Question

A) A compound composed of sulfur and fluorine is found to contain 62.79% by mass of sulfur. If the molar mass of the compound is 102.13 g/mol, what is its molecular formula?

1-S₂F₂

2-S₂F

3-SF

4-S₂F₄

5-S₃F

show your work (explian)

B)A hypothetical element consists of two isotopes of masses 74.95 amu and 76.95 amu with abundances of 34.9% and 65.1%, respectively. What is the average atomic mass of this element?

1-76.95 amu

2-74.95 amu

3-75.95 amu

4-75.6 amu

5-76.3 amu

show your work (explian)

Answer 1

A) it contains S and F, and we know that 62.79% belongs to S so:

MW = 102.13 g/mol

mass of Sulfur = 102.13 * 0.6279 = 64.13 g/mol

This means that the rest belongs to fluorine:

mass of fluorine = 102.13 - 64.13 = 38 g/mol

The atomic weight of Fluor is 19 g/mol and for Sulfur is 32 g/mol so:

n1 = 64.13 / 32 = 2.004

n2 = 38 / 19 = 2

This means that the formula should be S₂F₂

B) let's calculate the actual quantities of these isotopes:

74.95 * 0.349 = 26.15755

76.95 * 0.651 = 50.09445

the average mass would be:

50.09445 + 26.15755 = 76.252 or simply 76.3

Hope this helps