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
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