Answer:
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)
