The Normal Distribution & Peak Shapes — Maths for Analytical Chemistry

The bell curve, everywhere

From random error to molecular speeds

The normal (Gaussian) distribution is the symmetric bell curve that random processes tend to produce — the central limit theorem guarantees it whenever a measurement is the sum of many small independent effects. It is fixed by just two numbers: the mean $μ$ (the centre) and the standard deviation $σ$ (the width).

f (x) = \frac{1}{σ 2 π} e^{- (x - μ)^{2} / (2 σ^{2})}

The normalisation constant $1/ (σ 2 π)$ ensures the total area under the curve equals one (it is a probability density). In chemistry, random measurement errors are Gaussian by virtue of many small contributing sources (thermal fluctuations, readout noise, mechanical vibration). That underpins every confidence interval and t-test used in analytical chemistry.

The practical 68–95–99.7 rule states that approximately 68 %, 95 % and 99.7 % of values from a normal distribution fall within $\pm 1 σ$ , $\pm 2 σ$ and $\pm 3 σ$ of the mean, respectively. Knowing which band a result falls into is the basis of statistical hypothesis testing.

Key results and applications

Gaussian formula

f (x) = \frac{1}{σ 2 π} e^{- (x - μ)^{2} /2 σ^{2}}

68–95–99.7 rule

P (∣ x - μ ∣ < kσ)

for

k = 1, 2, 3

: 68.3 %, 95.4 %, 99.7 %.

FWHM (spectroscopy)

FWHM = 2 2 ln 2 σ \approx 2.355 σ

z-score

z = (x - μ) / σ

— measures how many standard deviations from the mean.

Standard normal

μ = 0, σ = 1

. Tables of the cumulative distribution

Φ (z)

give exact probabilities.

Maxwell–Boltzmann

Molecular speeds follow a skewed bell; the most probable speed

v_{p} = 2 RT / M

grows with

T

The z-score standardises any normal variable: $z = (x - μ) / σ$ . From a standard normal table (or the 68–95–99.7 rule), the fraction of values above a given z is the upper-tail probability $P (Z > z) = 1 - Φ (z)$ . Key values to memorise:

Tail probabilities

z = 1.645

: 5 % one-tail |

z = 1.960

: 2.5 % one-tail

z = 2.576

: 0.5 % one-tail |

z = 3.0

: 0.135 % one-tail

Area within ±kσ

±1σ: 68.3 % | ±2σ: 95.45 %
±3σ: 99.73 % | ±1.96σ: 95.00 %

For a spectroscopic peak modelled as a Gaussian, the full width at half maximum relates to σ as $FWHM = 2 2 ln 2 σ \approx 2.355 σ$ . Because FWHM is directly measurable from a spectrum while σ is a fitted parameter, instruments and textbooks always quote FWHM. Two peaks of the same FWHM are resolved by the Rayleigh criterion when their separation exceeds the FWHM.

The three characteristic speeds of the Maxwell–Boltzmann distribution are ordered as:

v_{p} = \frac{2 RT}{M} < \overset{v}{ˉ} = \frac{8 RT}{π M} < v_{rms} = \frac{3 RT}{M}

The ratio between them is fixed: $v_{p} : \overset{v}{ˉ} : v_{rms} = 1 : 1.128 : 1.225$ . Unlike a Gaussian, the Maxwell–Boltzmann is right-skewed (zero at v = 0, longer tail at high v), so the mean slightly exceeds the mode.

Common pitfalls

Forgetting to standardise. Do not look up a probability for a raw x value — always convert to a z-score first: $z = (x - μ) / σ$ .
One-tail vs two-tail confusion. "What fraction lies above 7.08?" is a one-tail question; "what fraction lies outside ±2σ?" is two-tail (double the one-tail value). Mixing them gives the wrong probability.
Reporting FWHM as σ (or σ as FWHM). FWHM is 2.355σ, not σ. Equating them underestimates the peak width by a factor of 2.355.
Applying the Gaussian formula to the Maxwell–Boltzmann directly. MB is a distinct, skewed distribution. The most probable, mean and rms speeds are different quantities with different formulae.

