Question

A 44.0mg sample of carbon reacts with sulfur to form 122 mg of the compound.

Answer 1

Answer:

Empirical formula = C₃S₂

Explanation:

Given data:

Mass of carbon = 44.0 mg (44/1000 = 0.044 g)

Mass of sulfur = 122 mg - 44.0 mg = 78 mg = 78/1000 = 0.078 g)

Empirical formula = ?

Solution:

First of all we will calculate the number of moles.

Number of moles of carbon = mass / molar mass

Number of moles of carbon = 0.044 g/ 12.01 g/mol

Number of moles of carbon = 0.0037 mol

Number of moles of sulfur:

Number of moles = mass / molar mass

Number of moles = 0.078 g/ 32,066 g/mol

Number of moles  = 0.0024 mol

Now we will compare the moles:

                C                            :                  S

             0.0037/0.0024        :             0.0024/0.0024

                  1.5                         :               1

C : S = 2(1.5 : 1)

C : S = 3 : 2

Empirical formula = C₃S₂