A complete guide to data presentation, interpretation, and analysis in chemistry experiments
In chemistry experiments, the type of data and its presentation can vary widely based on the nature of the experiment. This guide covers common types of chemistry experiments and best practices for presenting collected and transformed data.
Carefully note any colour changes, precipitate formation, gas evolution, temperature changes, etc.
Match the observations with expected results for specific reactions to identify the substance or determine its concentration.
For quantitative experiments, perform necessary calculations using formulas and relationships (e.g., stoichiometry, molarity, titration equations).
Based on the data and calculations, conclude the identity or concentration of the substance being tested.
Objective: To detect methyl ketones or ethanol in a sample.
Procedure: Add iodine and sodium hydroxide to the sample.
Data Collected: Observations of the formation of a yellow precipitate.
Interpretation:
Example Conclusion: If a yellow precipitate forms, the sample contains a methyl ketone or ethanol.
Volumes and concentrations.
Summarize raw data and calculations, such as initial and final volumes, and concentrations.
Clearly show the steps for any calculations (e.g., molarity).
| Trial | Initial Burette Reading (cm³) | Final Burette Reading (cm³) | Volume of NaOH used (cm³) | Concentration of HCl (M) |
|---|---|---|---|---|
| 1 | 0.00 | 24.80 | 24.80 | 0.0992 |
| 2 | 0.00 | 24.90 | 24.90 | 0.0996 |
| 3 | 0.00 | 24.85 | 24.85 | 0.0994 |
| Average Concentration: | 0.0994 M | |||
Given: NaOH concentration = 0.100 M, Volume of HCl = 25.0 cm³
Formula: M₁V₁ = M₂V₂ (for 1:1 stoichiometry)
Trial 1: (0.100 M) × (24.80 cm³) = M₂ × (25.0 cm³)
M₂ = (0.100 × 24.80) ÷ 25.0 = 0.0992 M
Average: (0.0992 + 0.0996 + 0.0994) ÷ 3 = 0.0994 M
Remember: “M₁V₁ = M₂V₂” is your best friend in titration calculations. Always check the stoichiometry first!
Trick: The endpoint is when the indicator changes color; the equivalence point is when moles of acid = moles of base.
Common Mistakes: Forgetting to convert cm³ to dm³ when using molarity (M = mol/dm³).
Concentration vs. time.
Record time and corresponding concentrations or other kinetic data.
Concentration vs. Time Graphs: Show how concentration changes over time for reactants and products.
| Time (s) | [Reactant] (M) | ln[Reactant] | 1/[Reactant] (M⁻¹) |
|---|---|---|---|
| 0 | 1.00 | 0.000 | 1.00 |
| 50 | 0.61 | -0.494 | 1.64 |
| 100 | 0.37 | -0.994 | 2.70 |
| 150 | 0.22 | -1.51 | 4.55 |
| 200 | 0.14 | -1.97 | 7.14 |
Remember: Rate = k[A]^m[B]^n. The order is not necessarily the coefficient from the balanced equation!
Graph Tricks:
Common Experiment: Clock reaction (iodine clock) is a classic kinetics experiment.
Temperature changes, heat capacities, enthalpy changes.
List initial and final temperatures, masses, specific heats, and calculated enthalpy changes.
Show the steps for calculating enthalpy changes or heat capacities.
| Measurement | Value | Unit |
|---|---|---|
| Mass of water | 100.0 | g |
| Initial temperature | 22.5 | °C |
| Final temperature | 28.3 | °C |
| Specific heat capacity of water | 4.18 | J/g°C |
| Temperature change (ΔT) | 5.8 | °C |
Step 1: Calculate heat absorbed by water:
q = m × c × ΔT = 100.0 g × 4.18 J/g°C × 5.8°C = 2424.4 J = 2.424 kJ
Step 2: Heat released by reaction = Heat absorbed by water = 2.424 kJ
Step 3: If 0.02 moles of reactant were used:
ΔH = -q ÷ moles = -2.424 kJ ÷ 0.02 mol = -121.2 kJ/mol
Conclusion: The reaction is exothermic with ΔH = -121.2 kJ/mol
Remember: q = m × c × ΔT is the key formula. For exothermic reactions, ΔH is negative; for endothermic, ΔH is positive.
Trick: “Calorimetry measures heat flow” – if temperature increases, reaction is exothermic; if temperature decreases, reaction is endothermic.
