Chemistry Lab Report Results Section Examples

Explore diverse examples of chemistry lab report results sections for better understanding.
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Examples of Example of a Chemistry Lab Report Results Section

Example 1: Analysis of Acid-Base Titration Results

This example illustrates the results from an acid-base titration experiment, where the concentration of hydrochloric acid (HCl) was determined using sodium hydroxide (NaOH) as the titrant.

In this experiment, a 25.00 mL sample of HCl was titrated with NaOH of known concentration (0.100 M). The endpoint was reached at 22.50 mL of NaOH added. The calculations for the concentration of HCl are as follows:

  • Moles of NaOH used = Volume (L) × Concentration (mol/L) = 0.02250 L × 0.100 mol/L = 0.00225 mol
  • Moles of HCl = Moles of NaOH (1:1 ratio) = 0.00225 mol
  • Concentration of HCl = Moles / Volume = 0.00225 mol / 0.025 L = 0.090 M

The final concentration of the hydrochloric acid solution was determined to be 0.090 M.

Notes: Ensure to conduct multiple trials for accuracy. Variations in titrant concentration may affect results.

Example 2: Determining the Rate of Reaction for Sodium Thiosulfate and Hydrochloric Acid

This example reports the results of an experiment aimed at measuring the rate of reaction between sodium thiosulfate and hydrochloric acid, using a colorimetric method to determine the time taken for a precipitate to form.

In this experiment, 50 mL of sodium thiosulfate solution (0.1 M) was mixed with 5 mL of hydrochloric acid (1 M). The time taken for the solution to turn opaque (indicating the formation of sulfur) was recorded under varying temperatures:

  • Room Temperature (20°C): Time = 60 seconds
  • Hot Water Bath (50°C): Time = 30 seconds
  • Ice Bath (0°C): Time = 120 seconds

The results indicate that the rate of reaction increases with temperature. The relationship can be summarized as follows:

Temperature (°C) Time to Precipitate (seconds)
0 120
20 60
50 30

Notes: Conducting this experiment in controlled environments can help minimize variability. Consider using a stopwatch for precise timing.

Example 3: Quantitative Analysis of Copper Content in an Unknown Sample

This example presents the results from a quantitative analysis of copper in an unknown sample using gravimetric analysis. The sample was dissolved in nitric acid, and copper was precipitated as copper (II) phosphate.

The unknown sample weighed 1.000 g, and after performing the necessary reactions and filtration, the precipitate was dried and weighed:

  • Weight of dried precipitate = 0.450 g
  • Molar mass of copper (II) phosphate (Cu3(PO4)2) = 380.58 g/mol
  • Moles of copper (II) phosphate = Weight / Molar Mass = 0.450 g / 380.58 g/mol ≈ 0.00118 mol
  • Moles of Cu = 3 × Moles of Cu3(PO4)2 = 3 × 0.00118 mol ≈ 0.00354 mol
  • Mass of Cu = Moles × Molar Mass of Cu = 0.00354 mol × 63.55 g/mol ≈ 0.225 g

The copper content in the original sample was determined to be approximately 22.5%.

Notes: Ensure that all reagents are of analytical grade to avoid impurities affecting the results. Repeating the experiment can validate findings.