Gravimetric Analysis Homework: Mass % of Barium Ions in Mixture

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Discussion Overview

The discussion revolves around a homework problem involving gravimetric analysis to determine the mass percent of barium ions in an impure mixture. Participants analyze the method of calculating the mass percent based on the precipitation of barium ions as barium sulfate.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The initial problem statement includes the mass of the impure mixture and the mass of dried barium sulfate obtained after precipitation.
  • One participant presents a balanced equation for the reaction and calculates the number of moles of barium sulfate produced.
  • The molar mass of barium sulfate is calculated, and the number of moles is used to find the mass of barium ions in the mixture.
  • Another participant confirms the approach but points out a typographical error in the initial calculations regarding the mass of barium sulfate.
  • A later reply emphasizes that the molar ratio between barium and barium sulfate is 1:1, suggesting that this ratio is sufficient for solving the problem without converting to moles.

Areas of Agreement / Disagreement

Participants generally agree on the method of using stoichiometric ratios to solve the problem, but there are differing opinions on the necessity of converting to moles for the calculations.

Contextual Notes

There are unresolved aspects regarding the initial calculations, including the typographical error noted by one participant, which may affect the accuracy of the results presented.

leah3000
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Homework Statement


An aqueous solution containing 1.728g of an impure mixture was analysed for barium ions by adding sufficient sulphuric acid solution to completely precipitate the barium ions as its sulphate. The precipitate was filtered, washed and dried to constant mass and finally weighed. What is the mass percent of barium ions in the mixture, if 0.408g of dried barium sulphate was obtained?


Homework Equations



I used this as the balanced equation but somehow it looks funny to me...

BaX + H2SO4 ------> BaSO4 + HX
1.728g 0.1408g

The Attempt at a Solution



no of mols BaSO4 produced = mass/Mr

Mr of BaSo4= (137.3)+ (32.1) + (16x4)
= 233.4

mols= 0.1408/233.4
=6.033x10^-4

BaX : BaSO4
1 : 1

mass Ba^2+ = molsx Mr
=(6.033x10^-4) x 137.3
=0.0828g

% of Ba^2+ present in mixture= (0.0828/1.728) x100
= 4.79%

Am i on the right track? Plz help!
 
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Hi leah3000! :smile:

You are very much on the right track!

Only one thing: you seem to have made a typo followed by the corresponding calculation errors when you wrote 0.1480 instead of 0.480.
 
oops. thanks for the help
 
The only thing that matters is that molar ratio between barium and barium sulfate is 1:1. This is enough to solve the problem.

Also note that while it is perfectly correct to convert everything to moles and back to masses, it is not necessary. See stoichiometric calculations using ratios for details.
 

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