Chemistry Calculating the empirical formula from a percentage composition

Click For Summary
SUMMARY

The empirical formula calculation for sodium sulfide oxide, represented as NaxSyOz, can be simplified by determining the number of moles of each element based on their mass percentages. Given the mass percentages of Na (29.11g), S (40.51g), and O (30.38g), the moles are calculated using their molar masses: MNa = 23 g/mol, MS = 32 g/mol, and MO = 16 g/mol. The resulting mole ratios of Na, S, and O are approximately 1.27, 1.27, and 1.90, respectively, which simplifies to the integer ratio of 2:2:3, leading to the empirical formula Na2S2O3.

PREREQUISITES
  • Understanding of mole calculations in chemistry
  • Familiarity with molar mass concepts
  • Basic knowledge of empirical formulas
  • Ability to perform ratio simplifications
NEXT STEPS
  • Learn how to calculate moles from mass using the formula n = m/M
  • Study the concept of empirical vs. molecular formulas
  • Explore methods for determining molar ratios in chemical compounds
  • Investigate the significance of empirical formulas in chemical reactions
USEFUL FOR

Chemistry students, educators, and professionals involved in chemical analysis or synthesis will benefit from this discussion on calculating empirical formulas from percentage compositions.

Hamiltonian
Messages
296
Reaction score
193
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##
 
Physics news on Phys.org
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.
 
  • Like
Likes Hamiltonian