Question

Question ❓✔️❓
1) 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 . ​

Answer 1

Answer :

• Molecular Formula of the compound will be CH₄N₂O .

• Emperical Formula of the hydrated salt will be FeSO₄.7H₂O₃

Explanation :

Let the mass of compounds be 100g.

It is given that an organic compounds contains oxygen , carbon , hydrogen and nitrogen .

• Mass of Carbon = 20g
• Mass of hydrogen = 6.7g
• Mass of nitrogen = 46.67

And the remaining will be the mass of the oxygen .

→ Mass of oxygen = 100 -(20+6.7+46.67)

→ Mass of Oxygen = 100- 73.37g

→ Mass of Oxygen = 26.63g

Now , calculating the number of moles in the compound.

• No. of moles = Give Mass/molecular mass

For Carbon :

No.of moles = 20/12

= 1.6

For Hydrogen :

No . of moles = 6.7/1

= 6.7

For Oxygen :

No. of moles = 26.63/16

= 1.6

For Nitrogen :

No. of moles = 46.67/14

= 3.3

Now, calculating the simplest ratio in the given compound .

By dividing 1.6 we get

C H N O

1 ⠀4⠀2 ⠀1 ⠀⠀⠀⠀⠀⠀⠀ (approx)

Empirical formula of the compound .

= CH₄N₂O

As it is given the compound has 60g molecular formula .So, checking the emperical Formula weight of compound.

Emperical Formula weight = 1×12 + 4×1 + 14×2 + 1×16

= 60g

• Hence, the molecular formula of the compound will be CH₄N₂O .

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It is given that 1.00 g of hydrated salt contains 0.2014 g of iron 0.1153 g of sulphur 0.2301 g of oxygen and 0.4532 g of water of crystallisation .we have to calculate the empirical formula .

Firstly we calculate the moles of the elements present in the hydrated salt .

Points to Remember {molar(atomic) mass of the elements present in the salt }

• Iron (Fe) = 56
• Sulphur (S) = 32
• Oxygen (O) = 16

For iron :

No. of moles = 0.2014/56

= 0.0036

For Sulphur :

No. of moles = 0.1153/32

= 0.0036

For Oxygen :

No. of moles = 0.2301/16

= 0.0143

For Water :

Molecular mass of water (H₂O) = 2×1 + 16 = 18

No. of moles = 0.4532/18

= 0.025

Calculating simplest ratio by dividing 0.0036 we get

→ Fe = 0.0036/0.0036 = 1

→ S = 0.0036/0.0036 = 1

→ O = 0.0143/0.0036 = 3.9722 ≈ 4

→ H₂O = 0.025/0.0036 = 6.944 ≈ 7

So , the emperical Formula of the hydrated salt will be

= FeSO₄.7H₂O₃

Answer 2

Required Answer:

❶

‣ Molecular formula of compound $\mapsto\:{\boxed{\tt{CH_{4}N_{2}O}}}$

❷

‣ Empirical formula of hydrated salt $\mapsto\:{\boxed{\tt{FeSO_{4}.7H_{2}O}}}$

Step - By - Step explanation:

❶

Given information,

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.

⚘ As percentage is given so we will take total mass = 100g. Mass of each element is equal to it's percentage;

Hence,

• Mass of C = 20g
• Mass of H = 6.7g
• Mass of N = 46.67g

⚘ Mass of O ::

➻ Total mass - Sum (Masses of other elements)

➻ 100 - (20 + 6.7 + 46.67)

➻ 100 - 73.37g

➻ 26.63g

• Mass of O = 26.63g

⚘ Converting all masses of into moles;

We know that,

$\bf{\dag}\:{\boxed{\underline{\underline{\bf{\red{No.\:of\:moles = \dfrac{Given\:mass}{Molar\:mass}}}}}}}$

Hence,

$\\ \longrightarrow\:\tt Moles\:of\:C = {\cancel{\dfrac{20}{12}}} = {\boxed{\tt{1.67\:moles}}}$

$\\ \longrightarrow\:\tt Moles\:of\:H = \dfrac{6.7}{1} = {\boxed{\tt{6.7\:moles}}}$

$\\ \longrightarrow\:\tt Moles\:of\:N = {\cancel{\dfrac{46.67}{14}}} = {\boxed{\tt{3.33\:moles}}}$

$\\ \longrightarrow\:\tt Moles\:of\:O = {\cancel{\dfrac{26.63}{16}}} = {\boxed{\tt{1.6\:moles}}}$

⚘ Finding mole ratio by dividing each value of moles by smallest value of moles calculated;

$\\ \longrightarrow\:\tt For\:C = {\cancel{\dfrac{1.67}{1.67}}} = {\boxed{\tt{1}}}$

$\\ \longrightarrow\:\tt For\:H = {\cancel{\dfrac{6.7}{1.67}}} = {\boxed{\tt{4}}}$

$\\ \longrightarrow\:\tt For\:N = {\cancel{\dfrac{3.33}{1.67}}} = {\boxed{\tt{2}}}$

$\\ \longrightarrow\:\tt For\:O = {\cancel{\dfrac{1.6}{1.67}}} = {\boxed{\tt{1}}}$

‣ Ratio of C:H:N:O = 1:4:2:1

‣ Empirical formula of compound $\mapsto\:{\boxed{\tt{CH_{4}N_{2}O}}}$

Now,

• Equivalent weight = 12(1) + 1(4) + 14(2) + 16(1) = 12 + 4 + 28 + 16 = 60g/mole

⚘ Molecular formula ::

$\\ \longrightarrow\:\tt n = \frac{Molecular\:weight\:of\:metal}{Equivalent\:of\:metal} = {\cancel{\frac{60}{60}}} = {\boxed{\tt{1}}}$

$\tt \longrightarrow\:Molecular\:formula = 1\:\times\:Empirical\:formula$

$\tt \longrightarrow\:Molecular\:formula = 1\:\times\:CH_{4}N_{2}O$

$\tt \longrightarrow\:Molecular\:formula = {\boxed{\tt{CH_{4}N_{2}O}}}$

❷

Given information,

1.00g of hydrated salt contains 0.2014g of iron 0.1153g of sulphur 0.2301g of oxygen and 0.4532g of water of crystallisation find empirical formula.

★ Finding no. of moles ::

• ${\boxed{\underline{\underline{\bf{\purple{No.\:of\:moles = \dfrac{Given\:mass}{Molar\:mass}}}}}}}\:\bf{\dag}$

➨ For Fe = 0.2014/56 = 0.0036

➨ For S = 0.1153/32 = 0.0036

➨ For O = 0.2301/16 = 0.0143

➨ For $\sf H_{2}O$ = 0.4532/18 = 0.025

⚘ Finding mole ratio by dividing each value of moles by smallest value of moles calculated;

➨ For Fe = 0.0036/0.0036 = 1

➨ For S = 0.0036/0.0036 = 1

➨ For O = 0.0143/0.0036 ≈ 4

➨ For $\sf H_{2}O$ = 0.025/0.0036 ≈ 7

‣ Empirical formula of compound $\mapsto\:{\boxed{\tt{FeSO_{4}7H_{2}O}}}$

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