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What's the empirical formula of 36.48% Na 25.41% S 38.11% O

Sagot :

MsARMY
a compound is found to contain 36.48% Na, 25.41% S, and 38.11% O. find its empirical formula. % in such problems are by weight, & can be changed to grams out of a 100 grams total using molar mass, find moles: 
36.48g Na @ 23 g/mol = 1.6 moles
Na 25.41g S, @ 32 g/mole = 0.8 mol
S 38.11g O. @ 16 g/mol = 2.4 mol
O ratio the moles, by dividing all by the smallest: 1.6 mol Na / 0.8 = 2 mol Na 0.8 mol S / 0.8 = 1 mole
 S 2.4 mol O / 0.8 = 3 mol O 

your empirical formula os Na2SO3 

It was hard when I took chemistry too! I passed with a B

empirical formula is the simplest ratio of whole numbers of components making up a compound

the percentages have been given so we are finding masses for 100 g of the compound

Na S O

mass 36.48 g 25.41 g 38.11 g

number of moles

36.48 g/23 g/mol 25.41 g /32 g/mol 38.11 g / 16 g/mol

= 1.6 = 0.8 = 2.4

divide by the least number of moles

1.6 / 0.8 = 2 0.8/0.8 = 1 2.4 / 0.8 = 3

the number of atoms of each element

Na - 2

S - 1

O - 3

empirical formula is - Na₂SO₃

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