Question

2.746 g of a compound on analysis gave 1.94 g of silver, 0.268 g of sulphur and 0.538 g of oxygen. Find the empirical formula of the compound.​

Answer 1

Answer:

Answer :

Empirical formula is Ag2SO4

Hope it helps you :)

Answer 2

Step 1 : convert given masses into moles

$\small{Moles \: of \: Ag =\frac{\text{ given mass of Ag}}{\text{ molar mass of Ag}}= \frac{1.94g}{108g/mole}=0.018 \: moles }$

$\small{Moles \: of \: S =\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{0.268g}{32g/mole}=0.008moles }$

$\small{Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.538g}{16g/mole}=0.033moles}$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

$For Ag = \frac{0.018}{0.008}=2$

$For S = \frac{0.008}{0.008}=1$

$For O = \frac{0.033}{0.008}=4$

The ratio of Ag: S: O= 2: 1: 4

Hence the empirical formula is $Ag_2SO_4$