# Periodic table message question

This problem requires the use of algebraic reasoning to derive and solve a fraction linear equation based on the concepts of moles and conservation of...

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Mole Fraction Revisited A. Mancott Queensborough Community College Bayside. NY 11364

This problem requires the use of algebraic reasoning to derive and solve a fraction linear equation based on the concepts of moles and conservation of moles. Inaddition, the concepts of sum of moles and mole fraction are required. This problem is applicable for a first-semester general chemistry student. Quesilon

The following data pertains to a mixture of KC1, NaC1, and LiCI. Given: (1)the weight of KC1 + NaCl = 2.370 g; (2)the weight of NsCl + LiCl = 1.290g; and (3)the weight of AgCl derived from the mixture = 6.435 g. Formula weights: KC1 = 75.00,NaCl = 58.00, LiCI = 42.00,AgCI = 143.0.

JOHN J. ALEXANDER University of Cincinnati Cincinnati. Ohio 45221

ing students' grasp of trends in periodicity demanding intellectual skill a t the applications level. Clearly, the nature of the message, which may or may not apply directly to chemistry, is limited by the restrictions that no element can be represented by more than two letters and that, once a letter combination is used, it cannot be repeated elsewhere in the table. Furthermore, it is desirable to impose the additional restriction that no word he continued from one row (period) to the next. Despite these limitations, i t has been possible to integrate some very interesting messages, chemical and nouchemical, into this format. Quesilon

Complete the following abbreviated periodic tahle using the fictitious elements discussed below:

Calculate the mole fraction of LiCl in the mixture. Acceptable Solutlon

Let X equal the amount, in grams, of KC1 in the mixture. Therefore, 2.370 - X = g NaCl in mixture X - 1.080 = g LiCl in mixture mole of KC1 = X175.00 mole of NaCl = (2.370- XV58.00 mole of LiCl = ( X - 1.080)142.00 mole of AgCl = 6.4321143.0 mole of KC1 NaCl LiCl = mole of AgCl [X/75.00] [(2.370- X)/58.00] [(X - 1.080)142.00] = [6.435/ 143.01

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Solve for X.

X = 1.500 g = g KC1 in mixture 2.370 - X = 2.370 - 1.500 = 0.8700g = g NaCl in mixture X - 1.080 = 1.500 - 1.080 = 0.4200g = g LiCl in mixture mole of KC1 = 1.500175.00= 0.0200 mole of NaCl = 0.8700158.00= 0.0150 mole of LiCl = 0.4200142.00= 0.0100 sum of male = 0.0200 0.0150 0.0100 = 0.0450 mole fraction of LiCl in mixture = 0.010010.0450= 0.222

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Element Fa has a is1 electron configuration. Elements It,V, and In form -1 ions. Element I t is more electronegative than element V but is less electronegative than element In. Elements Is, Oh, FI, and I are arranged in order of increasing ionization potential and comprise the group . . of elemints that form +1 ions. Elements Er,. Y.. G,. S, and Me compose the group . - of inert gases and are arranged in order of decreasing size. Elements Fo, Ee, and Sc combine with element V to give ionic compounds of general formula XV2. The atomic number of Sc is higher than that of Ee but is lower than that of Fo. Elements T, Ur, and Re are transition metals and are listed in order of increasing metallic character.

Acceptable Solutlon

Periodic Table Message Question Milton J. Wleder Metropolitan State College Denver, CO 80204

The structure of the periodic tahle is a topic of fundamental importance in general chemistry. This question presents an interesting and somewhat whimsical technique for test320

Journal of Chemical Education

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