Question

Hãy tìm công thức hóa học của những oxit có thành phần khối lượng như sau S:50%; C:42,8%; Mn:49,6%; Pb:86,6%

Answer 1

Answer:

Find the chemical formulas of oxides that contain the following composition:

S:50%; C:42,8%; Mn:49,6%; Pb:86,6%

Explanation:

In the given oxide of sulphur,

50%S is present that means remaining 50% is O.

Divide the % with atomic mass:

S                                     O

50/32= 1.5625            50/16=3.125

divide with smallest ratio:

1.5625/1.5625=1              3.125/1.5625=2

Hence, the formula of oxide is $SO_2$.

Similarly:

Carbon is 42.8% means the reamining % will be O that is 57.2%.

C                                                    O

42.8/12=3.575                         57.2/16=3.575

Divide with smallest ratio:

3.575/3.575=1                            3.575/3.575=1

Hence, the formula of oxide is CO.

Mn is 49.6% means the remining is % is O that is 50.4%.

First divide with atmic mass of respective element, then get the smallest ratio.

That gives the mole ratio of each constituent atom.

Mn                                 O

49.6/55.0=0.90            50.4/16=3.15

0.90/0.90=1                   3.15/0.90=3.5

Multiply with two

2                                        7

Hence, the formula becomes:

$Mn_2O_7$

Pb is 86.6% means ---remaining 13.4% is O.

Pb                                               O

86.6/207 = 0.418 mol                13.4/16=0.85

0.418/0.418=1                               0.85/0.418=2.0

Hence, the formula is:$PbO_2$