Question

what’s the empirical and molecular formula for a compound with 20.2% sodium, 37.6% sulfur, 42.2%oxygen and a molar mass of 682.8g/mol?

Answer 1

Answer:

Empirical formula: Na3S4O9

Molecular formula: Na6S8O18

Explanation:

1) To find the empirical formula, which is the simplest whole number ratio of each element that makes up this unknown compound, we first divide the portion of each element by it's respective atomic mass.

Na= 23g/mol, S= 32g/mol, O=16g/mol

Na = 20.2/23 = 0.878mol

S = 37.6/32 = 1.175mol

O = 42.2/16 = 2.638mol

Next, we divide each number of moles by the smallest value (0.878mol)

Na = 0.878/0.878 = 1

S = 1.175/0.878 = 1.34

O = 2.638/0.878 = 3.004

Next, we multiply each decimal fraction by 3 to make it whole no:

Na = 3

S= 4.02 ~ 4

O= 9.012 ~ 9

The simplest ratio between Na, S and O is 3:4:9 hence, the empirical formula is Na3S4O9

2) To get the molecular formula, we use the molecular weight of each element in the empirical ratio, and the overall molar mass of the compound i.e.

Molar mass of compound= 682.8g/mol

molecular weight of Na3S9O4 = (3 x 23g/mol) + (4 x 32 g/mol) + (9 x 16 g/mol)

= 69 + 128 + 144

= 341 g/mol

Hence, (Na3S4O9)n = 682.8g/mol

= (341)n = 682.8

n = 682.8/341

n = 2.002

n~ 2

Hence, molecular formula = (Na3S4O9)2

= Na6S8O18