Answer:
The answer to your question is Empirical formula InCl₃
Explanation:
Data
InCl = 0.5 g
Cl = 0.2404 g
Empirical formula = ?
Process
1.- Calculate the mass of Indium
Total mass = mass of Indium + mass of Chlorine
0.50 = mass of Indium + 0.2404
mass of Indium = 0.50 - 0.2404
mass of Indium = 0.2596 g
2.- Calculate the moles of Indium and Chloride
Atomic mass Indium = 115 g
Atomic mass Chlorine = 35.5 g
115 g of In ------------------ 1 mol
0.2596 g of In ------------- x
x = (0.2596 x 1) / 115
x = 0.0023 moles of Indium
35.5 g of Cl ------------- 1 mol
0.2404 g --------------- x
x = (0.2404 x 1) / 35.5
x = 0.0068 moles of Cl
3.- Divide by the lowest number of moles
Indium 0.0023 / 0.0023 = 1
Chlorine 0.0068 / 0.0023 = 3
4.- Write the empirical formula
InCl₃
The empirical formula of the indium compound is [tex]InCl_3[/tex]
Mass of indium = 0.500-0.2404
= 0.2596g
And,
Mass of chlorine = 0.2404g
Now
Divide each mass by atomic mass
[tex]In = 0.2596\div 114.818 = 0.002261\\\\Cl = 0.2404\div 35.453 = 0.00678[/tex]
Divide by smaller:
In = 1
Cl = 3
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