Need Help in Calculating MW of Unknown

  • Thread starter Relentless
  • Start date
In summary, the conversation discusses calculating the molecular weight and percentage of iron in an unknown iron compound. The weight of the unknown is 0.2g, and the number of moles of Fe(II) is 0.0005. The calculation for the mass of Fe(II) is 0.027925g, and the calculation for the percentage of Fe in the unknown is (0.027925g/0.2g) x 100. The conversation also mentions using the molecular weight of iron to solve for other parts of the question and provides guidance on how to do so.
  • #1
Relentless
1
0
This is my first time posting here, I hope I get some help:-)
Well here is my Information i have:


Weight of Unknown= 0.2g
# of mols of Fe(II)= 0.0005 mol

Question: Assuming that the number of atoms in Fe in the unknown compound is one and the MW(Fe)= 55.85 g/mol calculate the molecular weight og the unknown iron compound.

I've tried many things but so far I'm stuck
Any help would be great!
R

*edit*
I also have to calculate the Mass of Fe(II) I was wondering if this is the correct method I'm using:
0.0005 mol of Fe(II) x 55.85 g/mol (Fe)= 0.027925 g?

Also need to find the % Fe in the Unknown, now would the calculation be:
(0.027925 g /0.2 g) x 100?

Thanx again
 
Last edited:
Chemistry news on Phys.org
  • #2
Well everything you have said looks good. Now you just have to think if you have a total % of the Fe in the mass. Then you know what percentage of the MW is due to the Fe. Thus you can then accommodate to figure out the other element(s) percentage of the MW and figure it out.

hope this helps

pete
 
  • #3
Okay, since they tell you how many moles of iron are in 0.2 g of the compound, and they tell you that each molecule of that compound has only one iron atom in it, you also know how many moles of the molecule you have (0.0005). The molecular weight of iron is irrelevant to the first part of the question (you needed it for the later parts of the questions).

So far then, you know that for this compound, there are 0.2g per 0.0005 mol. All you need to do now is multiply numerator and denominator by a number that will set the denominator to 1 and you'll have the # of g/mol. I'll leave you to do that for yourself.
 

FAQ: Need Help in Calculating MW of Unknown

What is the MW of an unknown molecule?

The MW (molecular weight) of an unknown molecule is the sum of the atomic weights of all the atoms present in the molecule. It is measured in units of g/mol.

How do I calculate the MW of an unknown molecule?

To calculate the MW of an unknown molecule, you will need to know the number of atoms of each element present in the molecule. You can then use a periodic table to determine the atomic weight of each element and add them together to get the total MW. Alternatively, you can use online tools or software programs to calculate the MW for you.

What is the importance of calculating the MW of an unknown molecule?

Knowing the MW of an unknown molecule is important in many scientific fields, such as chemistry, biochemistry, and pharmacology. It helps in determining the molecular structure and composition of a substance, which is essential for understanding its properties and behavior. It also plays a crucial role in identifying and characterizing new compounds and developing new drugs.

Can the MW of an unknown molecule vary?

Yes, the MW of an unknown molecule can vary depending on factors such as isotope composition, molecular structure, and environmental conditions. Isotopes are atoms of the same element with different numbers of neutrons, which can affect the overall MW of a molecule. Additionally, the presence of different functional groups or molecular bonds can also alter the MW.

How can I verify the accuracy of my calculated MW for an unknown molecule?

You can verify the accuracy of your calculated MW by using different methods, such as mass spectrometry or nuclear magnetic resonance (NMR) spectroscopy. These techniques can provide more precise and accurate measurements of the MW of a molecule. Additionally, comparing your results with known values for similar molecules can also help confirm the accuracy of your calculation.

Back
Top