For parts a and b, explain why the substances follow (or do not follow) the law of multiple proportions. take m(o) = 16 g/mol, m(mn) = 55 g/mol, and m(i) = 127 g/mol and derive the empirical formulae of all mentioned compounds. part a. sample a of manganese (mn) oxide contains 330 mg of mn and 144 mg of o, sample b — 330 mg of mn and 336 mg of o. part b. sample a of iodine (i) oxide contains 508 mg of i and 128 mg of o, sample b — 508 mg of i and 144 mg of o.

2. part a. calculate the mass of barium chloride dihydrate required to prepare a 1.500 l solution of 0.1500 mol/l ba2+. m(bacl2·2h2o) = 244.26 g/mol. part b. calculate the mass of copper(ii) sulfate pentahydrate required to prepare a 2.500 l solution of 0.2500 mol/l cu2+. m(cuso4·5h2o) = 249.69 g/mol.

Answers are in the explanation.

Explanation:

The law of multiple propotions says that "when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers". Using this law it is possible to obtain empirical formula of the compounds, thus:

1. A, sample A.

0,330g Mn× = 6x10⁻³ moles

0,144g O× = 9x10⁻³ moles

Ratio O:Mn = 9x10⁻³ / 6x10⁻³ = 1,5

To explain this ratio, empirical formula must be:

Mn₂O₃

1. A, sample B.

0,330g Mn× = 6x10⁻³ moles

0,336g O× = 2,1x10⁻² moles

Ratio O:Mn = 2,1x10⁻² / 6x10⁻³ = 3,5

To explain this ratio, empirical formula must be:

Mn₂O₇

1. B, sample A.

0,508g I× = 4x10⁻³ moles

0,128g O× = 8x10⁻³ moles

Ratio O:Mn = 8x10⁻³ / 4x10⁻³ = 2

To explain this ratio, empirical formula must be:

I₂O

1. B, sample B.

0,508g I× = 4x10⁻³ moles

0,144g O× = 9x10⁻³ moles

Ratio O:Mn = 9x10⁻³ / 4x10⁻³ = 2,25

To explain this ratio, empirical formula must be:

I₉O₄

2. Part A.

1,5L solution×× = 54,96g of BaCl₂.2H₂O

2. Part B.

2,5L solution×× = 156,1g of CuSO₄.5H₂O

I hope it helps!

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