Question

6. + -19 points What is the empirical formula of a compound that is 44.25% Cu, 22.33% S, and 33.47% O? Write the elements in the order Cu, S, and O. 7. + -19 points A 4.523-g sample of an oxide of Ag contains 4.211 g Ag. What is the empirical formula of the oxide?

Answer 1

Answer 6 -

Given,

Percentage of Cu = 44.25 %

Percentage of S = 22.33 %

Percentage of O = 33.47 %

Molar mass of Cu = 63.546 u

Molar mass of S = 32.065 u

Molar mass of O = 15.999 u

Empirical Formula = ?

Let the total mass of compound is 100 g

So, mass of Cu = 44.25% of 100g = 0.4425 * 100g = 44.25 g

mass of S = 22.33% of 100g = 0.2233 * 100g = 22.33 g

mass of O = 33.47% of 100g = 0.3347 * 100g = 33.47 g

Now,

moles = mass / molar mass

Moles of Cu = 44.25 g / 63.546 u = 0.6963 mol

Moles of S = 22.33 g / 32.065 u = 0.6964 mol

Moles of O = 33.47 g / 15.999 u = 2.09 mol

Divide the moles of each by the smallest among them,

For Cu = 0.6963 mol / 0.6963 mol = 1

For S = 0.6964 mol / 0.6963 mol = 1

For O = 2.09 mol / 0.6963 mol = 3 approx.

So, Empirical Formula = CuSO₃ [ANSWER]

Answer 7-

Given,

Mass of Oxide of Ag = 4.523 g

Mass of Ag in Oxide = 4.211 g

Molar Mass of Ag = 107.8682 u

Molar Mass of O = 15.999 u

Empirical formula = ?

Mass of O in Oxide = Mass of Oxide of Ag - Mass of Ag

Mass of O in Oxide = 4.523 g - 4.211 g = 0.312 g

Now,

Moles = Mass / Molar Mass

Moles of Ag = 4.211 g / 107.8682 u = 0.03904 mol

Moles of O = 0.312 g / 15.999 u = 0.0195 mol

Divide the moles of each by the smallest among them,

For Ag = 0.03904 mol / 0.0195​​​​​​​ mol = 2 approx.

For O = 0.0195​​​​​​​ mol / 0.0195​​​​​​​ mol = 1

So, Empirical Formula = Ag₂O [ANSWER]