moles of Na needed $= 0.1 \times 4 = 0.4 \text{ mol}$
• Step 4: Mass of Na $= 0.4 \times 23 = 9.20 \text{ g}$

Example 2: Product mass calculation

Using the same equation, what mass of $Na_2O$ is produced from 0.4 mol Na?

Solution:

• From the equation: 4 mol Na produces 2 mol $Na_2O$
• So 0.4 mol Na produces $0.4 \times \frac{2}{4} = 0.2 \text{ mol } Na_2O$
• $M_r$ of $Na_2O = (23 \times 2) + 16 = 62$
• Mass of $Na_2O = 0.2 \times 62 = 12.4 \text{ g}$

2. Percentage Yield Calculations

In real-life chemistry, reactions rarely give you 100% of the product you expect. Some product is always lost during the process.

Percentage yield tells you how efficient a reaction is.

Formula:

Where:

• Actual yield = what you actually got in the experiment
• Theoretical yield = what you should get according to the equation (if everything was perfect)

Example:

When 2.50 g of $SO_2$ is heated with excess oxygen, 2.50 g of $SO_3$ is obtained.

Equation: $2SO_2 + O_2 \rightarrow 2SO_3$

Calculate the percentage yield.

Solution:

• $M_r$ of $SO_2 = 32 + (16 \times 2) = 64$
• $M_r$ of $SO_3 = 32 + (16 \times 3) = 80$
• From equation: 2 mol $SO_2$ (128 g) gives 2 mol $SO_3$ (160 g)
• So 2.5 g $SO_2$ should give: $\frac{2.5 \times 160}{128} = 3.125 \text{ g } SO_3$ (theoretical yield)
• Actual yield = 2.50 g
• $\% \text{ Yield} = \frac{2.50}{3.125} \times 100 = 80\%$

3. Volumes of Gases

Gases are special because, at the same temperature and pressure, equal volumes of different gases contain the same number of molecules.

At room temperature and pressure (RTP):

• 1 mole of any gas occupies 24 dm³ (or 24,000 cm³)

Formula:

or for cm³:

To find moles from volume:

or

Example 1: Volume of gas produced

Calculate the volume of $CO_2$ produced when 10.01 g of calcium carbonate decomposes at RTP:

$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$

Solution:

• $M_r$ of $CaCO_3 = 40.1 + 12 + (3 \times 16) = 100.1$
• Moles of $CaCO_3 = \frac{10.01}{100.1} = 0.1 \text{ mol}$
• From equation: 1 mol $CaCO_3$ produces 1 mol $CO_2$
• So moles of $CO_2 = 0.1 \text{ mol}$
• Volume of $CO_2 = 0.1 \times 24 = 2.4 \text{ dm}^3$

Example 2: Mass from gas volume

What mass of potassium chlorate(V) decomposes to produce 100.0 cm³ of oxygen gas at RTP?

$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$

Solution:

• Moles of $O_2 = \frac{100}{24000} = 0.00417 \text{ mol}$
• From equation: 3 mol $O_2$ comes from 2 mol $KClO_3$
• So moles of $KClO_3 = 0.00417 \times \frac{2}{3} = 0.00278 \text{ mol}$
• $M_r$ of $KClO_3 = 39.1 + 35.5 + (3 \times 16) = 122.6$
• Mass = $0.00278 \times 122.6 = 0.341 \text{ g}$

Important: When gases react together at the same temperature and pressure, the volume ratio equals the mole ratio from the equation.

For example: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$

If you start with 100 cm³ of $H_2$ and 200 cm³ of $Cl_2$, you'll use 100 cm³ of each and produce 200 cm³ of HCl (total gas volume = 300 cm³).

4. Volumes and Concentrations of Solutions

Concentration tells you how much substance is dissolved in a certain volume of solution.

Units:

• Concentration is measured in mol/dm³ (also written as mol dm⁻³ or M)
• Volume should be in dm³
• To convert cm³ to dm³: divide by 1000

Key Formula: $n = C \times V$

Where:

• n = number of moles
• C = concentration in mol/dm³
• V = volume in dm³

To find concentration: $C = \frac{n}{V}$

Example 1: Finding moles in a solution

Calculate the number of moles in 50.0 cm³ of a 0.0500 mol/dm³ solution of $FeCl_3$.

Solution:

• First convert cm³ to dm³: $50.0 \text{ cm}^3 = \frac{50}{1000} = 0.050 \text{ dm}^3$
• Moles of $FeCl_3 = 0.0500 \times 0.050 = 2.5 \times 10^{-3} \text{ mol}$
• Each $FeCl_3$ contains 3 chloride ions
• So moles of $Cl^- = 3 \times 2.5 \times 10^{-3} = 7.5 \times 10^{-3} \text{ mol}$

Example 2: Finding concentration

If 10.00 g of NaOH is dissolved in water and made up to 200.0 cm³, calculate the concentration in mol/dm³.

