Respuesta :
Answer:
The empirical formula of the compound is [tex]SO_3[/tex].
Explanation:
Suppose in 100 grams of compound there is 40% of sulfur and 60% of oxygen by weight.
Mass of sulfur = 40% of 100 g = [tex]\frac{40}{100}\times 100=40 g[/tex]
Mass of oxygen = 60% of 100 g = [tex]\frac{60}{100}\times 100=60 g[/tex]
Moles of sulfur = [tex]\frac{40 g}{32 g/mol}=1.25 mol[/tex]
Moles of oxygen = [tex]\frac{60 g}{16 g/mol}=3.75 mol[/tex]
In order to determine its empirical formulas divide smallest number of moles by each number of moles of elements:
Sulfur = [tex]\frac{1.25 mol}{1.25 mol}=1[/tex]
Oxygen = [tex]\frac{3.75 mol}{1.25 mol}=3[/tex]
Empirical formula = [tex]S_1O_3[/tex]
The empirical formula of the compound is [tex]SO_3[/tex].
The empirical formula of the compound is [tex]SO_3[/tex]
What is the Empirical formula?
Empirical formula is the simplest whole number ratio of atoms of each element in a compound.
Calculation:
Given, the amount of sulfur is 40%
The amount of oxygen is 60%
- Step 1:, we calculate the mass of the elements
Mass of sulfur [tex]= \dfrac{40}{100} \times100= 40 g[/tex]
Mass of Oxygen[tex]= \dfrac{60}{100} \times100= 60 g[/tex]
- Step 2:, we calculate moles: mass/ atomic mass
Moles of sulfur [tex]\dfrac{40g}{32g/mol}= 1.25\; mol[/tex]
Moles of Oxygen [tex]\dfrac{60g}{16g/mol}= 3.75\; mol[/tex]
- Step 3:, To calculate the empirical formula, divide the mole of each element by the smallest number of mole :
Sulfur [tex]= \dfrac{1.25mol}{1.25mol} = 1[/tex]
Oxygen[tex]= \dfrac{3.75mol}{1.25mol} = 3[/tex]
Thus, the empirical formula of the compound is [tex]SO_3[/tex].
Learn more about empirical formula here:
https://brainly.com/question/11588623
