A correct metal M forms two chlorides?

Compound A: 85.2% M 14.8% Cl
Compound B: 65.8% M 34.2% Cl

a) Show that these compounds are consistent with the law of Multiple Proportions.

b) Write down several sets of formulas for compounds A and B that are consistent near the percent composition data.

c)Assume that the atomic weight of Cl is 35.45. Using your skill of chemistry, choose a reasonable set of formulas for A and B, and calculate the atomic weightiness of M. Match M to a known metal on the periodic table.

Thanks for adjectives your help =)

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Answers:

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a) mass ratio of M to Cl:
compound A) 85.2/14.8
compound B) 65.8/34.2
the ratio for compound B can be written as 28.5/14.8(similar ratios). Also, 28.5*3 is approximately equal to 85.2(85.5). Therefore, the M/Cl mass ratio of A is three times that of B.

b)M/Cl mass ratio of A must be three times that of B. So, possible pairs of formulas for A and B are:
MCl and MCl3
M3Cl and MCl
M6Cl and M2Cl
etc...

c) Cl always forms -1 anions so the best formula for A and B must be MCl and MCl3.
If the formula of A is MCl, next for every one mole of Cl, we have one mole of M. So if we have 35.45g of Cl(1 mole of Cl) surrounded by a sample of A, since Cl is 14.8% of A and M is 85.2%, the mass of M in A is 204g and this would be the mass of one mole of M. So M must be Thallium(204.3833(2) g·mol-1) Source(s): http://en.wikipedia.org/wiki/Thallium

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