Inorganic chemistry and the periodic table

Inorganic Chemistry and the Periodic Table: Gas Mixture Calculations

Welcome to this study guide for The Ozone Story (OZ)! In this chapter of the syllabus, our inorganic chemistry focus is completely centered on one essential quantitative skill: composition-by-volume calculations for gas mixtures, specifically working with percentages (\(\%\)) and parts per million (\(\text{ppm}\)).

Cross-reference note: Atmospheric reactions, radicals, and ozone depletion mechanisms are explored in detail in the accompanying chapters Sustainability and Reaction mechanisms.

Don't worry if maths in chemistry sometimes feels daunting. Once you understand the straightforward relationship between gas volumes, percentages, and parts per million, these calculations become some of the most reliable marks you can score in your exam!

1. Understanding Gas Volumes in Mixtures

The atmosphere is a huge mixture of different gases, such as nitrogen, oxygen, argon, carbon dioxide, and trace amounts of ozone and pollutants. When dealing with mixtures of gases at the same temperature and pressure, there is a helpful scientific principle: equal volumes of gases contain equal numbers of moles (particles).

Because the volume of an ideal gas is directly proportional to the amount of gas in moles, we express the concentration of gases in air in terms of composition by volume. When we say that dry air is approximately \(78\%\) nitrogen by volume, it means that in any sample of \(100\text{ dm}^3\) of air, \(78\text{ dm}^3\) is nitrogen gas.

Key Takeaway: For gas mixtures at constant temperature and pressure, a fraction by volume is identical to a fraction by moles.

2. Calculating Percentage by Volume (\(\%\))

Percentage by volume describes how many parts of a specific gas are present in every \(100\) parts of total gas mixture.

To calculate the percentage by volume of a gas:

\(\% \text{ by volume} = \left( \frac{\text{volume of component gas}}{\text{total volume of gas mixture}} \right) \times 100\)

Essential Rule: Always ensure that the volume of the component gas and the total volume of the mixture are in the exact same units (e.g., both in \(\text{cm}^3\), both in \(\text{dm}^3\), or both in \(\text{m}^3\)) before calculating!

Worked Example 1:
A sealed sample of polluted urban air has a total volume of \(2.50\text{ dm}^3\). Laboratory analysis shows that it contains \(0.00105\text{ dm}^3\) of carbon monoxide (\(\text{CO}\)). Calculate the percentage by volume of \(\text{CO}\) in the air sample.

Step-by-step solution:
1. Check the units: Both volumes are in \(\text{dm}^3\), so no conversion is needed.
2. Apply the formula:
\(\% \text{ by volume of CO} = \left( \frac{0.00105\text{ dm}^3}{2.50\text{ dm}^3} \right) \times 100\)
3. Compute the result:
\(\% \text{ by volume} = 0.000420 \times 100 = 0.0420\%\) (or \(4.20 \times 10^{-2}\%\)).

3. What is "Parts Per Million" (\(\text{ppm}\)) and Why Do We Use It?

In atmospheric chemistry—especially in the ozone layer and troposphere—many crucial gases occur in very tiny amounts. For example, harmful pollutants or protective stratospheric ozone may exist in concentrations far below \(0.01\%\).

Writing numbers like \(0.000035\%\) can be clumsy, and it is easy to miscount zeroes. Scientists solve this by using parts per million (\(\text{ppm}\)). Instead of asking "how many parts are there in \(100\) parts?", we ask "how many parts are there in \(1\,000\,000\) (\(10^6\)) parts?".

Analogy: Imagine a huge stadium containing one million supporters. If exactly three people are wearing a yellow hat, the concentration of yellow-hat wearers is \(3\text{ ppm}\)!

To calculate parts per million by volume (\(\text{ppm}\)):

\(\text{ppm by volume} = \left( \frac{\text{volume of component gas}}{\text{total volume of gas mixture}} \right) \times 10^6\)

Worked Example 2:
A \(500\text{ cm}^3\) air sample taken from the stratosphere contains \(4.50 \times 10^{-3}\text{ cm}^3\) of ozone (\(\text{O}_3\)). Calculate the concentration of ozone in \(\text{ppm}\) by volume.

Step-by-step solution:
1. Check the units: Both volumes are given in \(\text{cm}^3\).
2. Apply the formula:
\(\text{ppm} = \left( \frac{4.50 \times 10^{-3}\text{ cm}^3}{500\text{ cm}^3} \right) \times 10^6\)
3. Compute the fraction:
\(\frac{4.50 \times 10^{-3}}{500} = 9.00 \times 10^{-6}\)
4. Multiply by \(10^6\):
\(\text{ppm} = 9.00 \times 10^{-6} \times 10^6 = 9.00\text{ ppm}\).

4. Converting Between \(\%\) and \(\text{ppm}\)

Examiners love testing your ability to switch smoothly between percentage by volume and \(\text{ppm}\). A simple mathematical relationship connects them:

• A percentage is "parts per hundred" (\(10^2\)).
• Parts per million is "parts per million" (\(10^6\)).
• The difference between \(100\) and \(1\,000\,000\) is a factor of \(10\,000\) (\(10^4\)).

Therefore:

\(1\% = 10\,000\text{ ppm} = 10^4\text{ ppm}\)
\(100\% = 1\,000\,000\text{ ppm} = 10^6\text{ ppm}\)

This gives us two easy conversion rules:

• To convert from \(\%\) to \(\text{ppm}\): multiply by \(10\,000\) (or \(\times 10^4\)).
• To convert from \(\text{ppm}\) to \(\%\): divide by \(10\,000\) (or \(\div 10^4\)).

