Question

The % composition of a substance is p =32.63, 0 =16.82 & S = 50.55. The V.D of
substance is 190. Find the molecular formula of the substance.
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Answer 1

Explanation:

ANSWER

Element

Moles

Least ratio

C

12

54.55

=4.54

2.27

4.54

=2

H

1

9.09

=9.09

2.27

9.09

=4

O

16

36.36

=2.27

2.27

2.27

=1

From the above calculations, the empirical formula =C

2

H

4

O.

Empirical formula weight =12×2+1×4+16=44 g.

Vapour density (VD) =44 g.

Molecular weight =2×VD=2×44=88 g.

n=

Empirical formula weight

Molecular weight

=

44

88

=2

Molecular Formula =2× Empirical formula=2×(C

2

H

4

O)=C

4

H

8

O

2

.

Answer 2

Your answer:

Find the ratio between the molar mass and the percentage of each element

P = $\frac{32.63}{31}$ = 1

O = $\frac{16.82}{16}$ = 1

S = $\frac{50.55}{32}$ = 1.5

By taking ratio: 1: 1: 1.5 = 2: 2: 3

So, the empirical formula will be: P₂O₂S₃

Molar mass of P₂O₂S₃ = 190

Molar mass of a compound = 2 x V.D

= 2 x 190 = 380

divide 380 by 190

You will get 2

Now, multiply the number of atoms in P₂O₂S₃ by 2

=> P₄O₄S₆ is your answer

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