Calculating the empirical formula from a percentage composition

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Discussion Overview

The discussion revolves around calculating the empirical formula from a percentage composition of a compound, specifically focusing on the formula ##Na_x S_y O_z##. Participants explore different methods for determining the empirical formula based on given mass percentages of the elements involved.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant proposes a method involving setting up equations based on the mass percentages and molar masses of the elements to solve for ##x, y, z##.
  • Another participant suggests a simpler approach of calculating the number of moles for each element and finding the lowest integer ratio of these moles.
  • A later reply clarifies the notation ##n(\cdot)## as representing 'moles of' and provides a calculation for moles of sodium based on its mass and molar mass.
  • There is a discussion about assuming a total mass of 100g for the compound to simplify calculations, leading to specific mole values for each element.
  • One participant notes that the resulting mole ratio can be simplified to derive the empirical formula, suggesting that the empirical formula is ##Na_2S_2O_3## based on the calculated ratios.

Areas of Agreement / Disagreement

Participants express different methods for calculating the empirical formula, with some favoring a more complex equation-based approach while others advocate for a simpler mole ratio method. The discussion does not reach a consensus on the preferred method, as multiple approaches are presented.

Contextual Notes

Participants assume a total mass of 100g for the calculations, which may limit the generalizability of the results. The discussion also involves some unresolved steps in deriving the empirical formula from the mole ratios.

Who May Find This Useful

This discussion may be useful for students or individuals interested in chemistry, particularly those learning about empirical formulas and stoichiometry.

Hamiltonian
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Homework Statement
An inorganic salt gave the following %composition: Na = 29.11% S = 40.51% and O = 30.38%. Calculate the empirical formula of the salt.
Relevant Equations
N/A
Let the empirical formula be ##Na_x S_y O_z##
then we can concude
$$\frac {29.11}{100} = \frac{xM_{Na}}{xM_{Na} + yM_S + z M_O}$$
$$\frac {40.51}{100} = \frac{yM_S}{xM_{Na} + yM_S + z M_O}$$
$$\frac {30.38}{100} = \frac{zM_O}{xM_{Na} + yM_S + z M_O}$$
where ##M_{Na} = 23, M_S = 32,M_O = 16##
solving these equations for ##x,y## and ##z## is awfully hard can someone suggest a better way to solve this problem?

the answer is supposed to be: ##Na_2S_2O_3##
 
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The easiest way to do it is just to calculate ##n(Na)##, ##n(S)## and ##n(O)##, and then determine the lowest integer ratio of moles.
 
etotheipi said:
The easiest way to do it is just to calculate ##n(Na)##, ##n(S)## and ##n(O)##, and then determine the lowest integer ratio of moles.
what is ##n(Na)##, ##n(S)## and ##n(O)##?
 
Oh, sorry. I'm just using ##n(\cdot)## to denote 'moles of'. For instance,$$n(Na) = \frac{m(Na)}{M_{Na}} = \frac{29.11 \text{g}}{23 \mathrm{gmol^{-1}}} \approx 1.27 \text{mol}$$Do the same for the other two, and then then look at the ratio of the moles of each.
 
etotheipi said:
Oh, sorry. I'm just using ##n(\cdot)## to denote 'moles of'. For instance,$$n(Na) = \frac{m(Na)}{M_{Na}} = \frac{29.11 \text{g}}{23 \mathrm{gmol^{-1}}} \approx 1.27 \text{mol}$$Do the same for the other two, and then then look at the ratio of the moles of each.
So are you assuming there to be say 100g of ##Na_x S_y O_z## hence out of this 100g there will be 29.11g of ##Na##, 40.51g of S and 30.38g of O hence the relative number of moles of these substances will be 1.266(moles of Na), 1.266(moles of S) and 1.898(moles of oxygen). but is still don't see how you could get ##x,y,z## from this.
 
Hamiltonian299792458 said:
So are you assuming there to be say 100g of ##Na_x S_y O_z## hence out of this 100g there will be 29.11g of ##Na##, 40.51g of S and 30.38g of O hence the relative number of moles of these substances will be 1.266(moles of Na), 1.266(moles of S) and 1.898(moles of oxygen). but is still don't see how you could get ##x,y,z## from this.

The ratio 1.266: 1.266: 1.898 is equivalent to the ratio 2:2:3, when you multiply through by a constant. That's to say, for every 2 moles of Na in your sample, there are 2 moles of S and 3 moles of O. Hence, Na2S2O3.
 
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