Respuesta :
Answer:
[tex]Ag_2SO_4[/tex]
Explanation:
Formula for the calculation of no. of Mol is as follows:
[tex]mol=\frac{mass\ (g)}{molecular\ mass}[/tex]
Molecular mass of Ag = 107.87 g/mol
Amount of Ag = 5.723 g
[tex]mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol[/tex]
Molecular mass of S = 32 g/mol
Amount of S = 0.852 g
[tex]mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol[/tex]
Molecular mass of O = 16 g/mol
Amount of O = 1.695 g
[tex]mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol[/tex]
In order to get integer value, divide mol by smallest no.
Therefore, divide by 0.02657
[tex]Ag, \frac{0.05305}{0.02657} \approx 2[/tex]
[tex]S, \frac{0.02657}{0.02657} \approx 1[/tex]
[tex]O, \frac{0.10594}{0.02657} \approx 4[/tex]
Therefore, empirical formula of the compound = [tex]Ag_2SO_4[/tex]
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
