Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Massimo on July 02, 2020, 04:57:50 PM
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Is it possible to calculate the number of moles (n), the molar mass (M) AND the mass (m) of a substance, if you don't know what the substance (Let's call it SUBSTANCE X) is? Example; I have 0.387 grams of CHLORIDE within 1 gram of a solution. What is the name and molar mass of the molecule that's taking up the remaining mass of the gram (0.613 grams)? I have been told to use the stoichiometric coefficients to work backwards to the solution... Which is 1(substance X)2Cl2
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Your problem has many solutions (XCl, XCl2 ...). You can limit them on the basis of salt solubility (e.g. you will never get such a concentration of NaCl).
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Ok. Thank you for that. Do you know how I go about calculating the molar mass of the unknown substance, (M)?
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You need to collect the solubility data of different chlorides at the same temperature (your choice). Then you need to convert the solubility into the chloride content. All solutions that have higher content in the saturated state can be immediately discarded - maybe 2 or 3 will remain and for them, you need to complete the calculations. You know the mass of chlorides, and therefore the number of moles - that's a simple stoichiometry.
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Example; I have 0.387 grams of CHLORIDE within 1 gram of a solution. What is the name and molar mass of the molecule that's taking up the remaining mass of the gram (0.613 grams)?
1 gram of SOLUTION or 1 gram of SOLUTE? Note that in the first case all you know is that the mass of the metal is lower than 0.613 g, you don't know what it really is.
As AWK wrote, this question can have no single answer.
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Excuse my ignorance, I'm very new to chemistry. So, from what I know 9based on my very limited knowledge) the Cl2 is soluble according to solubility rules. Oh ok, so look at the solubility rules and see what my options are? The question says that M(Unknown substance) is a metal ion.
This is the information. I need to find the NUMBER OF MOLES OF CL-, NUMBER OF MOLES OF (M), THE MASS OF M AND THE MOLAR MASS OF M.
Some of the formulations include CaCl2. An observant worker noticed that one batch from the supplier appeared to have powder flakes with a different appearance and the judging from the volume of material in the container, had a density much greater than usual. Suspecting that it was not CaCl2 at all but a different salt, further investigation was carried out.
Analysis of an aqueous solution made up from this material determined that 1.000 g of the solid contains 0.387 g of chloride. Assume that the solid sample contains only a salt of the form MCl2, where the unknown metal cation M accounts for the remaining mass. This data may be used to infer the identity of M as follows.
Q3(a) 0.387 g of chloride ions corresponds to how many moles? Show your working and enter your answer with units in the table below.
Q3(b) Consider what is the (stoichiometric) ratio of metal ions to chloride ions in the salt MCl2. Based on that and on your answer to Q3(a), how many mole of metal ions are present in 1.000 g of sample? Enter your answer with units in the table.
Q3(c) What is the molar mass of the metal ion?
hint – you need the number of mol from Q3b, along with the mass of the metal ions in a 1.000 g sample of MCl2. Enter your answer with units in the table.
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According to forum rules, you should show your attempt.
Start from Q3.
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Sorry!
So question 3a:
n (number of moles) = m (mass of chloride present)/ M (molar mass of chloride).
n(cl-)= 0.387/70.9=0.005
3b:-0.387 grams of Cl2
-2 mol of Cl- is 70.90
-2 mole of Cl-
-1 mole of M.
Therefore;
.387grams x 1mol/70.90 x 1(mol of Cl-)/1( mole of M) = 0.005
3c: The mass of Cl- is 0.387 grams.
So 1.000 gram - 0.387 grams= 0.613 grams (the remaining mass of M)
3d: I am stuck
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No intermediate results are rounded off. An experienced student can do it reliably, but then he would leave four significant digits.
In addition, you have chlorides in the solution, not chlorine. A short cut can lead to a swamp.
From your first post, we have the information: "I have 0.387 grams of CHLORIDE within 1 gram of a solution", so you have chlorides, metal cation, and water in the solution.
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Thank you for this. I am definitely not an experienced student!
