Question

An 85.00 g sample of an unknown compound contains 24.26 g molybdenum, 24.33 g sulfur, and the rest is oxygen. How many moles of molybdenum are present in the compound

Answer 1

Answer:

A) 0.253 mol B) 0.7603 mol C) 2.276 mol D) MoS₃O₉ or Mo(SO₃)₃ E) molybdenum(VI) sulfite

Explanation:

For all the Mole problems you want to find divide grams of the element in compound by the molar mass of the element.

24.26 g Mo / 95.95 g = 0.2528

24.33 g S / 32.06 g = 0.7589 mol S

36.41 g O / 16 g = 2.276 mol O

S /  Mo =  0.7587 mol  / 0.2529 mol  =  3/ 1

O / Mo =  2.276 mol  / 0.2529 mol  =  9/ 1

The empirical formula of the compound is MoS₃O₉ or Mo(SO₃)₃.

Answer 2

The number of mole of molybdenum present in the compound is 0.253 mole

From the question given above, the following data were obtained:

Mass of compound = 85 g

molybdenum = 24.26 g

Sulphur = 24.33 g

The number of mole of molybdenum in the compound can be obtained as follow:

Mass of molybdenum = 24.26 g

Molar mass of molybdenum = 95.95 g/mol

Mole of molybdenum =?

Mole = mass / molar mass

Mole = 24.26 / 95.95

Mole of molybdenum = 0.253 mole

Therefore, the number of mole of molybdenum present in the compound is 0.253 mole

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