Peaks and speeds

Random measurement errors are Gaussian (the basis of confidence intervals); spectroscopic line shapes are often Gaussian with FWHM = 2.355σ; and the Maxwell–Boltzmann speed distribution is a close cousin. Open the Gaussian explorer to move and widen the bell curve and see how σ controls peak width.

Gaussian explorer

Shift μ to move the peak; adjust σ to narrow or broaden it. The shaded ±1σ region always holds about 68% of the area.

Shift and widen the bell curve

f (x) = \frac{1}{σ 2 π} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}}

mean μ8.0

standard deviation σ1.5

shaded ±1σ holds≈ 68%

FWHM = 2.355σ3.53

The same bell describes random measurement error, the Maxwell–Boltzmann spread of molecular speeds, and the shape of many spectroscopic lines. σ is the single knob that sets how wide it is.

Worked example 1Applying the 68–95–99.7 rule to replicate measurements

A burette reading is modelled as normally distributed with $μ = 25.00 mL$ and $σ = 0.05 mL$ . What fraction of readings lies between 24.90 and 25.10 mL?

Worked example 2FWHM of a UV–vis absorbance peak

A Gaussian absorbance peak centred at $λ_{0} = 520 nm$ has a standard deviation of $σ = 12 nm$ . Find its FWHM, and determine the wavelengths at which the peak reaches half its maximum height.

Worked examples

z-scores, peak resolution and the Maxwell–Boltzmann link — try each before revealing the solution.

Worked example 3z-score and probability for a pH measurement

A pH meter gives readings that are normally distributed with $μ = 7.00$ and $σ = 0.04$ . What is the probability of a reading above 7.08?

Worked example 4Resolving two overlapping NMR peaks

Two ¹H NMR peaks in the same spectrum each have $σ = 0.30 ppm$ (so $FWHM = 2.355 \times 0.30 = 0.71 ppm$ ). They are centred at $δ = 3.20$ and $δ = 3.90 ppm$ .

Are they resolved (separation > FWHM)?

Worked example 5Maxwell–Boltzmann: most probable speed and the Gaussian link

The Maxwell–Boltzmann speed distribution has its mode (most probable speed) at $v_{p} = 2 RT / M$ . For N₂ ( $M = 0.028 kg mo l^{- 1}$ ) at 298 K, find $v_{p}$ , and state how the distribution changes on heating to 600 K.

Worked example 6Mean speed, rms speed and the three Maxwell–Boltzmann speeds for O₂

For O₂ ( $M = 0.032 kg mo l^{- 1}$ ) at 300 K, calculate the most probable speed $v_{p}$ , the mean speed $\overset{v}{ˉ}$ , and the rms speed $v_{rms}$ . Verify the ratio $1 : 1.128 : 1.225$ .

ChallengeChallenge — fraction of N₂ molecules above an activation speed

In a simplified Lindemann model, a reaction proceeds when two N₂ molecules collide with a relative kinetic energy exceeding $E^{*} = 40 kJ mo l^{- 1}$ at 500 K. Treating the one-dimensional speed distribution as a Gaussian with $σ_{v} = RT / M$ ( $M = 0.028 kg mo l^{- 1}$ ), estimate the fraction of molecules with kinetic energy exceeding $E^{*}$ in the x-direction. (Use the equivalent speed $v^{*} = 2 E^{*} / M$ and the tail probability $P (v > v^{*}) = 1 - Φ (v^{*} / σ_{v})$ . Take $Φ (3.28) = 0.9995$ .)

Check yourself

Four quick questions on the normal distribution, FWHM and the Maxwell–Boltzmann connection.

Question 1 of 4 · Score 0

For a normal distribution, roughly what fraction of values lie within ±1 standard deviation of the mean?

Choose an answer.