Identify 2 Unknown Elements in a Chemical Equation?

In summary, By using the given information and applying the equation MW(MX)=MW(M)+MW(X), it can be concluded that the metallic element M is silver (Ag) and the element X is chlorine (Cl). This is based on the molar masses of MX and M, which were calculated to be 143 g/mol and 107.8 g/mol, respectively. From this, the molar mass of X was determined to be 35.23 g/mol, which matches the molar mass of chlorine (35.453 g/mol). Therefore, the identities of M and X are silver and chlorine, respectively.
  • #1
Lori

Homework Statement


A silly sample of a metallic element M, reacts completely with 0.0237 mols of X2 to form 6.78 g MX. What are the identities of M and X?

Homework Equations



mols = grams/molar mass[/B]

The Attempt at a Solution


(kinda have no clue what to do because I'm not sure what to do. There aren't any elements to compare the unknown with !)

Here's the equation i got

M + X2 > MX
balancing it: 2M + X2 > 2MX

6.78g MX * (?)

I honestly don't know what to start with and would appreciate if someone gave me hints ! :(
 
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  • #2
Hello,

As long as the reaction is complete and you know the moles of one of the substances, you automatically know the moles of the other substances.

The mass of one of them is given, so you can calculate its molar mass. This can lead you to the solution (search in periodic table).
 
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  • #3
DoItForYourself said:
Hello,

As long as the reaction is complete and you know the moles of one of the substances, you automatically know the moles of the other substances.

The mass of one of them is given, so you can calculate its molar mass. This can lead you to the solution (search in periodic table).
I get that MX has a molar mass of 143 g/mol but that's not for separate elements!

Actually, i just figured out that i was given 5.11 grams of X?

Would M be Ag? I found its molar mass to be 107.8g/mol

I also found X to be Ga because i found its molar mass is 70.46 g/mol
(Im pretty sure i messed up! .-. I'll post my work when i wake up~ tomorrow xD)
 
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  • #4
Lori said:
Actually, i just figured out that i was given 5.11 grams of X?
That helps.
Lori said:
Would M be Ag? I found its molar mass to be 107.8g/mol I also found X to be Ga because i found its molar mass is 70.46 g/mol
How did you get these values?
 
  • #5
mfb said:
That helps.How did you get these values?
I was actually given 5.11 grams of M
 
  • #6
0.0237 mols X2 * (2 mol MX / 1 mol X2)
=0.0484 mol MX

6.78g MX/0.0484 mol Mx =143 g/mol

0.0237 mol x2 * (2 mol M/2 mol X) = ?? Mols M

5 .11g M / ?? Mols M = 107.8 g/mol
 
  • #7
Lori said:
I was actually given 5.11 grams of M
Okay, that makes more sense.
I agree with silver then.

The molar mass of X2 that you probably calculated earlier is fine as well - what does that imply for an atom of X?
 
  • #8
Lori said:
0.0237 mols X2 * (2 mol MX / 1 mol X2)
=0.0484 mol MX

6.78g MX/0.0484 mol Mx =143 g/mol

0.0237 mol x2 * (2 mol M/2 mol X) = ?? Mols M

5 .11g M / ?? Mols M = 107.8 g/mol

You calculated the molar mass of MX to be 143 g/moland the molar mass of M to be 107.8 g/mol.
So you can calculate the molar mass of X directly from these data.

Beware: It is 0.0474 and not 0.0484 mols MX.
 
  • #9
Lori said:
I also found X to be Ga because i found its molar mass is 70.46 g/mol

Beware: X2.
 
  • #10
Lori said:
Would M be Ag? I found its molar mass to be 107.8g/mol

That must be why the problem statement said: 'A silly sample of a metallic element M.'
silly = silvery :oldeyes:
 
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  • #11
I like Serena said:
That must be why the problem statement said: 'A silly sample of a metallic element M.'
silly = silvery :oldeyes:
Actually, i ready 5.11 g as Silly ! XD Look~
https://imgur.com/a/OdEcl
OdEcl
 
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  • #12
Funny, as 5.11 g is not necessary to solve the problem.
 
  • #13
Since there is 5.11 g of M and 6.78 g of MX in the products, can i subtract the two to get grams of X2?

So i can do 0.0237 mols X2 * (2moles X / 1 mole X2) = 0.0474 moles X

6.78 g - 5.11 g = 1.67 g X2

1.67 g X2 / (.0237 moles X2) = 70.46 g/mols X2 /2 = 35.23 g/mol of X = Chlorine (35.453 g/mol)
 
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  • #14
DoItForYourself said:
You calculated the molar mass of MX to be 143 g/moland the molar mass of M to be 107.8 g/mol.
So you can calculate the molar mass of X directly from these data.

Beware: It is 0.0474 and not 0.0484 mols MX.
How do you use just the molar masses to figure out X?
 
  • #15
Lori said:
How do you use just the molar masses to figure out X?
As long as MW(MX)=143 and MW(M)=107.8, you can use the equation MW(MX)=MW(M)+MW(X). The only unknown is MW(X).

From the MW (molecular weight) of X, you can define the chemical element X directly.
 
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  • #16
DoItForYourself said:
As long as MW(MX)=143 and MW(M)=107.8, you can use the equation MW(MX)=MW(M)+MW(X). The only unknown is MW(X).

From the MW (molecular weight) of X, you can define the chemical element X directly.
Oh ! ! that makes sense! Thanks! This confirms my answer too!
 

1. What is the process for identifying unknown elements in a chemical equation?

The process for identifying unknown elements in a chemical equation involves several steps. First, you need to balance the equation by making sure that the number of atoms of each element is the same on both sides. Then, you can use the atomic mass of each element to calculate the total mass of each side of the equation. Next, you can compare the masses of the two sides and identify any discrepancies. Finally, you can use this information to determine the identity of the unknown element.

2. How do you determine the atomic mass of an element?

The atomic mass of an element is determined by adding the number of protons and neutrons in the nucleus of an atom. This can be found on the periodic table, as the atomic mass is listed below the element's symbol. It is important to note that the atomic mass may be an average of different isotopes of the same element.

3. Can an element have more than one atomic mass?

Yes, an element can have more than one atomic mass. This is because some elements have multiple isotopes, meaning they have different numbers of neutrons in their nucleus. This can affect the overall atomic mass of the element, resulting in different values.

4. What if the masses of the two sides of the equation are not equal?

If the masses of the two sides of the equation are not equal, it could indicate the presence of an unknown element. This could also mean that there was an error in the balancing of the equation or the calculation of the masses. It is important to double check all steps to ensure accuracy in identifying the unknown element.

5. Are there any limitations to identifying unknown elements in a chemical equation?

There are a few limitations to identifying unknown elements in a chemical equation. First, the equation must be balanced and accurate in order to determine the identity of the unknown element. Additionally, if the masses of the two sides of the equation are very close, it may be difficult to accurately identify the unknown element. In these cases, further testing and analysis may be necessary.

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