Question

The percent composition of a compound was found to be 63.5 % silver, 8.2% nitrogen and 28.3% oxygen. Determine the compound's empirical formula with solution

Answer 1

$\large{\bf{\gray{\underline{\underline{\orange{Given:}}}}}}$

✠ % composition of Ag = 63.5%

✠ % composition of N = 8.2%

✠ % composition of O = 28.3%

$\large{\bf{\gray{\underline{\underline{\green{To\:Find:}}}}}}$

We have to find Empirical formula of the compound.

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➝ The empirical formula is the simplest whole number ratio of atoms present in a compound.

◈ No. of moles :

⟶ Ag = 63.5/108 = 0.58

⟶ N = 8.2/14 = 0.58

⟶ O = 28.3/16 = 1.76

◈ Mole ratio :

➠ Ag = 0.58/0.58 = 1

➠ N = 0.58/0.58 = 1

➠ O = 1.76/0.58 = 3

◈ Whole no. ratio :

➢ Ag : N : 0 = 1 : 1 : 3

Empirical Formula ⇒ AgNO_3

Answer 2

Step by step explanation:

Given: Compound with $Ag=63.5$%, $N=8.2$%  and $O=28.3$%.

To find: Compound's empirical formula.

For calculation of empirical formula,

We know that the empirical formula is the simplest whole-number ratio of atoms present in a compound.

i. Number of moles:

⇒ $Ag= \frac{63.5}{108} = 0.58$

⇒ $N=\frac{8.2}{14} = 0.58$

⇒ $O=\frac{28.3}{16}= 1.76$

ii. Ratio of moles:

⇒ $Ag=\frac{0.58}{0.58} = 1$

⇒ $N=\frac{0.58}{0.58} = 1$

⇒ $O= \frac{1.76}{0.58} = 3$

∴ whole number ratio,

⇒ $Ag:N:O= 1:1:3$

⇒ $AgNO_{3}$

The compound's empirical formula is $AgNO_{3}$.