Calculating the empirical formula from a percentage composition

  • #1
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##
 

Answers and Replies

  • #2
etotheipi
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.
 
  • #3
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)##?
 
  • #4
etotheipi
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.
 
  • #5
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.
 
  • #6
etotheipi
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 Hamiltonian299792458

Related Threads on Calculating the empirical formula from a percentage composition

  • Last Post
Replies
2
Views
3K
Replies
2
Views
3K
  • Last Post
Replies
1
Views
8K
Replies
3
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
8
Views
3K
Replies
3
Views
32K
  • Last Post
Replies
3
Views
3K
Replies
1
Views
3K
Top