Unit 3: Quantities, Units & Measuring Instruments

Master the fundamental measurements and precision instruments essential for quantitative chemistry.

3.4 Important Quantities and Units

Quantitative chemistry relies on precise measurements of fundamental quantities. Each physical quantity has both an SI (International System) unit and commonly used laboratory units. Understanding unit conversions is crucial for accurate calculations and experimental work. The most important quantities in chemistry include mass, volume, temperature, time, and amount of substance, each serving specific roles in chemical analysis and reactions.

Quantity SI Unit Symbol Common Lab Unit Key Conversions
Mass kilogram kg gram (g) 1 kg = 1000 g
Volume cubic meter m³ cm³, dm³, L 1 m³ = 10⁶ cm³ = 1000 dm³
Temperature kelvin K Celsius (°C) T(K) = T(°C) + 273.15
Time second s minute, hour 1 min = 60 s, 1 h = 3600 s
Amount mole mol millimole (mmol) 1 mol = 1000 mmol
Solved Examples:
  1. Convert 300 K to °C and explain the significance of absolute zero.
    Solution: T(°C) = T(K) - 273.15 = 300 - 273.15 = 26.85°C ≈ 27°C. The Kelvin scale starts at absolute zero (-273.15°C), where all molecular motion theoretically stops, making it essential for gas law calculations.
  2. A sample has a mass of 2.50 kg. Express this in grams and explain why grams are preferred in laboratory work.
    Solution: 2.50 kg × 1000 g/kg = 2500 g. Grams are preferred because typical laboratory samples range from milligrams to hundreds of grams, making calculations more convenient than using fractional kilograms.
  3. Convert 75.0 cm³ to m³ and explain when each unit is appropriate.
    Solution: 75.0 cm³ × (1 m³ / 10⁶ cm³) = 7.50 × 10⁻⁵ m³. cm³ is used for small laboratory volumes, while m³ is used for large-scale industrial processes or gas calculations under standard conditions.
  4. A reaction takes 2.5 hours. Express this in seconds and explain the importance of consistent time units.
    Solution: 2.5 h × 3600 s/h = 9000 s. Consistent time units are crucial for rate calculations, where rates are typically expressed as concentration change per second.
  5. Convert 0.750 dm³ to cm³ and mL, explaining the relationship between these units.
    Solution: 0.750 dm³ × 1000 cm³/dm³ = 750 cm³ = 750 mL (since 1 cm³ = 1 mL). This equivalence makes volume calculations straightforward in laboratory work.

3.5 Base and Derived Quantities

The SI system is built upon seven base quantities that are fundamentally independent and cannot be expressed in terms of other quantities. All other physical quantities are derived quantities, formed by mathematical combinations of base quantities. Understanding this hierarchy is essential for dimensional analysis and unit conversions in chemistry calculations.

Quantity Type SI Unit Symbol Base Unit Expression
Mass Base kilogram kg kg
Length Base meter m m
Time Base second s s
Temperature Base kelvin K K
Amount of substance Base mole mol mol
Volume Derived cubic meter m³ m³
Density Derived kg per m³ kg m⁻³ kg m⁻³
Pressure Derived pascal Pa kg m⁻¹ s⁻²
Energy Derived joule J kg m² s⁻²
Solved Examples:
  1. Derive the base units for volume and explain the relationship.
    Solution: Volume = length × length × length = m × m × m = m³. This shows volume is a derived quantity from the base quantity of length, cubed to represent three-dimensional space.
  2. Express pressure (Pa) in base units and explain the physical meaning.
    Solution: Pressure = force/area. Force = mass × acceleration = kg × m s⁻² = kg m s⁻². Area = length² = m². Therefore, Pa = (kg m s⁻²)/m² = kg m⁻¹ s⁻². This represents force per unit area.
  3. Derive the base units for energy (J) and relate it to chemical processes.
    Solution: Energy = force × distance = (kg m s⁻²) × m = kg m² s⁻². In chemistry, this represents the energy required to break bonds or the energy released in reactions.
  4. Why is amount of substance (mol) considered a base quantity?
    Solution: The mole represents a fundamental counting unit for particles (atoms, molecules, ions) and cannot be expressed in terms of other base quantities. It bridges the atomic scale with macroscopic measurements.
  5. Express the units of concentration (mol/dm³) in base units.
    Solution: Concentration = amount/volume = mol/m³. Since 1 dm³ = 10⁻³ m³, then mol/dm³ = mol/(10⁻³ m³) = 10³ mol m⁻³. This shows concentration as a derived quantity from two base quantities.

