Calculating the empirical formula from a percentage composition

• Chemistry
• Hamiltonian
In summary, the empirical formula for the given substance is ##Na_2S_2O_3## and the easiest way to determine this is by calculating the moles of each element and finding the lowest integer ratio.

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##

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.

Hamiltonian

1. What is the purpose of calculating the empirical formula from a percentage composition?

The purpose of calculating the empirical formula from a percentage composition is to determine the simplest whole number ratio of elements present in a compound. This information is important for identifying the chemical formula and understanding the chemical properties of the compound.

2. How is the empirical formula calculated from a percentage composition?

The empirical formula is calculated by converting the percentage of each element in the compound into moles, finding the smallest whole number ratio of moles, and using this ratio to determine the subscripts in the chemical formula.

3. What information is needed to calculate the empirical formula from a percentage composition?

To calculate the empirical formula, you need the percentage composition of each element in the compound and the molar mass of the compound.

4. Can the empirical formula be the same as the molecular formula?

Yes, the empirical formula can be the same as the molecular formula if the compound is made up of simple whole number ratios of elements. However, if the compound has a more complex structure, the empirical formula may be different from the molecular formula.

5. What are some limitations of calculating the empirical formula from a percentage composition?

One limitation is that the empirical formula does not give information about the actual number of atoms in a compound, only the ratio of elements. Additionally, it may not accurately reflect the true structure of the compound if it has a more complex arrangement of atoms.