Question

2.746 g of a compound gave an analysis of 1.94g of silver, 0.26g of sulphur, o.538g of oxygen. Find its empirical formula

Answer 1

The empirical formula is ag2so4

Answer 2

The empirical formula is $Ag_2SO_4$

Explanation:

Step 1 : convert given masses into moles

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

Moles of S =$\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{0.26g}{32g/mole}=0.008moles$

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 $Ag_2SO_4$

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