Question

Two oxides of metal contains 27.6% and 30% oxygen respectively. If the formula of the first oxide is M₃O₄.find that of second.​

Answer 1

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Given that formula of first oxide = M₃O₄

Let mass of the metal = x

$\% \: of \: \: metal \: \: in \: M₃O₄ \: = \: \frac{3x}{3x \: + \: 64} \times 100$

But as given % age = (100 - 27.6) = 72.4%

$so \: , \: \frac{3x}{3x \: + \: 64} \times 100 \: = \: 72.4 \: \: or \: x \: = \: 56$

In 2nd oxide ,

Given , oxygen = 30% , So metal = 70%

$So \: \: the \: \: ratio \: \: is \: \: M : O \: = \: \frac{70}{56} \: : \: \frac{30}{16} \\ \\ ✰ \: \: 1.25 \: : \: 1.875 \\ \\ ✰ \: \: 2 : 3</em></p><p><em>So \: \: the \: \: ratio \: \: is \: \: M : O \: = \: \frac{70}{56} \: : \: \frac{30}{16} \\ \\ ✰ \: \: 1.25 \: : \: 1.875 \\ \\ ✰ \: \: 2 : 3$

So , 2nd oxide is M²O³

Answer 2

Answer:

Given :-

$\leadsto$ Two oxide of metal contains 27.6% and 30% respectively.

$\leadsto$ The formula of the first oxide is M₃O₄.

To Find :-

$\leadsto$ What is the formula of second oxide.

Solution :-

$\sf\bold{\purple{\underline{\bigstar\: \: In\: the\: first\: oxide\: :-}}}$

Given :

• Oxygen = 27.6%

So,

$\implies \sf Percentage\: of\: metal =\: 100 - 27.6$

$\implies \bf Percentage\: of\: metal =\: 72.4\: of\: mass$

Now, given that :

• Formula for the first oxide = M₃O₄

So here contains that :

➳ 3 atoms of metal

➳ 4 atoms of oxygen

$\sf\bold{\purple{\underline{\bigstar\: In\: the\: second\: oxide\: :-}}}$

Given :

• Oxygen = 30%

So, similarly :

$\implies \sf Percentage\: of\: metal =\: 100 - 30$

$\implies \bf Percentage\: of\: metal =\: 70\: of\: mass$

So, now we have to find the number of atoms of metals :

$\dashrightarrow \sf \dfrac{3}{72.4} \times 70$

$\dashrightarrow \sf \dfrac{210}{72.4}$

$\dashrightarrow \sf\bold{\green{2.90\: atoms\: of\: metal}}$

Again, we have to find the number of atoms of oxygen :

$\dashrightarrow \sf \dfrac{4}{27.4} \times 30$

$\dashrightarrow \sf \dfrac{120}{27.4}$

$\dashrightarrow \sf\bold{\green{4.37\: atoms\: of\: oxygen}}$

Now, we have to find the ratio of metal to oxygen :

$\longrightarrow \sf Ratio\: of\: M : O_{(Second\: Oxide)} =\: 2.90 : 4.37$

$\longrightarrow \sf Ratio\: of\: M : O_{(Second\: Oxide)} =\: 1 : 1.5$

$\longrightarrow \sf\bold{\red{Ratio\: of\: M : O_{(Second\: Oxide)} =\: 2 : 3}}$

Hence, the formula of second oxide is M₂O₃.

$\therefore$ The formula of second oxide is M₂O₃ .

EXTRA IMPORTANT INFORMATION :-

1) Carbon dioxide :- CO₂

2) Magnesium oxide :- MgO

3) Sodium chloride :- NaCl

4) Copper Sulphate :- CuSO₄

5) Sulphuric Acid :- H₂SO₄

6) Nitric Acid :- HNO₃

7) Sodium Hydroxide :- NaOH

8) Sulphur Dioxide :- SO₂

9) Water :- H₂O

10) Methane :- CH₄