Common Mistake: Forgetting the negative sign for exothermic reactions when calculating ΔH.
Objective: To identify the presence of aldehydes in a sample.
Procedure: Perform tests such as the Tollens’ test, Fehling’s test.
| Test | Positive Result | Negative Result | Interpretation |
|---|---|---|---|
| Tollens’ Test | Formation of a silver mirror | No change | Confirms presence of an aldehyde |
| Fehling’s Test | Formation of a red precipitate | No change | Confirms presence of an aldehyde |
| Iodoform Test | Yellow precipitate | No precipitate | Methyl ketone or ethanol present |
Systematic Approach: Always follow: Observe → Record → Compare → Calculate → Conclude.
Key Differences:
Common Error: Confusing the iodoform test result – it’s positive for ethanol too, not just ketones!
Answer: In an acid-base titration, a solution of known concentration (titrant) is gradually added to a solution of unknown concentration (analyte) until the reaction is complete. The endpoint is determined using an indicator that changes color at or near the equivalence point (when stoichiometrically equivalent amounts of acid and base have reacted). Common indicators include phenolphthalein (colorless in acid, pink in base) and methyl orange (red in acid, yellow in base).
Answer: The equivalence point is the theoretical point at which stoichiometrically equivalent amounts of acid and base have reacted. The endpoint is the point at which the indicator changes color, which should be as close as possible to the equivalence point. They may differ slightly due to indicator characteristics, which is why choosing the right indicator for a particular titration is important.
Answer: To prepare a standard solution:
Answer: Using M₁V₁ = M₂V₂ (for 1:1 stoichiometry of NaOH and HCl):
M₁ × 30.0 cm³ = 0.1 M × 45.0 cm³
M₁ = (0.1 M × 45.0 cm³) ÷ 30.0 cm³ = 0.15 M
The molarity of NaOH solution is 0.15 M.
Answer: Reaction rate is the change in concentration of reactants or products per unit time. It can be experimentally determined by:
Answer: Using q = m × c × ΔT:
q = 50 g × 4.18 J/g°C × 10°C = 2090 J = 2.09 kJ
The heat absorbed is 2090 J or 2.09 kJ.
1. Calculate the molarity of HCl if 30 cm³ of 0.1 M NaOH neutralizes 25 cm³ of the HCl solution.
Solution: Using M₁V₁ = M₂V₂ (NaOH and HCl react 1:1):
MHCl × 25 cm³ = 0.1 M × 30 cm³
MHCl = (0.1 × 30) ÷ 25 = 0.12 M
Answer: 0.12 M
2. Determine the concentration of acetic acid (CH₃COOH) in vinegar if 50 cm³ of 0.2 M NaOH is required to titrate 10 cm³ of the vinegar solution.
Solution: CH₃COOH + NaOH → CH₃COONa + H₂O (1:1 ratio)
Macid × Vacid = Mbase × Vbase
Macid × 10 cm³ = 0.2 M × 50 cm³
Macid = (0.2 × 50) ÷ 10 = 1.0 M
Answer: 1.0 M acetic acid
3. Find the molarity of a FeSO₄·7H₂O solution if 20 cm³ of 0.025 M KMnO₄ is required to titrate 50 cm³ of the FeSO₄·7H₂O solution.
Solution: The redox reaction is: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Mole ratio: 1 mol MnO₄⁻ : 5 mol Fe²⁺
Moles of KMnO₄ = 0.025 M × 0.020 dm³ = 0.0005 mol
Moles of Fe²⁺ = 5 × 0.0005 = 0.0025 mol
Molarity of FeSO₄ = 0.0025 mol ÷ 0.050 dm³ = 0.05 M
Answer: 0.05 M
4. Calculate the molarity of a sodium thiosulphate (Na₂S₂O₃) solution if 30 cm³ of it is required to titrate 25 cm³ of 0.05 M iodine solution.
Solution: The reaction is: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻
Mole ratio: 1 mol I₂ : 2 mol S₂O₃²⁻
Moles of I₂ = 0.05 M × 0.025 dm³ = 0.00125 mol
Moles of S₂O₃²⁻ = 2 × 0.00125 = 0.0025 mol
Molarity of Na₂S₂O₃ = 0.0025 mol ÷ 0.030 dm³ = 0.0833 M
Answer: 0.0833 M
Test your knowledge with these 20 multiple choice questions. Select your answer and click submit to check your score.