Solving a Chemistry Equation: Question 10

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Discussion Overview

The discussion revolves around solving a chemistry equation from a homework assignment, specifically focusing on question 10 from a provided exam paper. Participants are exploring methods to derive equations based on atomic masses and empirical formulas.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant describes their approach of creating two equations in terms of variables x and y, using the total atomic mass of 571.5 as a basis for one of the equations.
  • Another participant questions the charge of the resulting ion after removing certain elements from the formula, indicating a potential exploration of ionic charges.
  • Some participants inquire about calculating percentages of the total mass, specifically asking for 14.04% of 571.5, which is later confirmed by one participant as part of their solution process.
  • A participant reports finding a value for x based on their calculations, indicating a close approximation to one of the answer options, and expresses gratitude for assistance received.

Areas of Agreement / Disagreement

The discussion does not show clear consensus, as participants are exploring different aspects of the problem, including calculations and theoretical considerations.

Contextual Notes

Limitations include the lack of clarity on how to derive the second equation and the dependence on the definitions of atomic masses and empirical formulas. Some mathematical steps remain unresolved.

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Homework Statement



http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s11_qp_12.pdf

question 10

Homework Equations



The Attempt at a Solution


The way I am doing it, is to get two equations in terms of x and y and then equate them. For the first equation I used the fact that all of the element's individual atomic masses will add up to 571.5. So i got 27x + 28.1y = 172
My problem now is how to get the second equation? I think it has something to do with the empirical formula, but I can't figure it out. Help!
 
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If you remove the Al and Si from the formula, what is the charge on the resulting ion?
 


What is 14.04% of 571.5?
 


Borek said:
What is 14.04% of 571.5?

I think I got it, thanks to you. So for x we find 14.17% of 571.5. That answer we divide by 27, (because we know that the mole of aluminium is 1), and the x i got is 2.999(pretty close to one of the options). Same for y. Thanks!
 

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