Question

Question: Calculate the weight of each element and water in 13 g of ferric alum (NH4)2SO4.Fe2(SO4)3. 24H2O.

Answer:
N = 0.38
H = 0.11
S = 1.73
Fe = 1.52
O = 3.46
H2O = 5.84
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Answer 1

Q. Calculate the weight of each element and water in 13 g of ferric alum (NH4)2SO4.Fe2(SO4)3. 24H2O.

Ans :-

2

Molar weight of ferric alum = $(NH_4)_2SO_4.Fe_2(SO_4)_3. 24H_2O$

= $2 \times N + 56 \times H+ 4 \times S + 40 \times O + 2 \times Fe$

Atomic mass of elements:-

N = 14u

H = 1u

S = 32u

O = 16u

Fe = 56u

Now ,

Molar weight = $2 \times 14 + 56\times 1 + 4 \times 32 + 40\times 16 +2 \times 56$

$28 + 56 + 128 + 610 + 112$

Molar weight = 954g.

Now,

954g of ferric alum contain 28 g of nitrogen.

13 g of ferric alum will contain = 28/954 × 13

= 364 /954

= 0.38 g

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954 g of ferric alum contain 128 g of sulphur.

13 g of ferric alum will contain =

128/954 × 13

= 1664 /954

= 1.74 g

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954 g of ferric alum contain 112 g of iron.

13 g of ferric alum contain = 112/954 × 13

= 1456 /954

= 1.52 g

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Now,

954 g of ferric alum contain 8g hydrogen.

13 g of ferric alum will contain=

8/954 × 13

= 104/954

= 0.109 ~ 0.11g

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954 g of ferric alum contain 256 oxygen

13 g of ferric alum will contain=

256/954 × 13

= 3328/954

= 3.48 ~ 3.46

__________________________________________

Now,

954 g of ferric alum contain 432 g water.

13 g of ferric alum contain = 432/954 × 13

= 5616/954

= 5.84 g ans..

Note :- Don't add hydrogen and oxygen of water when calculating their actual weight.

Answer 2

Solution :

Explanation:

Given mass = 13 g

We know mass of different element as

N = 14 u

H = 1 u

S = 32 u

O = 16 u

Fe = 55.8 u

Molecular mass of Ferric alum =

$\large 36+32+64+112+228+432\\\\\\ \large \implies960 \ g$

First let us find numbers of mole

We know formula for numbers of mole

$\large \text{Numbers of mole = \dfrac{Given \ mass}{Molar \ mass} }$

Putting value here we get

$\large \text{Numbers of mole = \dfrac{13}{960} }\\\\\\\large \text{Numbers of mole = 0.0135 }$

Now mass of Nitrogen ( N ) in Ferric alum =

$\large \text{Numbers of mole = \dfrac{Requrie \ mass \ of \ N}{Molar \ mass \ of \ N} }\\\\\\\large \text{Rewrite as}\\\\\\\large Requrie \ mass \ of \ N=Numbers \ of \ mole \times Molar \ mass \ of \ N$

1.

$\large Requrie \ mass \ of \ N=0.0135\times28\\\\\\\large \text{ Requrie \ mass \ of \ N=0.378 \ g About 0.38 g}$

2.

$\large Requrie \ mass \ of \ H=0.0135\times8 \ g\\\\\\\large \text{ Requrie \ mass \ of \ H=0.108 \ g About 0.11 g}$

3.

$\large Requrie \ mass \ of \ S=0.0135\times4\times32 \ g\\\\\\\large \text{ Requrie \ mass \ of \ S=1.728 \ g About 1.73 g}$

4.

$\large Requrie \ mass \ of \ Fe=0.0135\times56\times2 \ g\\\\\\\large \text{ Requrie \ mass \ of \ Fe=1.516 \ g Anout 1.52 g}$

5 .

$\large Requrie \ mass \ of \ O=0.0135\times16\times16 \ g\\\\\\\large \text{ Requrie \ mass \ of \ O=3.456 \ g About 3.46 g}$

6.

$\large Requrie \ mass \ of \ H_2O=0.0135\times24\times18 \ g\\\\\\\large \text{ Requrie \ mass \ of \ H_2O=5.832 \ g About 5.83 g }$

Thus we get all answer.