Answered

Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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

Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.