Calculating Charge on 1 g N3- Ions: Where Did I Go Wrong?"

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Discussion Overview

The discussion revolves around calculating the charge in coulombs on 1 gram of N3- ions, addressing the participants' attempts at a solution and identifying potential errors in their calculations. The scope includes homework-related problem-solving and mathematical reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the number of N3- ions in 1 gram of nitrogen based on the atomic mass and Avogadro's number, concluding a charge of 2.065 × 10^4 C, but questions the correctness of their answer.
  • Another participant points out that the atomic mass of nitrogen is 14 Da, not 14 g, and calculates the molecular mass of N3- as 42 Da, suggesting that the charge on one N3- ion should be the charge of a single electron.
  • A different participant emphasizes the need for clarification between nitride and azide ions, indicating a potential misunderstanding in the initial problem statement.
  • One participant proposes that the calculation should be for 1 g-ion, explaining that 1 ion of nitride has a charge of 1.6 × 10^-19 C multiplied by 3, leading to a total charge of 2.89 × 10^5 C for Avogadro's number of ions.
  • Another participant corrects a unit error, clarifying that Coulombs and Amperes are different units, but acknowledges a previous misstatement regarding the charge calculation.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the charge on N3- ions, with no consensus reached on the correct method or final answer. Multiple competing models and interpretations of the problem are present.

Contextual Notes

Some participants highlight potential misunderstandings regarding the definitions of ions and their charges, as well as the distinction between different types of nitrogen ions. There is also confusion regarding the units used in the calculations.

Alioth
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Homework Statement


What is the charge in coulomb on 1 gram of N3- ions

2. The attempt at a solution
the atomic mass of nitrogen is 14 gram.
hence 14 g nitrogen will contain 6.022*10^23 ions.
therefore 1 g nitrogen will contain n=6.022*10^23/14 no. of ions.
one N3- ion has a charge of q=4.8*10^-19 C.
so n ions of N3- will carry a charge of n*q C= 2.065*10^4 C
But this answer is wrong. Where am I going wrong?
Does 14 g nitrogen contain Avogadro's no. of atoms?
 
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Alioth said:

Homework Statement


What is the charge in coulomb on 1 gram of N3- ions

2. The attempt at a solution
the atomic mass of nitrogen is 14 gram.
hence 14 g nitrogen will contain 6.022*10^23 ions.
therefore 1 g nitrogen will contain n=6.022*10^23/14 no. of ions.
one N3- ion has a charge of q=4.8*10^-19 C.
so n ions of N3- will carry a charge of n*q C= 2.065*10^4 C
But this answer is wrong. Where am I going wrong?
Does 14 g nitrogen contain Avogadro's no. of atoms?

Homework Statement


Homework Equations


The Attempt at a Solution


The atomic mass of N is 14Da (it is NOT 14g!). But the molecular mass of N3- is 14*3 = 42Da.

Another mistake: the charge on one N3- ion is just e (a single electron charge).

Hence your answer overestimates by a factor of 9.
 
Last edited:
Lousy typing gives lousy results. N3- and N3- are different things.
 
Borek said:
Lousy typing gives lousy results. N3- and N3- are different things.

Very good point. He has to clarify if he meant nitride or azide (I assumed the latter).
 
Your solution is right but the question you ask for might be for 1 g-ion.
  • 1 g-ion = (weight in grams of sum of atomic weight of atoms making the ion.)
  • Here N^3- has 14 amu so 1g-ion 14 gm.
  • So here we have to find for 14g that is 1mole of nitride ion.
1 ion of nitride has =1.6 ×10^-19 × 3 C
Therefore 6.022 × 10^23 ions has = 6.022 × 10^23 ×1.6 ×10^-19 × 3 C = 2.89 × 10^5 C.
 
Last edited:
Welcome to the PF @Strawberry1711 - just in case, please read the forum rules.

Strawberry1711 said:
6.022 × 10^23 ×1.6 ×10^-19 × 3 C = 2.89 × 10^5 A.
I hope that helps...

Stating that Coulomb equals Ampere definitely doesn't help. These are two different units.
 
Sorry
@Borek
It's 2.89 × 10^5 C.
And Thank you for telling me that I have corrected it.
 
Last edited:

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