Calculating the empirical formula from a percentage composition

[Hamiltonian]

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: $Na_2S_2O_3$

 

[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.

 

[Hamiltonian]

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 $n(O)$?

 

[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.

 

[Hamiltonian]

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.

 

[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.