3.5 Measuring Instruments in Chemistry

Accurate measurements are fundamental to quantitative chemistry. Different instruments serve specific purposes, each with characteristic precision, accuracy, and uncertainty. Understanding when to use each instrument and their limitations is crucial for obtaining reliable experimental data. The choice of instrument depends on the required precision, volume range, and type of measurement.

Instrument Volume Range Typical Uncertainty Main Purpose Key Features
Measuring Cylinder 10-1000 cm³ ±0.5-1.0 cm³ Approximate measurements Graduated scale, wide opening
Pipette (volumetric) 1-50 cm³ ±0.02-0.06 cm³ Precise volume delivery Single volume, high accuracy
Burette 10-50 cm³ ±0.05-0.10 cm³ Variable volume delivery Fine graduations, stopcock
Volumetric Flask 25-2000 cm³ ±0.02-0.30 cm³ Standard solution preparation Single volume, narrow neck
Gas Syringe 10-100 cm³ ±0.1-1.0 cm³ Gas volume measurement Movable piston, gas-tight
Micropipette 0.1-1000 μL ±0.1-3.0% Microscale measurements Digital display, disposable tips
Solved Examples:
  1. Which instrument should be used to measure exactly 25.0 cm³ of NaOH solution for a titration? Justify your choice.
    Solution: A 25.0 cm³ volumetric pipette should be used. It provides the highest accuracy (±0.06 cm³) for this specific volume, essential for quantitative analysis. The uncertainty is only 0.24%, compared to 2-4% for a measuring cylinder.
  2. Explain why a burette is preferred over a measuring cylinder in titrations.
    Solution: Burettes have finer graduations (0.1 cm³) and lower uncertainty (±0.05 cm³) compared to measuring cylinders (±1.0 cm³). The stopcock allows precise control of flow rate, and readings can be taken to 0.05 cm³, crucial for determining accurate end points.
  3. A student needs to prepare 250 cm³ of 0.100 mol/dm³ HCl solution. What instrument should be used and why?
    Solution: A 250 cm³ volumetric flask should be used. It ensures exactly 250 cm³ at the calibrated temperature with high accuracy (±0.15 cm³). The narrow neck allows precise filling to the graduation mark, essential for accurate concentration.
  4. In a gas evolution experiment, 78.5 cm³ of CO₂ is collected. What is the uncertainty if a gas syringe is used?
    Solution: For a 100 cm³ gas syringe with ±1.0 cm³ uncertainty, the percentage uncertainty is (1.0/78.5) × 100% = 1.27%. This is acceptable for most gas volume measurements in quantitative experiments.
  5. Compare the percentage uncertainty when measuring 10 cm³ using a 10 cm³ pipette versus a 25 cm³ measuring cylinder.
    Solution: 10 cm³ pipette: (0.04/10.00) × 100% = 0.4%. 25 cm³ measuring cylinder: (0.5/10.0) × 100% = 5.0%. The pipette provides 12.5 times better precision, demonstrating the importance of choosing appropriate instruments.

3.7 Measuring Density

Density is an intensive property that relates mass to volume, providing insight into molecular packing and composition. It's calculated using the relationship $\rho = \frac{m}{V}$, where ρ (rho) represents density, m is mass, and V is volume. Accurate density measurements require precise determination of both mass and volume, with different techniques for solids, liquids, and gases. Understanding density helps identify substances and calculate concentrations. 

