Question

Find the empirical formula for a compound which contains 0.463 g Tl (#81), 0.0544 g of carbon, 0.00685 g of hydrogen and 0.0725 g oxygen by finding its empirical formula.

Answer 1

Answer:

The answer to your question is    TlC₂H₃O₂

Explanation:

Data

Tl  = 0.463g

C  = 0.0544 g

H = 0.00685 g

O = 0.0725 g

Process

1.- Convert the mass of each element to moles

Thalio

                         204.4g ----------------  1 mol

                         0.463 g ---------------   x

                         x = (0.463 x 1) / 204.4

                         x = 0.0023 moles

Carbon

                          12 g ------------------  1 mol

                          0.0544 g ----------  x

                          x = (0.0544 x 1) / 12

                          x = 0.0045 moles

Hydrogen

                           1 g ------------------  1 mol

                         0.00685 g --------   x

                          x = (0.00685 x 1) / 1

                          x = 0.00685 moles

Oxygen

                           16 g -------------- 1 mol

                       0.0725 g ----------  x

                            x = (0.0725 x 1) / 16

                            x = 0.0045 moles

2.- Divide the number of moles by the lowest number of moles

Thalio        0.0023 / 0.0023 = 1

Carbon      0.0045 /  0.0023 = 1.9 ≈ 2

Hydrogen  0.0069 /  0.0023 = 3

Oxygen     0.0045 /  0.0023 = 1.9 ≈ 2

3.- Write the empirical formula

                               TlC₂H₃O₂

Answer 2

Answer:

The empirical formula is C2H3O2Tl

Explanation:

Step 1: Data given

Mass of Tl = 0.463 grams

Mass of carbon = 0.0544 grams

Mass of hydrogen = 0.00685 grams

Mass of oxygen = 0.0725 grams

Molar mass of Tl = 204.38 g/mol

Molar mass of C = 12 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate total mass of the compound

Total mass = 0.463 grams + 0.0544 grams + 0.00685 grams + 0.0725 grams =  0.59675 grams

Step 3: Calculate moles

Moles = mass / molar mass

Moles Tl = 0.463 grams / 204.38 g/mol = 0.00227 moles

Moles C = 0.0544 grams / 12.0 g/mol = 0.00453 moles

Moles H = 0.00685 grams / 1.01 g/mol = 0.00678 moles

Moles O = 0.0725 grams / 16 g/mol = 0.00453 moles

Step 4: Calculate mol ratio

We divide by the smallest number of moles

Tl = 0.00227/0.00227 = 1

C = 0.00453 / 0.00227 = 2

H = 0.00678/0.00227 = 3

O = 0.00453 / 0.00227 = 2

The empirical formula is C2H3O2Tl