Avogadro's Number — Question 4

Section 19-1 · Physics pages 3038–3039

Question 4 An isotope mixture can be inferred from its mean molar mass Use NA = 6.02214076 × 1023 mol−1, exact. A hypothetical element has two isotopes with assigned molar masses 20.0 and 22.0 g/mol. The measured mixture molar mass is 20.6 g/mol. A sample has mass 4.12 g. Let x be the number fraction of the heavier isotope. isotope 20 isotope 22 Separate samples or species for comparison. (a) Find x and the count of each isotope. (b) Find the heavier isotope's mass fraction and explain why it differs from x.
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Question 4 - Solution An isotope mixture can be inferred from its mean molar mass isotope 20 isotope 22 (a) Method and calculation 20(1 −x) + 22x = 20.6 ⇒x = .300. n = 4.12/20.6 = .200 mol, N22 = .0600NA = 3.613284456 × 1022. N20 = .140NA = 8.430997064 × 1022. (b) Interpretation and further calculation m22 = .0600(22) = 1.32 g, w22 = 1.32/4.12 = .320388. w22 = xM22 (1 −x)M20 + xM22 > x. Each heavier-isotope atom contributes more mass than each lighter-isotope atom. Physical check The inferred fraction lies between zero and one, and the average molar mass lies between the isotope values.

Question 4

An isotope mixture can be inferred from its mean molar mass

Use NA=6.02214076×1023mol−1N_A=6.02214076\times10^{23}\,\mathrm{mol^{-1}}, exact. A hypothetical element has two isotopes with assigned molar masses 20.0 and 22.0 g/mol. The measured mixture molar mass is 20.6 g/mol. A sample has mass 4.12 g. Let xx be the number fraction of the heavier isotope.

Diagram shown in the original SVG worksheet.

  1. Find xx and the count of each isotope.

  2. Find the heavier isotope's mass fraction and explain why it differs from xx.

Question 4 - Solution

An isotope mixture can be inferred from its mean molar mass

Diagram shown in the original SVG worksheet.

(a) Method and calculation

20(1−x)+22x=20.6⇒x=.300.20(1-x)+22x=20.6\ \Rightarrow\ x=.300.n=4.12/20.6=.200mol,N22=.0600NA=3.613284456×1022.n=4.12/20.6=.200\,\mathrm{mol},\quad N_{22}=.0600N_A=3.613284456\times10^{22}.N20=.140NA=8.430997064×1022.N_{20}=.140N_A=8.430997064\times10^{22}.

(b) Interpretation and further calculation

m22=.0600(22)=1.32g,w22=1.32/4.12=.320388.m_{22}=.0600(22)=1.32\,\mathrm g,\quad w_{22}=1.32/4.12=.320388.w22=xM22(1−x)M20+xM22>x.w_{22}=\frac{xM_{22}}{(1-x)M_{20}+xM_{22}}>x.Each heavier-isotope atom contributes more mass than each lighter-isotope atom.

Physical check

The inferred fraction lies between zero and one, and the average molar mass lies between the isotope values.