Question

A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.

Answer 1

Answer:

$Ag_2SO_4$

Explanation:

Formula for the calculation of no. of Mol is as follows:

$mol=\frac{mass\ (g)}{molecular\ mass}$

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

$mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol$

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

$mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol$

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

$mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol$

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

$Ag, \frac{0.05305}{0.02657} \approx 2$

$S, \frac{0.02657}{0.02657} \approx 1$

$O, \frac{0.10594}{0.02657} \approx 4$

Therefore, empirical formula of the compound = $Ag_2SO_4$

Answer 2

Answer:

The empirical formula is Ag2SO4

Explanation:

Step 1: Data given

A compound contains Ag, S and O

Mass of Ag = 5.723 grams

Mass of S = 0.852 grams

Mass of O = 1.695 grams

Molar mass of Ag = 107.87 g/mol

Molar mass of S = 32.065 g/mol

Molar mass of O = 16 g/mol

Total mass = Mass of Ag + mass of S + mass of O = 5.723g + 0.852g + 1.695g = 8.27 grams

Step 2: Calculate moles

Moles = mass / molar mass

Moles of Ag = 5.723 grams / 107.87 g/mol = 0.05305 moles

Moles S = 0.852 grams / 32.065 g/mol = 0.0266 moles

Moles O = 1.695 grams / 16.00 g/mol = 0.1059 moles

Step 3: Calculate mole ratio

We divide by the smallest amount of moles

Ag: 0.05305/0.0266 = 2

S: 0.0266/0.0266 = 1

O: 0.1059 / 0.0266 = 4

The empirical formula is Ag2SO4