Question

A compound containing sodium, Sulphur, hydrogen and oxygen gave the following results on analysis: Na = 14.28%, S = 9.92%, H = 6.20% Calculate the molecular formula of the anhydrous compound. If all the hydrogen in the compound is present in combination with oxygen as water of crystallization, what is the formula of the crystalline salt? The molecular mass of the crystalline salt is 322.

Answer 1

Answer:

N2SO4. 10H2O

Explanation:

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Answer 2

Answer:

The molecular formula of the anhydrous compound is $Na_{2} SO_{4} .10H_{2} O$.

Explanation:

From the given data, the percentage of oxygen content is missing. Thus, it can be calculated as follows.

Percentage of oxygen, $O$ = $100$ % $-$ $( Na + S + H)$ %

$= 100 - 14.28 - 9.92 - 6.20$

$= 69.60$ %.

Now, in order to identify the compound, we calculate atomic ratios and then the least ratios of each, as follow,

• Atomic ratios (Percentage of content is divided by molecular weight)

For $Na = 14.28/23 = 0.62$

For $S = 9.92/32 = 0.31$

For $H = 6.20/1 = 6.20$

For $O = 69.60/16 = 4.35$

• Least ratios ($0.31$ is the least value obtained, so we divide each atomic ratio with $0.31$)

For $Na = 0.62/0.31 = 2$

For $S = 0.31/0.31 = 1$

For $H = 6.20/0.31 = 20$

For $O = 4.35/0.31 = 14$

That implies, the anhydrous compound to be : $Na_{2} SH_{20} O_{14}$.

But, it was given that hydrogen in the compound is present in combination with oxygen, as water. Thus, we have $10H_{2} O$.

(The molecular mass of is $(2$ x $23) + 32 + (20$ x $1) + (14$ x $16) = 322$).

Therefore, the final compound is $Na_{2} SO_{4} .10H_{2} O$.