Asked by Jessy Luna
A metal M forms 2 different chloride. if 12.7g of chloride A and 16.5g of chloride B show that the figures agree with the law of multiple proportion and write their formular
Answers
Answered by
Damon
maybe one of them is MCl and the other is MCl2
a mol of Cl is about 35.5 grams
so when we added a Cl to the molecule, the mass went up by (16.5 - 12.7) =3.8 grams which is 3.8/35.5 = 0.107 mols of Cl
so we must have had 0.107 mols of M in the MCl
3.8 grams of the original MCl was Cl so
12.7 - 3.8 = 8.9 grams of M is 0.107 mols of M
so molar mass of M is 8.9 g / 0.107 mols = 83.2 grams/mol
Looks like Pb, lead :)
PbCl and PbCl2 ????
a mol of Cl is about 35.5 grams
so when we added a Cl to the molecule, the mass went up by (16.5 - 12.7) =3.8 grams which is 3.8/35.5 = 0.107 mols of Cl
so we must have had 0.107 mols of M in the MCl
3.8 grams of the original MCl was Cl so
12.7 - 3.8 = 8.9 grams of M is 0.107 mols of M
so molar mass of M is 8.9 g / 0.107 mols = 83.2 grams/mol
Looks like Pb, lead :)
PbCl and PbCl2 ????
Answered by
Damon
I have a hunch that there is more given in the problem statement than you gave us. It could be MCl2 and MCl4 or whatever However that is the general idea.
Answered by
DrBob222
I intended to let this go but since Damon made a stab at it I will be brave. Frankly, I don't know how to handle this because I can get any answer I want. Let's do an example where we know what M is.
100 g C combines with 133 g oxygen to form 233 g of an oxide.(It's CO)
100 g C combines with 266 g oxygen to form 366 g of a different oxide.(CO2)
The Law of Multiple Proportions says that for a fixed mass of an element (100 g C) then the ratio of the oxygen to form the two compounds will be in the ratio of small whole numbers (and they are because 266/133 = 2:1.
BUT suppose we don't know the identity of M AND we don't know the mass of M. All we know is MClx = 12.7 g and MCly = 16.5 g. We can assume M = a mass of say 5 g, then 12.7-5 = 7.7 for mass Cl in A and 16.5-5 = 11.5 g Cl in B. 11.5/7.7 = 1.5. Yea!! That's a ratio of 2:3 and yes small whole numbers. BUT, what if we pick another constant number for M, say 10 g?
Then 12.7-10 = 2.7 and 16.5 - 10 = 6.5 and 6.5/2.7 = 2.4 which, with some stretching to allow for "experimental error" the ratio is 2:5. Suppose we pick 1 for M, then 12.7 - 1 =11.7 and 16.5-1 = 15.5 and the ratio is 1:1.32 for a ratio of 3:4. All of these give small whole numbers but they aren't consistent. Therefore,
I think we might say it obeys the Law of M P but I don't think I could write an equation for it.
100 g C combines with 133 g oxygen to form 233 g of an oxide.(It's CO)
100 g C combines with 266 g oxygen to form 366 g of a different oxide.(CO2)
The Law of Multiple Proportions says that for a fixed mass of an element (100 g C) then the ratio of the oxygen to form the two compounds will be in the ratio of small whole numbers (and they are because 266/133 = 2:1.
BUT suppose we don't know the identity of M AND we don't know the mass of M. All we know is MClx = 12.7 g and MCly = 16.5 g. We can assume M = a mass of say 5 g, then 12.7-5 = 7.7 for mass Cl in A and 16.5-5 = 11.5 g Cl in B. 11.5/7.7 = 1.5. Yea!! That's a ratio of 2:3 and yes small whole numbers. BUT, what if we pick another constant number for M, say 10 g?
Then 12.7-10 = 2.7 and 16.5 - 10 = 6.5 and 6.5/2.7 = 2.4 which, with some stretching to allow for "experimental error" the ratio is 2:5. Suppose we pick 1 for M, then 12.7 - 1 =11.7 and 16.5-1 = 15.5 and the ratio is 1:1.32 for a ratio of 3:4. All of these give small whole numbers but they aren't consistent. Therefore,
I think we might say it obeys the Law of M P but I don't think I could write an equation for it.
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