Density Formula: ρ = m/V
Where: ρ = density (kg/m³ or g/cm³)
       m = mass (kg or g)  
       V = volume (m³ or cm³)
Sample Type Mass Measurement Volume Measurement Typical Method
Regular Solid Electronic balance Geometric calculation Direct measurement
Irregular Solid Electronic balance Water displacement Displacement method
Liquid Balance (container + liquid) Volumetric glassware Direct measurement
Solution Balance Volumetric flask/pipette Concentration analysis
Gas Mass difference method Gas syringe/container STP conditions preferred
Solved Examples:
  1. A metal sample has a mass of 87.6 g and occupies 12.4 cm³. Calculate its density and suggest possible identity.
    Solution: ρ = m/V = 87.6 g / 12.4 cm³ = 7.06 g/cm³. This density is close to zinc (7.14 g/cm³) or tin (7.31 g/cm³), suggesting the metal could be one of these elements.
  2. Convert a density of 2.70 g/cm³ to kg/m³ and explain when each unit is used.
    Solution: 2.70 g/cm³ × (1 kg/1000 g) × (10⁶ cm³/1 m³) = 2.70 × 10³ kg/m³ = 2700 kg/m³. g/cm³ is used for laboratory samples, while kg/m³ is the SI unit used in engineering and large-scale calculations.
  3. An irregular solid displaces 15.8 cm³ of water and has a mass of 42.3 g. Will it float in water?
    Solution: ρ = 42.3 g / 15.8 cm³ = 2.68 g/cm³. Since this is greater than water's density (1.00 g/cm³), the solid will sink. Objects float when their density is less than the fluid they're placed in.
  4. Calculate the volume occupied by 150 g of ethanol (density = 0.789 g/cm³).
    Solution: Rearranging ρ = m/V gives V = m/ρ = 150 g / 0.789 g/cm³ = 190 cm³. This demonstrates how density can be used to predict volume from mass.
  5. A solution has a density of 1.15 g/cm³. What is the mass of 250 cm³ of this solution?
    Solution: Rearranging ρ = m/V gives m = ρ × V = 1.15 g/cm³ × 250 cm³ = 288 g. This calculation is important for preparing solutions by mass rather than volume.

Knowledge Check (20 Questions)

Answer: Kilogram (kg). The kilogram is one of the seven SI base units and is defined by the Planck constant.

Answer: T(°C) = 373 - 273.15 = 99.85°C ≈ 100°C. This is the boiling point of water at standard atmospheric pressure.

Answer: Cubic centimeter (cm³) or its equivalent, milliliter (mL), because laboratory samples typically involve small volumes.

Answer: 2.5 m³ × 10⁶ cm³/m³ = 2.5 × 10⁶ cm³. The conversion factor 10⁶ comes from (100 cm/m)³ = 10⁶ cm³/m³.

Answer: A base quantity is fundamental and cannot be expressed in terms of other quantities. Examples: mass (kg), time (s), temperature (K), length (m), and amount of substance (mol).

Answer: kg m⁻¹ s⁻². Pressure = force/area = (mass × acceleration)/area = (kg × m s⁻²)/m² = kg m⁻¹ s⁻².

Answer: A 25.0 cm³ volumetric pipette, with typical uncertainty of ±0.04 cm³ (0.16% error).

Answer: 50.0 cm³, graduated in 0.1 cm³ divisions for precise titration measurements.

Answer: Safety hazard - chemicals could be ingested or inhaled. Always use a pipette filler or safety bulb to create suction.

Answer: kg/m³ (SI unit). Relationship: 1 g/cm³ = 1000 kg/m³. Laboratory uses g/cm³ for convenience with small samples.

Answer: ρ = m/V = 156 g / 65.0 cm³ = 2.40 g/cm³.

Answer: 3.20 g/cm³ × (1 kg/1000 g) × (10⁶ cm³/1 m³) = 3.20 × 10³ kg/m³ = 3200 kg/m³.

Answer: Volume is derived (length³ = m³). Mass, time, and temperature are base quantities in the SI system.

Answer: Gas syringe, which has a movable piston and gas-tight seal to accurately measure gas volumes.

Answer: V = m/ρ = 180 g / 1.54 g/cm³ = 117 cm³.

Answer: Burettes have finer graduations (0.1 cm³ vs 1.0 cm³) and can be read to ±0.05 cm³, compared to ±0.5-1.0 cm³ for measuring cylinders.

Answer: 3.75 kg × 1000 g/kg = 3750 g. Use kg for large quantities and industrial processes; use g for laboratory work and small samples.

Answer: kg m² s⁻². Energy = force × distance = (kg m s⁻²) × m = kg m² s⁻².

Answer: They provide a single, highly accurate volume (±0.15 cm³ for 250 cm³) with a narrow neck for precise filling to the graduation mark, essential for accurate concentrations.

Answer: Metal density = 340 g / 45.2 cm³ = 7.52 g/cm³. Since 7.52 < 13.6 g/cm³, the metal will float in mercury.
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