Question

The V.D. of an organic compound is 59. If the
percentage of C, O, H are 40.856%, 54.128% and
5.016% respectively, then empirical formula of the
hydrocarbon is
(1) C₂H₂O₂ (2) C₃H₂O2
(3) CH20
(4) C₂H₂O
ans is 1​

Answer 1

Answer:

What's the meaning of V.D?

Answer 2

Answer:

1. An organic compound was found on analysis to contain 47.37% carbon and 10.59% hydrogen.

The balance was presumed to be oxygen. What is the empirical formula of the compound?

Ans.: C3H8O2

C 47.37 g ×

1 mol

= 3.944 mol

3.944 mol

= 1.501 mol × 2 = 3.002 mol ≈ 3 mol

12.01 g 2.6275

H 10.59 g ×

1 mol

= 10.506 mol

10.506 mol

= 3.998 mol × 2 = 7.996 mol ≈ 8 mol

1.008 g 2.6275

O 42.04 g ×

1 mol

= 2.6275 mol

2.6275 mol

= 1.000 mol × 2 = 2.000 mol ≈ 2 mol

16.00 g 2.6275

2. A hydrocarbon containing 92.3% C and 7.74% H was found (by measuring its gas density) to

have a molar mass of approximately 79 g/mol. What is the molecular formula?

Ans.: C6H6

C 92.3 g ×

1 mol

= 7.69 mol

7.69 mol

= 1.00 mol ≈ 1 mol

12.01 g 7.68

H 7.74 g ×

1 mol

= 7.68 mol

7.68 mol

= 1.00 mol ≈ 1 mol

1.008 g 7.68

Molar Mass from Emp. Formula = 12.01 g/mol × 1 mol + 1.008 g/mol × 1 mol = 13.02 g/mol

Molec. Formula: (CH)n n = (79 g/mol) / (13.02 g/mol) ≈ 6

3. A 1.500-g sample of a compound containing only C, H, and O was burned completely. The only

combustion products were 1.738 g CO2 and 0.711 g H2O. What is the empirical formula of the

compound?

Ans.: C2H4O3

1.738 g CO2 ×

1 mol CO2

×

1 mol C

×

12.01 g C

= 0.4743 g C

44.01 g CO2 1 mol CO2 1 mol C

0.711 g H2O ×

1 mol H2O

×

2 mol H

×

1.008 g H

= 0.0796 g H

18.01 g CO2 1 mol H2O 1 mol H

mass of O = 1.500 g – 0.4743 g – 0.0796 g = 0.946 g

C 0.4743 g ×

1 mol

= 0.03949 mol

0.03949 mol

= 1.000 mol × 2 = 2.00 mol ≈2 mol

12.01 g 0.03949

H 0.0796 g ×

1 mol

= 0.07897 mol

0.07897 mol

= 2.000 mol × 2 = 4.00 mol ≈ 4 mol

1.008 g 0.03949

O 0.946 g ×

1 mol

= 0.0591 mol

0.0591 mo

= 1.50 mol × 2 = 3.00 mol ≈ 3 mol

16.00 g 0.03949

4. In the article on “Chemistry” in the Ninth Edition of the Encyclopaedia Britannica (published in

1878) the author (H. A. Armstrong) says that Mendeleev had recently proposed that uranium

be assigned the atomic weight 240 in place of the old value 120 that had been assigned to it

by Berzelius, but that he himself preferred 180. Mendeleev was right. The correct formula of

pitchblende, an important ore of uranium, is U3O8.What formula was written for pitchblende by

(a) Berzelius, (b) Armstrong?(From General Chemistry by L. Pauling)

Ans.: (a) U3O4; (b) UO2

Chemist

Accepted Atomic

Weight of

Uranium

Uranium-toOxygen Mole

Ratio

Accepted

Formula for

Pitchblende

Mendeleev 240 3:8 U3O8

Berzelius 120 6:8 U3O4

Armstrong 180 4:8 UO2

5. One of the earliest methods for determining the molar mass of proteins was based on chemical

analysis. A hemoglobin preparation was found to contain 0.335% iron. (a) If the hemoglobin

molecule contains 1 atom of iron, what is its molar mass? (b) If it contains 4 atoms of iron, what

is its molar mass?

Ans.: (a) 1.67104 g/mol;

(b) 6.67104 g/mol

(a)

g Fe

=

0.335 g

=

55.85 g

g protein 100 g X g

(100 g protein) (1 mole of protein)

X = 16,672 Molar mass of the protein = 1.67104 g/mol

(b)

g Fe

=

0.335 g

=

4×55.85 g

g protein 100 g Z g

(100 g protein) (1 mole of protein)

Z = 4X = 66,686 Molar mass of the protein = 6.67104 g/mol