Solution:

• $M_r$ of NaOH = 23 + 16 + 1 = 40
• Moles of NaOH = $\frac{10.00}{40} = 0.25 \text{ mol}$
• Volume in dm³ = $\frac{200}{1000} = 0.2 \text{ dm}^3$
• Concentration = $\frac{0.25}{0.2} = 1.25 \text{ mol/dm}^3$

Example 3: Titration calculation

25.0 cm³ of a sodium carbonate solution reacted with 35.8 cm³ of 0.100 mol/dm³ HCl.

Equation: $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$

Calculate the concentration of the sodium carbonate solution.

Solution:

• Moles of HCl = $0.100 \times \frac{35.8}{1000} = 3.58 \times 10^{-3} \text{ mol}$
• From equation: 2 mol HCl reacts with 1 mol $Na_2CO_3$
• Moles of $Na_2CO_3 = \frac{3.58 \times 10^{-3}}{2} = 1.79 \times 10^{-3} \text{ mol}$
• This is in 25.0 cm³ = 0.025 dm³
• Concentration = $\frac{1.79 \times 10^{-3}}{0.025} = 0.0716 \text{ mol/dm}^3$

5. Limiting Reagent and Excess Reagent

When you mix two reactants, usually one runs out first. This is called the limiting reagent (or limiting reactant). It limits how much product you can make. The other reactant is in excess – there's more than you need.

How to identify the limiting reagent:

1. Calculate the moles of each reactant
2. Use the equation to work out what each reactant needs
3. The one that runs out first is the limiting reagent

Example:

0.20 mol $AsCl_3$ reacts with 0.25 mol $H_2O$.

Equation: $4AsCl_3 + 6H_2O \rightarrow As_4O_6 + 12HCl$

Which is the limiting reagent?

Solution:

• From the equation, the ratio is 4:6, which simplifies to 2:3
• For every 2 mol $AsCl_3$, you need 3 mol $H_2O$
• For 0.20 mol $AsCl_3$, you need: $0.20 \times \frac{3}{2} = 0.30 \text{ mol } H_2O$
• But you only have 0.25 mol $H_2O$
• Water is the limiting reagent (it runs out first)
• $AsCl_3$ is in excess

Another example with masses:

0.72 g of titanium reacts with excess chlorine to form 2.85 g of a chloride.

What's the empirical formula?

Solution:

• Moles of Ti = $\frac{0.72}{47.9} = 0.015 \text{ mol}$
• Mass of Cl that reacted = $2.85 - 0.72 = 2.13 \text{ g}$
• Moles of Cl atoms = $\frac{2.13}{35.5} = 0.060 \text{ mol}$
• Ratio Ti:Cl = 0.015:0.060 = 1:4
• Empirical formula = TiCl₄

6. Back-Titrations (Advanced Application)

Sometimes you can't measure how much of a substance reacts directly. Instead, you:

1. Add a known excess of a reagent
2. Let it react
3. Measure how much of the excess is left over
4. Work backwards to find how much reacted

Example:

A 5.00 g fertilizer sample is reacted with 50.0 cm³ of 2.00 mol/dm³ NaOH. The excess NaOH required 31.2 cm³ of 1.00 mol/dm³ HCl for neutralization.

Equation: $(NH_4)_2SO_4 + 2NaOH \rightarrow 2NH_3 + Na_2SO_4 + 2H_2O$

Calculate the percentage purity of ammonium sulfate in the fertilizer.

Solution:

• Initial moles of NaOH = $2.00 \times \frac{50.0}{1000} = 0.100 \text{ mol}$
• Moles of HCl used = $1.00 \times \frac{31.2}{1000} = 0.0312 \text{ mol}$
• Moles of excess NaOH = 0.0312 mol (1:1 ratio with HCl)
• Moles of NaOH that reacted with fertilizer = $0.100 - 0.0312 = 0.0688 \text{ mol}$
• From equation: 2 mol NaOH reacts with 1 mol $(NH_4)_2SO_4$
• Moles of $(NH_4)_2SO_4 = \frac{0.0688}{2} = 0.0344 \text{ mol}$
• $M_r$ of $(NH_4)_2SO_4 = (2 \times 14) + (8 \times 1) + 32 + (4 \times 16) = 132$
• Mass of $(NH_4)_2SO_4 = 0.0344 \times 132 = 4.54 \text{ g}$
• Percentage purity = $\frac{4.54}{5.00} \times 100 = 90.8\%$