Name: Avogadro's Law and Its Uses
Uploaded: 2019-07-19
Duration: 10 min 28 s
Description: Avogadro's law is an experimental gas law relating the volume of a gas to the amount of substance of gas present. Let us learn more through this video.

Question

A crystalline salt on heating becomes anhydrous and loses

45.6 %

of its weight. Its percentage composition is

Al (10.50 %), K (15.1 %), S (24.96 %)

and

O (49.92 %)

. What is the formula of anhydrous and crystalline salt?

Accepted Answer

Suppose we have $100 g$ of the crystalline salt.

Therefore, the amount of $Al = 10.50 g, K = 15.1 g, S = 24.96 g and O = 49.92 g$

Number of moles $= \frac{Weight}{Molar mass}$

Number of moles of $Al$ $= \frac{10.5}{27} \approx 0.39$

Number of moles of $K$ $= \frac{15.1}{39} \approx 0.39$

Number of moles of $S$ $= \frac{24.96}{32} = 0.78$

Number of moles of $O$ $= \frac{49.92}{16} = 3.12$

To obtaining the simplest ratio of compound, divide it by the smallest number of moles.

Simple ratio of

$Al = \frac{0.39}{0.39} = 1 K = \frac{0.39}{0.39} = 1 S = \frac{0.78}{0.39} = 2 O = \frac{3.12}{0.39} = 8$

Therefore, the empirical formula for the compound $=$ ${KAlS}_{2} O_{8}$

Molecular mass $=$ $27 + 39 + 32 \times 2 + 16 \times 8 = 258$

On heating $100 g$ of crystalline salt it loses $45.6 g$ of its weight, this means the weight of anhydrous salt present $= 100 - 45.6 = 54.4$

Therefore, the crystalline compound contains $H_{2} O$ $=$ $\frac{45.6}{54.4} \times 258 = 216.26 g$

The molecules of $H_{2} O$ present $= \frac{216.26}{18} \approx 12$

Hence, the formula of crystalline salt $= {KAlS}_{2} O_{8} . 12 H_{2} O$

And the formula of anhydrous salt $=$ ${KAlS}_{2} O_{8}$

A crystalline salt on heating becomes anhydrous and loses $45.6 %$ of its weight. Its percentage composition is $Al (10.50 %), K (15.1 %), S (24.96 %)$ and $O (49.92 %)$ . What is the formula of anhydrous and crystalline salt?

Important Questions on Some Basic Concepts, Mole Concept and Stoichiometry

Learn and Prepare for any exam you want !

A crystalline salt on heating becomes anhydrous and loses 45.6% of its weight. Its percentage composition is Al 10.50%, K 15.1%, S 24.96% and O 49.92%. What is the formula of anhydrous and crystalline salt?

Important Questions on Some Basic Concepts, Mole Concept and Stoichiometry

Learn and Prepare for any exam you want !

A crystalline salt on heating becomes anhydrous and loses $45.6 %$ of its weight. Its percentage composition is $Al (10.50 %), K (15.1 %), S (24.96 %)$ and $O (49.92 %)$ . What is the formula of anhydrous and crystalline salt?