Easy GC problem

[Mamona]

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I can't figure out how to do it...... appreciate any help

Ammonia gas is a compound of nitrogen and hydrogen in the atomic ratio of 1:3. A sample of ammonia contains 7.933g N and 1.712g H. What is the atomic mass of N relative to H?



Answer: 13.9

 

[Mstoothlady2012]

I can't figure out how to do it...... appreciate any help

Ammonia gas is a compound of nitrogen and hydrogen in the atomic ratio of 1:3. A sample of ammonia contains 7.933g N and 1.712g H. What is the atomic mass of N relative to H?



Answer: 13.9

First thing we need to do is deterimine the # of mol of Hydrogen in Ammonia

1.712g H X 1mol/1g = 1.712 mol H

Now we need to figure out what mol of Nitrogen is present. Since the ratio of N:H is 1:3

mol of N = 1.712mol H/3 = 0.57 mol of N present.

Now, MW = 7.933g/0.57 mol = 13.91 g/mol is the atomic mass of N relative to H

I hope I did this right!

 

[Mamona]

Thank you very much it makes sense now!!!

 

[DentalDeac]

I still don't really get it. You divided the grams of N present by the mols of N so how is that relative to H?

 

[DDSguyLA]

think of it this way...

1.712 Hydrogen has 7.933 Nitrogen
if there are 3 H for every 1 N

Then you will have 3 x 7.933 = about 24 and then 24 divide by 1.7 = about 14

 

[Mstoothlady2012]

I still don't really get it. You divided the grams of N present by the mols of N so how is that relative to H?

because I found the mol of N relative to mol of H

 

[Mstoothlady2012]

think of it this way...

1.712 Hydrogen has 7.933 Nitrogen
if there are 3 H for every 1 N

Then you will have 3 x 7.933 = about 24 and then 24 divide by 1.7 = about 14

what you did doesn't make sense to me. You found g of H and then divided it by given g of H????

 

[harrygt]

First thing we need to do is deterimine the # of mol of Hydrogen in Ammonia

1.712g H X 1mol/1g = 1.712 mol H

Now we need to figure out what mol of Nitrogen is present. Since the ratio of N:H is 1:3

mol of N = 1.712mol H/3 = 0.57 mol of N present.

Now, MW = 7.933g/0.57 mol = 13.91 g/mol is the atomic mass of N relative to H

I hope I did this right!

Missthoothy, you got the right answer, but I see that you used the atomic mass of H at the part colored red. well, you are supposed to solve this without knowing the atomic mass of N and H. Otherwise, you could just get the atmoic masses from the periodic table and divide them by each other😀 Makes sense?
Try to solve it without using the atomic mass of H😀 Not hard to do that

 

[boxdesker]

"atomic mass of N relative to H"

The ratio of atomic mass of N relative to atomic mass of H is equal to the ratio of "(g of N)/(mol of N)" to "(g of H)/(mol of H)" which equals "(g of N/mol of N)/(g of H/mol of H)"

Plugging in the values:

[(7.933 g of N)/(1 mol N)]/[(1.712 g of H)/(3 mol H)] = 13.9

But if this problem came up on the DAT--you should automatically know that it's 14 without any given piece of information for the reason that harrygt gave.

 

[DDSguyLA]

what you did doesn't make sense to me. You found g of H and then divided it by given g of H????

It worked... didn't it??