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What is the empirical formula for a compound composed of 47.9 g zinc and 52.1 g chlorine?

A) Zn2Cl3
B) ZnCl2
C) ZnCl3
D) Zn3Cl2

User Ricco
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1 Answer

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Final answer:

To determine the empirical formula of a compound of zinc and chlorine, convert the masses to moles, find the simplest whole number ratio, and match to the formula ZnCl2.

Step-by-step explanation:

To determine the empirical formula for a compound composed of zinc and chlorine, we first need to calculate the moles of each element in the compound. The molar mass of zinc (Zn) is approximately 65.38 g/mol and the molar mass of chlorine (Cl) is approximately 35.45 g/mol. Using these molar masses, we can convert the given masses of the elements to moles:

  • Moles of zinc: 47.9 g Zn × (1 mol Zn / 65.38 g Zn) = 0.7325 moles Zn
  • Moles of chlorine: 52.1 g Cl × (1 mol Cl / 35.45 g Cl) = 1.470 moles Cl

We then divide the moles of each element by the smallest number of moles to get the ratio of atoms:

  • Zinc to chlorine ratio: 0.7325 moles Zn / 0.7325 = 1 Zn; 1.470 moles Cl / 0.7325 = 2 Cl

This results in a Zn to Cl ratio of 1:2, which corresponds to the empirical formula ZnCl2. Thus, the correct option is B) ZnCl2.

User Olubukola
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