Memory Trick: A \(\text{ppm}\) value should always be a much larger number than the percentage value for the same gas mixture. If your \(\text{ppm}\) number is smaller than your percentage, you divided when you should have multiplied!

Worked Example 3:
The concentration of carbon dioxide in a dry air sample is \(0.0415\%\) by volume. Express this concentration in \(\text{ppm}\) by volume.

Step-by-step solution:
\(\text{Concentration in ppm} = 0.0415 \times 10\,000 = 415\text{ ppm}\).

Worked Example 4:
In a photochemical smog episode, the ground-level ozone concentration reaches \(0.120\text{ ppm}\). Express this concentration as a percentage by volume.

Step-by-step solution:
\(\text{Concentration in } \% = \frac{0.120}{10\,000} = 0.0000120\% = 1.20 \times 10^{-5}\%\).

5. Rearranging the Formulas to Find Gas Volumes

Sometimes an exam question gives you the concentration (either in \(\%\) or \(\text{ppm}\)) and the total volume of the mixture, and asks you to find the volume of the specific gas component.

By rearranging our main formulas:

\(\text{Volume of component gas} = \frac{\% \text{ by volume} \times \text{total volume}}{100}\)

\(\text{Volume of component gas} = \frac{\text{ppm by volume} \times \text{total volume}}{10^6}\)

Worked Example 5:
A classroom has a total air volume of \(180\text{ m}^3\). The concentration of argon in the air is \(9340\text{ ppm}\) by volume. Calculate the volume of argon in the room in \(\text{m}^3\).

Step-by-step solution:
1. Identify the given values: \(\text{ppm} = 9340\), \(\text{Total volume} = 180\text{ m}^3\).
2. Apply the rearranged formula:
\(\text{Volume of argon} = \frac{9340 \times 180\text{ m}^3}{10^6}\)
3. Calculate:
\(\text{Volume of argon} = \frac{1\,681\,200}{1\,000\,000} = 1.68\text{ m}^3\) (to 3 significant figures).

6. Handling Unit Conversions Accurately

A classic student trap in OCR exams is mixing up units of volume. Always convert all volumes to the same unit before calculating!

Volume Conversion Refresher:
• \(1\text{ dm}^3 = 1000\text{ cm}^3\)
• To convert \(\text{cm}^3\) to \(\text{dm}^3\): divide by \(1000\)
• To convert \(\text{dm}^3\) to \(\text{cm}^3\): multiply by \(1000\)
• \(1\text{ m}^3 = 1000\text{ dm}^3 = 1\,000\,000\text{ cm}^3\)

Worked Example 6:
A gas canister contains \(2.00\text{ dm}^3\) of a test gas mixture. It contains \(0.040\text{ cm}^3\) of sulfur dioxide. Calculate the concentration of sulfur dioxide in \(\text{ppm}\) by volume.

Step-by-step solution:
1. Spot the unit mismatch: The component volume is in \(\text{cm}^3\), but the mixture volume is in \(\text{dm}^3\).
2. Convert the total volume into \(\text{cm}^3\):
\(2.00\text{ dm}^3 = 2.00 \times 1000 = 2000\text{ cm}^3\).
3. Substitute into the \(\text{ppm}\) formula:
\(\text{ppm} = \left( \frac{0.040\text{ cm}^3}{2000\text{ cm}^3} \right) \times 10^6\)
4. Calculate:
\(\frac{0.040}{2000} = 2.0 \times 10^{-5}\)
\(\text{ppm} = 2.0 \times 10^{-5} \times 10^6 = 20\text{ ppm}\).

7. Common Exam Pitfalls to Avoid

• Inconsistent volume units: Always check the units for each gas volume given in the question. If one is in \(\text{cm}^3\) and the other is in \(\text{dm}^3\) or \(\text{m}^3\), convert them to match first.
• Multiplying/dividing the wrong way: Remember that moving from \(\%\) to \(\text{ppm}\) makes the number larger (\(\times 10^4\)), while moving from \(\text{ppm}\) to \(\%\) makes the number smaller (\(\div 10^4\)).
• Significant figures: As required by OCR assessment guidelines, ensure your final answer is quoted to the appropriate number of significant figures, matching the least precise measurement given in the question data.
• Confusing \(\text{ppm}\) by volume with \(\text{ppm}\) by mass: In the context of atmospheric gas mixtures, always calculate composition using volumes unless specifically directed otherwise.

8. Quick Review Checklist

Before moving on, make sure you are confident with each of the following:

• Can you write down the formula for percentage by volume?
\(\% \text{ by volume} = \left( \frac{V_{\text{gas}}}{V_{\text{total}}} \right) \times 100\)

• Can you write down the formula for \(\text{ppm}\) by volume?
\(\text{ppm by volume} = \left( \frac{V_{\text{gas}}}{V_{\text{total}}} \right) \times 10^6\)

• Can you state the direct conversion factor between \(\%\) and \(\text{ppm}\)?
\(1\% = 10\,000\text{ ppm}\) (\(10^4\text{ ppm}\))

• Can you convert between \(\text{cm}^3\), \(\text{dm}^3\), and \(\text{m}^3\) before carrying out a calculation?