10.47 g of a compound contained 6.25 g of metal a and rest mom metal b calculate the empirical formula of the compound (at wt of a=207,b=35.5)
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Mass of a = 6.25 g
Mass of b = 10.47 g - 6.25 g = 4.22 g
Moles of a = (6.25 g) / (207 g/mol) = 0.03 mol
Moles of b = (4.22 g) / (35.5 g/mol) = 0.1 mol
Ratio of a and b is
a : b = 0.03 : 0.1 = 3 : 10
Empirical formula = a₃b₁₀
Mass of b = 10.47 g - 6.25 g = 4.22 g
Moles of a = (6.25 g) / (207 g/mol) = 0.03 mol
Moles of b = (4.22 g) / (35.5 g/mol) = 0.1 mol
Ratio of a and b is
a : b = 0.03 : 0.1 = 3 : 10
Empirical formula = a₃b₁₀
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