Question

In the Carius determination, 0.234 g of an organic compound gave 0.334 g of Barium Sulphate. Calculate the percentage of Sulphur in the given compound. (at.wt. of Ba=137, S = 32, O = 16)


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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Mass and Mass% has been used. We see that we are given the the mass of the organic compound as well as the mass of the product. We already are given the atomic masses of the given things. So firstly we can calculate the mass of Sulpher used in the product. From that we can take out the mass% .

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{Mass\;\%\;=\;\bf{\dfrac{Mass\;of\;Substance}{Total\;Mass\;of\;Compound}\;\times\;100}}}}$

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★ Solution :-

Given,

» Atomic Mass of Barium = 137 u

» Atomic Mass of Sulpher = 32 u

» Atomic Mass of Oxygen = 16 u

» Mass of Organic Compound = 0.234 g

» Mass of Barium Sulphate = 0.334 g

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~ For Molecular Mass of the Barium Sulphate ::

✒ Molecular Mass = 137 + 32 + 4(16)

✒ Molecular Mass = 233 g

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~ For Mass of Sulpher in the Barium Sulphate ::

We know that,

$\\\;\sf{\rightarrow\;\;\green{233\;g}\;of\;Barium\;Sulphate\;has\;\red{32\;g}\;of\;Sulpher}$

Then,

$\\\;\sf{\rightarrow\;\;\green{1\;g}\;of\;Barium\;Sulphate\;has\;=\;\red{\dfrac{32}{233}}\;g\;of\;Sulpher}$

Now we know that, Barium Sulphate formed here is of 0.334 g quantity.

So,

$\\\;\sf{\rightarrow\;\;\green{0.334\;g}\;of\;Barium\;Sulphate\;has\;=\;\red{0.13733\;\times\;0.334}\;g\;of\;Sulpher}$

$\\\;\sf{\rightarrow\;\;\green{0.334\;g}\;of\;Barium\;Sulphate\;has\;=\;\orange{0.0458}\;g\;of\;Sulpher}$

Hence,

$\\\;\bf{\mapsto\;\;Mass\;of\;Sulpher\;in\;Barium\;Sulphate\;=\;\orange{0.0458\;g}}$

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~ For Mass% of Sulpher in the given compound ::

We already have the mass of Sulpher in the Barium Sulphate. Now we have the mass of the given compound. So now let's calculate the mass percentage of sulpher in it.

By using the formula, we get

$\\\;\sf{:\Longrightarrow\;\;Mass\;\%\;=\;\bf{\dfrac{Mass\;of\;Substance}{Total\;Mass\;of\;Compound}\;\times\;100}}$

$\\\;\sf{:\Longrightarrow\;\;Mass\;\%\;of\;Sulpher\;=\;\bf{0.196\;\times\;100}}$

$\\\;\sf{:\Longrightarrow\;\;Mass\;\%\;of\;Sulpher\;=\;\bf{\blue{19.6\;\;\%}}}$

$\\\;\underline{\boxed{\tt{Mass\;\%\;of\;\:Sulpher\;\;=\;\bf{\purple{19.6\;\;\%}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;Molarity\;=\;\dfrac{No.\;of\;moles\;of\;solute}{Volume\;of\;solution\;in\;L}}$

$\\\;\sf{\leadsto\;\;Molality\;=\;\dfrac{No.\;of\;moles\;of\;solute}{Mass\;of\;solvent\;in\;Kg}}$

$\\\;\sf{\leadsto\;\;Density\;=\;\dfrac{Mass}{Volume}}$

$\\\;\sf{\leadsto\;\;Pressure\;=\;\dfrac{Force}{Area}}$

$\\\;\sf{\leadsto\;\;^{\circ}F\;=\;\dfrac{9}{5}\:(^{\circ}C)\;+\;32}$

Answer 2

Answer:

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Explanation:

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