Question

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.

Answer 1

Answer:

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×$\frac{1mol}{55g}$ = 6x10⁻³ moles

0,144g O×$\frac{1mol}{16g}$ = 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×$\frac{1mol}{55g}$ = 6x10⁻³ moles

0,336g O×$\frac{1mol}{16g}$ = 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×$\frac{1mol}{127g}$ = 4x10⁻³ moles

0,128g O×$\frac{1mol}{16g}$ = 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×$\frac{1mol}{127g}$ = 4x10⁻³ moles

0,144g O×$\frac{1mol}{16g}$ = 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×$\frac{0,15mol}{1L}$×$\frac{244,26g}{1mol}$ = 54,96g of BaCl₂.2H₂O

2. Part B.

2,5L solution×$\frac{0,25mol}{1L}$×$\frac{249,69g}{1mol}$ = 156,1g of CuSO₄.5H₂O

I hope it helps!