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An organic compound containing oxygen, carbon, hydrogen and nitrogen contains 20% carbon, 6.7 % hydrogen and 46.67 % nitrogen. Its molecular mass was found to be 60. Find the molecular formula of the compound. 2) 1.00 g of hydrated salt contains 0.2014 g of iron 0.1153 g of sulfhur 0.2301 g of oxygen and 0.4532 g of water of crystallisation find empirical formula .

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Answer 1

Answer:

Q.1

Molecular formula of compound

is

$CH _{4}N_{2} O$

Q.2

Empirical formula of compound is

$FeSO_{4}.7H_{2}O$

Step-by-step explanation:

Q.1

Percentage of Carbon = 20%

Percentage of Hydrogen = 6.7%

Percentage of Nitrogen = 46.67%

Percentage of oxygen = 100 -(20+6.7+46.67)

percentage of oxygen= 26.63%

Now,

% by mass ration of carbon =

$\frac{20}{12} = 1.6$

% by mass ratio of Hydrogen =

$\frac{6.7}{1} = 6.7$

% by mass ratio of Nitrogen=

$\frac{46.67}{14} = 3.3$

% by mass ratio of oxygen=

$\frac{26.63}{16} = 1.6$

Now,

C:H:N:O= 1.6 : 6.7 : 3.3 : 1.6

C:H:N:O = 1 : 4 : 2 :1

Therefore empirical formula of compound is

$CH _{4}N_{2} O$

and empirical formula mass is 60 and also molecular mass is 60

so n = 60/60 = 1

The molecular formula of compound is

$CH _{4}N_{2} O$

Q.2

Mass of Fe = 0.2014 g

Mass of S = 0.1153

Mass of O = 0.2301

Mass of Water = 0.4532

now

Moles of Fe =

$\frac{0.2014}{56} = 0.0036$

Moles of S =

$\frac{0.1153}{32} = 0.0036$

Moles of oxygen =

$\frac{0.2301}{16} = 0.0143$

Moles of Water =

$\frac{0.4532}{18} = 0.025$

Now

Ratio of

=Iron : Sulphur : oxygen : water

=0.0036 : 0.0036 : 0.0143 : 0.025

= 1 : 1 : 4 :7

so empirical formula of compound is

$FeSO_{4}.7H_{2}O$

Answer 2

Q.1

Molecular formula of compound

is

CH _{4}N_{2} OCH4N2O

Q.2

Empirical formula of compound is

FeSO_{4}.7H_{2}OFeSO4.7H2O

Step-by-step explanation:

Q.1

Percentage of Carbon = 20%

Percentage of Hydrogen = 6.7%

Percentage of Nitrogen = 46.67%

Percentage of oxygen = 100 -(20+6.7+46.67)

percentage of oxygen= 26.63%

Now,

% by mass ration of carbon =

\frac{20}{12} = 1.61220=1.6

% by mass ratio of Hydrogen =

\frac{6.7}{1} = 6.716.7=6.7

% by mass ratio of Nitrogen=

\frac{46.67}{14} = 3.31446.67=3.3

% by mass ratio of oxygen=

\frac{26.63}{16} = 1.61626.63=1.6

Now,

C:H:N:O= 1.6 : 6.7 : 3.3 : 1.6

C:H:N:O = 1 : 4 : 2 :1

Therefore empirical formula of compound is

CH _{4}N_{2} OCH4N2O

and empirical formula mass is 60 and also molecular mass is 60

so n = 60/60 = 1

The molecular formula of compound is

CH _{4}N_{2} OCH4N2O

Q.2

Mass of Fe = 0.2014 g

Mass of S = 0.1153

Mass of O = 0.2301

Mass of Water = 0.4532

now

Moles of Fe =

\frac{0.2014}{56} = 0.0036560.2014=0.0036

Moles of S =

\frac{0.1153}{32} = 0.0036320.1153=0.0036

Moles of oxygen =

\frac{0.2301}{16} = 0.0143160.2301=0.0143

Moles of Water =

\frac{0.4532}{18} = 0.025180.4532=0.025

Now

Ratio of

=Iron : Sulphur : oxygen : water

=0.0036 : 0.0036 : 0.0143 : 0.025

= 1 : 1 : 4 :7

so empirical formula of compound is

FeSO_{4}.7H_{2}OFeSO4.7H2O