Avogadro's Number — Question 9

Section 19-1 · Physics pages 3048–3049

Question 9 An exact counting constant does not remove sample uncertainty Use NA = 6.02214076 × 1023 mol−1, exact. A sample mass is m = 2.00 ± .02 g and its assigned molar mass is M = 40.0 ± .4 g/mol. Treat these as independent hard bounds. The desired entity count is N = mNA/M. N Nmin Nc Nmax Axes and labels specify the model; the sketch is schematic. (a) Find the central entity count and the tight allowed interval. (b) Compare the exact fractional bounds with the first-order sum of relative bounds. Why does the exact Avogadro constant not make the count exact?
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Question 9 - Solution An exact counting constant does not remove sample uncertainty N Nmin Nc Nmax (a) Method and calculation Nc = (2.00/40.0)NA = .0500NA = 3.01107038 × 1022. Nmin = 1.98 40.4NA, Nmax = 2.02 39.6NA. N/Nc ∈[.99/1.01, 1.01/.99] = [.980198, 1.020202]. (b) Interpretation and further calculation δN N ≲.02 2.00 + .4 40.0 = .0200. Exact deviations: −1.98020%, +2.02020%. The first-order approximation is nearly symmetric; the exact inverse-mass relation is not. Uncertainty in measured inputs remains even when the proportionality constant is exact. Physical check The smallest count pairs the smallest mass with the largest molar mass, rather than pairing both lower endpoints.

Question 9

An exact counting constant does not remove sample uncertainty

Use NA=6.02214076×1023mol−1N_A=6.02214076\times10^{23}\,\mathrm{mol^{-1}}, exact. A sample mass is m=2.00±.02m=2.00\pm.02 g and its assigned molar mass is M=40.0±.4M=40.0\pm.4 g/mol. Treat these as independent hard bounds. The desired entity count is N=mNA/MN=mN_A/M.

Diagram shown in the original SVG worksheet.

  1. Find the central entity count and the tight allowed interval.

  2. Compare the exact fractional bounds with the first-order sum of relative bounds. Why does the exact Avogadro constant not make the count exact?

Question 9 - Solution

An exact counting constant does not remove sample uncertainty

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Nc=(2.00/40.0)NA=.0500NA=3.01107038×1022.N_c=(2.00/40.0)N_A=.0500N_A=3.01107038\times10^{22}.Nmin=1.9840.4NA,Nmax=2.0239.6NA.N_{\min}=\frac{1.98}{40.4}N_A,\quad N_{\max}=\frac{2.02}{39.6}N_A.N/Nc∈[.99/1.01,1.01/.99]=[.980198,1.020202].N/N_c\in[.99/1.01,1.01/.99]=[.980198,1.020202].

(b) Interpretation and further calculation

|δNN|≲.022.00+.440.0=.0200.\left|\frac{\delta N}{N}\right|\lesssim\frac{.02}{2.00}+\frac{.4}{40.0}=.0200.Exact deviations: −1.98020%,+2.02020%.\text{Exact deviations: }-1.98020\%,\ +2.02020\%.The first-order approximation is nearly symmetric; the exact inverse-mass relation is not. Uncertainty in measured inputs remains even when the proportionality constant is exact.

Physical check

The smallest count pairs the smallest mass with the largest molar mass, rather than pairing both lower endpoints.