A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of the oxygen and gives a red binary compound that is 9.334% oxygen by mass. Strong heating drives off more oxygen and gives still another binary compound, which is only 7.168% oxygen by mass.

(a) Compute the mass of oxygen that is combined with 1.000 g of the metal in each of these three oxides.

(b) Assume that the empirical formula of the first compound is MO2 (where M represents the metal). Give the empirical formulas of the second and third compounds.

(c) Name the metal.

Question

Chemistry

Carpenter

20 March, 16:26

A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of the oxygen and gives a red binary compound that is 9.334% oxygen by mass. Strong heating drives off more oxygen and gives still another binary compound, which is only 7.168% oxygen by mass.

(a) Compute the mass of oxygen that is combined with 1.000 g of the metal in each of these three oxides.

(b) Assume that the empirical formula of the first compound is MO2 (where M represents the metal). Give the empirical formulas of the second and third compounds.

(c) Name the metal.

+4

Answers (1)

Know the Answer?

Iyana Schultz · Answer 1

a) Mass of O in compound A = 32.72 g

Mass of O in compound B = 21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A: M = m1 * 0.8662 O = m1 * 0.1338

compound B: M = m2 * 0.90666 O = m2 * 0.09334

compound C: M = m3 * 0.92832 O = m3 * 0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16) / 0.1338 m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48 ⇒ m2 = 227.74 g

0.92832m3 = 206.48 ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42 - 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 / 16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)