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Measurement and Data Processing - Uncertainties and errors in measurement

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Distinction between randomrandom errorserrors (caused by unpredictable fluctuations) and systematicsystematic errorserrors (caused by flaws in equipment or experimental design).

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AccuracyAccuracy refers to how close a measured value is to the literature or true value, while precisionprecision refers to how close a series of measurements are to each other.

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Significant figures: When multiplying/dividing, the result should have the same number of significant figures as the measurement with the fewestfewest significant figures.

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Uncertainty in readings: For analog instruments, uncertainty is typically ±0.5\pm 0.5 of the smallest division; for digital instruments, it is ±1\pm 1 of the smallest division.

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Propagation of uncertainty: When adding or subtracting, add absoluteabsolute uncertaintiesuncertainties. When multiplying or dividing, add percentagepercentage (relative) uncertaintiesuncertainties.

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Repeatable results indicate low randomrandom errorerror, while a consistent deviation from the true value indicates a systematicsystematic errorerror (often seen as a non-zero intercept on a graph that should pass through the origin).

📐Formulae

Percentage Error=∣literature value−experimental value∣literature value×100%\text{Percentage Error} = \frac{|\text{literature value} - \text{experimental value}|}{\text{literature value}} \times 100\%

Percentage Uncertainty=absolute uncertaintymeasured value×100%\text{Percentage Uncertainty} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100\%

Δy=Δa+Δb(for y=a±b)\Delta y = \Delta a + \Delta b \quad \text{(for } y = a \pm b\text{)}

Δyy=Δaa+Δbb(for y=a×b or y=ab)\frac{\Delta y}{y} = \frac{\Delta a}{a} + \frac{\Delta b}{b} \quad \text{(for } y = a \times b \text{ or } y = \frac{a}{b}\text{)}

Δyy=∣n∣Δaa(for y=an)\frac{\Delta y}{y} = |n| \frac{\Delta a}{a} \quad \text{(for } y = a^n\text{)}

💡Examples

Problem 1:

A student measures the mass of a sample as 2.50±0.01 g2.50 \pm 0.01\text{ g} and its volume as 10.0±0.2 cm310.0 \pm 0.2\text{ cm}^3. Calculate the density of the sample and its absolute uncertainty.

Solution:

Density ρ=mV=2.50 g10.0 cm3=0.250 g cm−3\rho = \frac{m}{V} = \frac{2.50\text{ g}}{10.0\text{ cm}^3} = 0.250\text{ g cm}^{-3}.

Percentage uncertainty in m=0.012.50×100=0.4%m = \frac{0.01}{2.50} \times 100 = 0.4\%. Percentage uncertainty in V=0.210.0×100=2.0%V = \frac{0.2}{10.0} \times 100 = 2.0\%. Total percentage uncertainty =0.4%+2.0%=2.4%= 0.4\% + 2.0\% = 2.4\%.

Absolute uncertainty in ρ=0.250×0.024=0.006 g cm−3\rho = 0.250 \times 0.024 = 0.006\text{ g cm}^{-3}.

Final value: 0.250±0.006 g cm−30.250 \pm 0.006\text{ g cm}^{-3}.

Explanation:

Since density is calculated by division, we must sum the percentage uncertainties of the mass and volume. The final absolute uncertainty is then calculated by applying this total percentage to the density value.

Problem 2:

The initial temperature of a reaction is 20.2±0.1∘C20.2 \pm 0.1 ^\circ\text{C} and the final temperature is 25.8±0.1∘C25.8 \pm 0.1 ^\circ\text{C}. Calculate the change in temperature (ΔT\Delta T) and its uncertainty.

Solution:

ΔT=Tfinal−Tinitial=25.8−20.2=5.6∘C\Delta T = T_{final} - T_{initial} = 25.8 - 20.2 = 5.6 ^\circ\text{C}.

Absolute uncertainty in ΔT=0.1+0.1=0.2∘C\Delta T = 0.1 + 0.1 = 0.2 ^\circ\text{C}.

Final value: 5.6±0.2∘C5.6 \pm 0.2 ^\circ\text{C}.

Explanation:

For subtraction (and addition), the absolute uncertainties of the individual measurements are added together to find the total uncertainty.

Uncertainties and errors in measurement Revision - Grade 11 Chemistry IB