Percentage of NaNO3 in 5.37g Mixture: 1.61g Sodium

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Discussion Overview

The discussion revolves around calculating the percentage by mass of NaNO3 in a mixture containing NaNO3 and Na2SO4, given the total mass of the mixture and the mass of sodium present. The scope includes mathematical reasoning and problem-solving related to chemistry.

Discussion Character

  • Mathematical reasoning, Homework-related, Technical explanation

Main Points Raised

  • One participant states the total mass of the mixture is 5.37g and that it contains 1.61g of sodium, prompting a calculation for the percentage of NaNO3.
  • Another participant questions the identification of sodium, seeking clarification on its formula.
  • A subsequent reply confirms that sodium is represented by Na.
  • One participant suggests that the problem can be solved by performing a division operation to find the percentage.
  • Another participant expresses confusion about what to divide, noting that both compounds contain sodium.
  • A participant provides a calculation for the total percentage of sodium in the mixture, arriving at approximately 29.98%.
  • Further, a participant outlines a method involving moles, proposing to set up an equation to determine the amounts of NaNO3 and Na2SO4 based on the given mass of sodium.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the method for calculating the percentage of NaNO3, with some expressing confusion and others suggesting different approaches.

Contextual Notes

There are unresolved assumptions regarding the distribution of sodium between the two compounds and the specific calculations needed to determine the percentage of NaNO3.

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a mixture of NaNO3 and Na2SO4 of mass 5.37g contains 1.61g of sodium, what is percentage by mass of NaNO3 in the mixture?
 
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Hmm. Sodium? Which formula is that?
 
Um...Na?
 
You have already solved the question, just do the dividing operation and multiply by 100 to learn the percentage.
 
Divide what by what? both compunds contain Na...
 
1.61/5.37*100=29.98% total sodium is present in the mixture.

Na=23, N=14, O=16, S=32 and NaNO3 is 48+14+23=85 g/mol, whereas Na2SO4 is 46+32+64=142 g/mol.

Devise an equation, in which x moles of nitrate and y moles of sulfate is present, and make it equal to 5.37. Use mole amounts to learn x and y by making them equal to 1.61. It shouldn't be hard...
 

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