A molecule contains 47.30 g mercury (Hg) and 16.70 (Cl). what is its empirical formula?
1 answer:
Answer: B) HgCl2
Explanation:
Convert grams to moles for both elements by dividing the mass by it’s molar mass (Mr).
Hg- 47.30/201= 0.23 mols
Cl- 16.70/35= 0.47 mols
Now, divide the moles of both elements with the smallest number. The smallest number here is 0.23.
Hg- 0.23/0.23= 1
Cl- 0.47/0.23= 2.04 (since there is a 0 after the decimal we will forget the decimal and write the number as a whole.)
Empirical Formula= Hg1Cl2
Explanation:
Convert grams to moles for both elements by dividing the mass by it’s molar mass (Mr).
Hg- 47.30/201= 0.23 mols
Cl- 16.70/35= 0.47 mols
Now, divide the moles of both elements with the smallest number. The smallest number here is 0.23.
Hg- 0.23/0.23= 1
Cl- 0.47/0.23= 2.04 (since there is a 0 after the decimal we will forget the decimal and write the number as a whole.)
Empirical Formula= Hg1Cl2
You might be interested in
Mass of Fe₂O₃ : 74.26 g
Volume of O₂ at STP = 15.624 L
<h3>Further explanation</h3>Reaction
4Fe (s) + 3O₂ (g) ==> 2Fe₂O₃ (s)
mol Fe :
Ar Fe = 56 g/mol
mol Fe₂O₃ :
mass Fe₂O₃ :
MW Fe₂O₃ : 159,69 g/mol
mol O₂ :
volume O₂ at STP(1 mol=22.4 L)