I have reached a point in the equation where I have 0.005 moles of M, and the mass is 0.613 grams. So when I do the equation 0.613/0.005=122.6, but 122.6 isn't a molar mass on the periodic table so that can't be the answer, right?
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You have 0.387 g of chloride in 1 gram of solution. What mass of anhydrous salt (LiCl, CsCl, MgCl2, CaCl2, etc.) can be obtained from this mass of chlorides? Water will supplement the mass up to 1 gram. For calculations, the stoichiometry of the chemical formula must be used. One of the salts mentioned by me can be excluded at this stage of calculations. Then you need to check the salt solubility at different temperatures. Probably none of these salts will form a solution with this chloride content at room temperature.
The calculation method based on the stoichiometry of the chemical formula can be found in every general chemistry textbook.
1.000 g (4 significant digits) even requires intermediate calculations with a minimum of 5 significant digits.
0.387/Cl = 0.010916 (0.010915860434942035) - your approximation at the beginning of the calculation gives almost a 10% error.
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Are you sure your problem is with the solution? Or maybe it concerns solid calcium chloride hydrate? Some CaCl2 hydrate contains 0.3874 g of chloride in 1 g of the hydrate.
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you are given this
MCl2
with
mass Cl = 0.387g
mass M = 1.000g - 0.387g = 0.613g
right?
and asked to find
(1) moles Cl
(2) moles M
(3) molar mass M = mass M / moles M
solution
********
0.387g Cl 1 mol Cl
mole Cl = ------------ x ------------ =____
1 35.45g Cl
___ mol Cl 1 mol M
mole M = ------------- x ----------- = _____
1 2 mol Cl
mass M 1.00g - 0.387g
molar mass M = ----------- = ----------------- = ___ g/mol
mol M ___ mol M
fill in the blanks
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mass M = 1.000g - 0.387g = 0.613g
right?
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mass M 1.00g - 0.387g
molar mass M = ----------- = ----------------- = ___ g/mol
mol M ___ mol M
You forgot about water of crystallization (or water in solution).
Such help is not worth much.
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mass M = 1.000g - 0.387g = 0.613g
right?
*******
mass M 1.00g - 0.387g
molar mass M = ----------- = ----------------- = ___ g/mol
mol M ___ mol M
You forgot about water of crystallization (or water in solution).
Such help is not worth much.
@AWK you are wrong because you read what the Massimo said in their initial post; if you read the full question as they listed later then MINO has given the correct way to solve the question Massimo was set.
@Massimo if you have a question make sure you ask the actual question fully not half of it, not your interpretation of it and not some made up bit.
The question you were given does not mention 1g of solution is says 1.000g of solid. By asking about a solution you sent AWK off down a rabbit hole.
The question you had also states to "Assume that the solid sample contains only a salt of the form MCl2, where the unknown metal cation M accounts for the remaining mass" by not giving this information in your OP you made the question you did ask and answers given more complicated than reality.
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It doesn't matter if it's a solution (as Massimo wrote in the first post) or a solid (as he wrote in the second post, though, he didn't write that this is a correction). In both cases, the initial stage of calculations is identical. Massimo calculated the task as You and MNIO suggested and came to the conclusion that it gives the wrong result. The resulting atomic mass does not reasonably correspond to any of the known mono-, bi- or trivalent metals. So the only possibility is a salt hydrate (or salt solution) - compound crystallized from water, which means the question is:
Hydrate (solution) of which metal chloride may contain 38.7% chlorine. Such hydrate exists and contains exactly 38.74% Cl, which gives less than 0.1% error.
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The resulting atomic mass does not reasonably correspond to any of the known mono-, bi- or trivalent metals.
Sure it does.
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A potential compound possibly matching the problem also forms hydrates, and Massimo never wrote that the sample was dried to constant weight.
The problem results from the imprecision of the question or the ambiguity of the translation.
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@AWK You are trying to apply actual real world knowledge and experience to a question that as actually fully given is solvable without it. I agree that it would not work like the question suggests in reality but so many homework questions fail that test. If Massimo had not asked an incorrect and incomplete question to begin with and then rounded up badly later it would have been much easier to guide them to the "correct" answer needed.