Determine the empirical formula of a compound whose percentage composition
is 50.05% S and 49.95% O by mass.
Guide Questions
1. What are the elements involved in the sample problem?
2. What will be solved in this problem?
3. What is empirical formula?​


Sagot :

Answer:

1. Sulfur (S) and Oxygen (O)

2. empirical formula (mole ratio)

3.  or Sulfur Dioxide

Solution:

G: 50.05% S ; 49.95%O

R: empirical formula

A: find the mole ratio

S:

assume 100g:

50.05% S = 50.05g S

49.95% O = 49.95g O

convert from mass to moles:

[tex]50.05g S(\frac{1 mol S}{32.07g S}) = 1.5605 mol S[/tex]

[tex]49.95g O(\frac{1 mol O}{16.00g O}) = 3.1219 mol O[/tex]

solve for mole ratio:

*1.5605mol is the lowest amongst the moles so use it as the divisor

[tex]\frac{1.5605 molS}{1.5605} = 1[/tex]

[tex]\frac{3.1219 molO}{1.5605} = 2[/tex]

use the mole ratio to write the formula:

[tex]SO_{2}[/tex]

A: The empirical formula is [tex]SO_{2}[/tex] more commonly known as sulfur dioxide