Atomic Structure Revision Notes
Subject: Chemistry | Level: A-Level | Exam Board: OCR
Master the fundamental building blocks of chemistry that underpin the entire A-Level course. This guide covers atomic structure, the mole concept, stoichiometry, and redox reactions—essential topics that examiners love to test and which are critical for success in future modules."
Revision Notes & Key Concepts
Worked Examples
Worked Example
Question: Calculate the relative atomic mass ($A_r$) of a sample of magnesium containing 78.99% $^{24}Mg$, 10.00% $^{25}Mg$, and 11.01% $^{26}Mg$. Give your answer to 2 decimal places. [2 marks]
Solution: Step 1: Write out the formula for weighted mean mass. $$A_r = \frac{\sum(\text{isotope mass} \times \text{abundance})}{100}$$ Step 2: Substitute the values. $$A_r = \frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100}$$ Step 3: Calculate the numerator. $$1895.76 + 250.00 + 286.26 = 2432.02$$ Step 4: Divide by total abundance (100). $$A_r = \frac{2432.02}{100} = 24.3202$$ Step 5: Round to 2 decimal places. $$A_r = 24.32$$ Final answer: **24.32**
Worked Example
Question: 2.65 g of sodium carbonate, $Na_2CO_3$, was dissolved in water to make 250.0 $cm^3$ of solution. Calculate the concentration of the solution in mol dm$^{-3}$. ($A_r$ values: $Na=23.0, C=12.0, O=16.0$) [3 marks]
Solution: Step 1: Calculate the molar mass ($M$) of $Na_2CO_3$. $$M = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0 \text{ g mol}^{-1}$$ Step 2: Calculate the amount in moles ($n$). $$n = \frac{\text{mass}}{M} = \frac{2.65}{106.0} = 0.0250 \text{ mol}$$ Step 3: Convert volume to $dm^3$. $$250.0 \text{ cm}^3 = 0.2500 \text{ dm}^3$$ Step 4: Calculate concentration ($c$). $$c = \frac{n}{V} = \frac{0.0250}{0.2500} = 0.100 \text{ mol dm}^{-3}$$ Final answer: **0.100 mol dm$^{-3}$**
Worked Example
Question: Magnesium reacts with hydrochloric acid: $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$. Calculate the volume of hydrogen gas, in $m^3$, produced at 100 kPa and 298 K when 0.486 g of magnesium reacts with excess acid. [4 marks]
Solution: Step 1: Calculate moles of Magnesium ($n_{Mg}$). $$n = \frac{\text{mass}}{M} = \frac{0.486}{24.3} = 0.0200 \text{ mol}$$ Step 2: Use stoichiometry to find moles of $H_2$. Ratio $Mg : H_2$ is $1:1$. $$n_{H_2} = 0.0200 \text{ mol}$$ Step 3: Rearrange ideal gas equation for Volume ($V$). $$pV = nRT \rightarrow V = \frac{nRT}{p}$$ Step 4: Substitute values (converting pressure to Pa). $$p = 100 \text{ kPa} = 100,000 \text{ Pa}$$ $$V = \frac{0.0200 \times 8.314 \times 298}{100,000}$$ $$V = \frac{49.55144}{100,000} = 4.96 \times 10^{-4} \text{ m}^3$$ Final answer: **$4.96 \times 10^{-4} \text{ m}^3$**
Practice Questions
Question: Define the term 'relative isotopic mass'. [2 marks]
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Question: Sodium reacts with water: $2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)$. Calculate the mass of sodium required to produce 0.500 $dm^3$ of hydrogen gas at RTP (24.0 $dm^3$ $mol^{-1}$). [3 marks]
Answer:
Question: A sample of hydrated calcium chloride, $CaCl_2 \cdot xH_2O$, has a molar mass of 147.1 g mol$^{-1}$. Calculate the value of $x$. ($A_r$: $Ca=40.1, Cl=35.5, H=1.0, O=16.0$) [3 marks]
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Question: Explain, in terms of electrons, what occurs during oxidation and reduction. [2 marks]
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Question: Identify the oxidising agent in the reaction: $Cl_2 + 2NaBr \rightarrow 2NaCl + Br_2$. Explain your answer using oxidation numbers. [3 marks]
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