Question

Page No
Date
a) A compound is made up of 19.3% of Na, 26.9%
of S and 53.8% of Oxygen whose molecular
mass is 238 Derive these molecular formula of the
compound​

Answer 1

Answer:

option c is the write answer for your question.

Answer 2

Solution :-

Given :-

Percentage composition of sodium (Na) = 19.3%

Percentage composition of sulfur (S) = 26.9%

Percentage composition of oxygen (O) = 53.85%

We know :-

Atomic mass of sodium (Na) = 23g

Atomic mass of sulfur (S) = 32g

Atomic mass of oxygen (O) = 16g

$\bullet \sf \ No \ of \ moles \ of \ sodium \ (Na) = \dfrac{19.3}{23}= 0.839 \\\\\bullet \ No \ of \ moles \ of \ sulfur \ (S) = \dfrac{26.9}{32} = 0.84 \\\\\bullet \ No \ of \ moles \ of \ oxygen \ (O) = \dfrac{53.8}{16} = 3.362$

Simplest ratio :-

$\longrightarrow \sf \dfrac{0.839}{0.839} \ : \ \dfrac{0.84}{0.839} \ : \ \dfrac{3.362}{0.839} \\\\\longrightarrow 1 \ : \ 1 \ : \ 4$

→ Empirical formula = $\sf NaSO_4$

→ Empirical formula mass = 23 + 32 + ( 16 × 4 )

                                          = 23 + 32 + 64

                                          = 119g

We know :-

$\bf Molecular \ formula = n \times Empirical \ formula$

Here,

$\longrightarrow \sf n = \dfrac{Molecular \ mass}{Emprical \ formula \ mass } \\\\\longrightarrow n = \dfrac{238}{119} \\\\\longrightarrow n = 2$

Molecular formula = $\sf 2 \times (NaSO_4)$

$\boxed{\bf Molecular \ formula = Na_2S_2O_8}$