Question

0.6g of a compound occupies 224cc at NTP it contains 6.67℅hydrogen 40℅ carbon and rest oxygen. calculate it's molecular mass of the compound, it's empirical formula, and molecular formula​

Answer 1

The %age of oxygen is: 100 - (6.67 + 40)

= 100 - 46.67 = 53.33

Now, take the ratio of the percentage of the composition and molar mass

Hydrogen ratio: 6.67/1 = 6.67

Carbon ratio: 40/12 = 3.33

Oxygen ratio: 53.33/16  = 3.33

now, taking ratio: 3.33 : 6.67 : 3.33

= 1 : 2 : 1

So, empirical formula is: CH₂O

mass of empirical formula = 30

Molar mass of the compound is 60

Now, divide 60 by empirical formula mass = 60/30 = 2

n=2

Now, the formula is: C₂H₄O = CH₃CHO

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Answer 2

Answer:

Molecular mass of the compound is 60 U

Empirical formula of the compound is $CH_{2} O$

Molecular formula of the compound is $C_{2} H_{4} O_{2}$

Explanation:

Percentage of Hydrogen = 6.67 %

Percentage of carbon = 40%

Percentage of oxygen =100−(6.67+40)

                                     =100−46.67

                                     =53.33

Percentage of oxygen = 53.33%

Moles:

• The mole is a unit of measurement or the primary source for determining the concentration of a chemical in a sample.
• A mole is a measurement of material quantity.

Formula:

Number of moles: Mass of substance / Mass of moles

Atomic mass ratio:

The average mass of atoms of a chemical element in a particular sample divided by the atomic mass constant is known as the atomic mass ratio.

Carbon

Element = C

Percentage = 40

Atomic mass = 12

Gram atom = $\frac{40}{12}$  = 3.33

Atomic ratio = $\frac{3.33}{3.33}$  = 1

Simplest ratio = 1

Hydrogen:

Element = H

Percentage = 6.67

Atomic mass = 1

Gram atom = $\frac{6.67}{1} = 6.67$  

Atomic ratio = $\frac{6.67}{3.33}$ = 2.0

Simplest ratio = 2

Oxygen

Element = O

Percentage = 53.33

Atomic mass = 16

Gram atom = $\frac{53.33}{16}$  = 3.33

Atomic ratio = $\frac3.33}{3.33}$ = 1

Simplest ratio = 1

Empirical formula of the compound = $CH_{2} O$

Molecular formula of the compound

224 cc of the vapours of the compound at N.T.P occupy mass = 0.6 g

22400 cc of the vapours at N.T.P occupy mass = $\frac{0.6}{224}$ × 22400

                                                                                   = 60.0 g

Molecular mass of the compound = 60 u

Empirical formula mass = 12 + 2× 1 + 16

                                       = 12+2+16

                                        = 30 u

Molecular formula = 2 × $CH_{2} O$

                               = $C_{2} H_{4} O_{2}$

Final answer:

Molecular mass of the compound is 60 U

Empirical formula of the compound is $CH_{2} O$

Molecular formula of the compound is $C_{2} H_{4} O_{2}$

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