Simple empirical formula question

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To determine the empirical formula of the manganese chloride compound, the reaction involves 1.36 g of manganese and produces 4.00 g of the compound, indicating that 2.64 g of chlorine is present. The calculations depend on whether chlorine is considered as Cl2 or Cl. When using Cl2, the resulting formula is Mn3Cl4, while using Cl gives Mn3Cl. The correct approach involves calculating the number of moles of manganese and chlorine to find their ratio. The discussion highlights the importance of accurate calculations in determining the empirical formula.
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Question:
The reaction of 1.36 g of manganese with excess cholrine produced 4.00 g of chloride compound of manganese. Determine the empirical formula of the product.

If I calculate this with cholrine as Cl_{2} I get Mn_{3}Cl_{4}.

But if I calculate it with cholrine as Cl I get Mn_{3}Cl.

Which way is correct?

Steve
 
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How do you get those two answers? Since you start with 1.36 g of manganese and end with 4 g of compound, there must be 4- 1.36 = 2.64 g of chlorine atoms in the compound. 1.36 g is how many manganese atoms? 2.64 g of chlorine is how many chlorine atoms? What is the ratio?
 
Doh! Dyslexia kicking in again. Thanks HallsofIvy, I found where I messed up now.

Steve
 

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