A sample of a compound contains only the elements sodium, sulfur, and oxygen. it is found by analysis to contain 0.979 g na, 1.365 g s, and 1.021 g o. determine its empirical formula.

First, convert the masses into moles using the molar weights: 23 g Na/mol, 32.06 g S/mol and 16 g O/mol.

Mol Na: 0.979/23 = 0.04256522

Mol S: 1.365/32.06 = 0.04257642

Mol O: 1.021/16 = 0.0638125

The least value is mol Na. Thus, divide all mole compositions to that of mol Na:

Mol Na: 0.04256522/0.04256522 = 1

Mol S: 0.04257642/0.04256522 = 1

Mol O: 0.0638125/0.04256522 = 1.5

The ratios must be whole numbers. To make 1.5 a whole number, multiply all ratios with 2 to keep it uniform.

Mol Na: 1*2 = 2

Mol S: 1*2 = 2

Mol O: 1.5*2 = 3

Thus, the empirical formula is Na₂S₂O₃.

To determine the empirical formula for the compound that contains 0.979 g Na, 1.365 g S, and 1.021 g O, we convert these to mole units. The molar masses to be used are:

Molar mass of Na = 23 g/mol

Molar mass of S = 32 g/mol

Molar mass of O = 16 g / mol

The number of moles is obtained using the molar mass for each element.

moles Na = 0.979 g Na / 23 g/mol Na = 0.04256

moles S = 1.365 g Na / 32 g/mol Na = 0.04265

moles O = 1.021 g O / 16 g/mol Na = 0.06326

We then divide each with the smallest number of moles obtained.

Na: 0.04256 / 0.04256 = 1

S: 0.04265 / 0.04256 = 1.002 ≈ 1

O: 0.06326 / 0.04256 = 1.49 ≈ 1.5

We then have an empirical formula of NaSO₁.₅. However, chemical formulas must have only integers as subscripts, thus, we multiply each to 2. The empirical formula is then Na₂S₂O₃ also known as sodium thiosulfate.