### Chemistry Objective Question: #### Question: Imagine a newly discovered element, named Imaginium (Im ), with an atomic number of 140. Imaginium readily reacts with chlorine to form a compound where Imaginium exhibits an oxidation state of +3. If a sample of this compound contains 70% chlorine by mass, determine the empirical formula of the Imaginium chloride compound. (Atomic masses: Im = 320, Cl = 35.5) (A) ImCI (B) ImCl₂ (C) ImCI, (D) Im₂Cl, #### Solution: To find the empirical formula, we'll first establish the mass percentage of each element in the compound and then convert these to moles to find the simplest whole number ratio. 1. Moles of Chlorine (CI): - Mass percent of Cl = 70% - In a 100 g sample, mass of Cl = 70 g - Moles of Cl = Mass of Cl / Atomic mass of Cl = 70 g / 35.5 g/mol = 1.97 mol 2. Moles of Imaginium (Im): - Mass percent of Im = 100% - 70% (of Cl) = 30% - In a 100 g sample, mass of Im = 30 g - Moles of Im = Mass of lm / Atomic mass of Im = 30 g/320 g/mol = 0.094 mol Next, we determine the simplest whole number ratio: - The mole ratio of Im to Cl in the compound is approximately 0.094: 1.97. To simplify this, we divide both numbers by the smaller number (0.094): Im = 0.094/0.094 = 1- Cl = 1.97/0.094 = 21 Since we cannot have fractional atoms in a formula and the ratio of Cl to Im is approximately 3:1 (21 is close to 3 times 7), the empirical formula is likely ImCl,, where three chlorine atoms balance the +3 oxidation state of one Imaginium atom. The correct answer is (C) ImCl3. This question is designed to be a mix of simple concepts (empirical formula calculation, oxidation states) in a more complex, theoretical context, involving a hypothetical element.
### Chemistry Objective Question: #### Question: Imagine a newly discovered element, named Imaginium (Im ), with an atomic number of 140. Imaginium readily reacts with chlorine to form a compound where Imaginium exhibits an oxidation state of +3. If a sample of this compound contains 70% chlorine by mass, determine the empirical formula of the Imaginium chloride compound. (Atomic masses: Im = 320, Cl = 35.5) (A) ImCI (B) ImCl₂ (C) ImCI, (D) Im₂Cl, #### Solution: To find the empirical formula, we'll first establish the mass percentage of each element in the compound and then convert these to moles to find the simplest whole number ratio. 1. Moles of Chlorine (CI): - Mass percent of Cl = 70% - In a 100 g sample, mass of Cl = 70 g - Moles of Cl = Mass of Cl / Atomic mass of Cl = 70 g / 35.5 g/mol = 1.97 mol 2. Moles of Imaginium (Im): - Mass percent of Im = 100% - 70% (of Cl) = 30% - In a 100 g sample, mass of Im = 30 g - Moles of Im = Mass of lm / Atomic mass of Im = 30 g/320 g/mol = 0.094 mol Next, we determine the simplest whole number ratio: - The mole ratio of Im to Cl in the compound is approximately 0.094: 1.97. To simplify this, we divide both numbers by the smaller number (0.094): Im = 0.094/0.094 = 1- Cl = 1.97/0.094 = 21 Since we cannot have fractional atoms in a formula and the ratio of Cl to Im is approximately 3:1 (21 is close to 3 times 7), the empirical formula is likely ImCl,, where three chlorine atoms balance the +3 oxidation state of one Imaginium atom. The correct answer is (C) ImCl3. This question is designed to be a mix of simple concepts (empirical formula calculation, oxidation states) in a more complex, theoretical context, involving a hypothetical element.
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:### Chemistry Objective Question: #### Question: Imagine a newly discovered element, named Imaginium (Im
), with an atomic number of 140. Imaginium readily reacts with chlorine to form a compound where Imaginium
exhibits an oxidation state of +3. If a sample of this compound contains 70% chlorine by mass, determine the
empirical formula of the Imaginium chloride compound. (Atomic masses: Im = 320, Cl = 35.5) (A) ImCI (B)
ImCI₂ (C) ImCI, (D) ImgCl, #### Solution: To find the empirical formula, we'll first establish the mass percentage
of each element in the compound and then convert these to moles to find the simplest whole number ratio. 1.
Moles of Chlorine (CI): - Mass percent of Cl = 70% - In a 100 g sample, mass of Cl = 70 g - Moles of Cl =
Mass of Cl / Atomic mass of Cl = 70 g / 35.5 g/mol = 1.97 mol 2. Moles of Imaginium (Im): - Mass percent of
Im = 100% - 70% (of Cl) = 30% - In a 100 g sample, mass of Im = 30 g - Moles of Im = Mass of lm /
Atomic mass of Im = 30 g/320 g/mol = 0.094 mol Next, we determine the simplest whole number ratio: - The
mole ratio of Im to Cl in the compound is approximately 0.094: 1.97. To simplify this, we divide both numbers
by the smaller number (0.094): - Im = 0.094/0.094 = 1- Cl = 1.97/0.094 = 21 Since we cannot have
fractional atoms in a formula and the ratio of Cl to Im is approximately 3:1 (21 is close to 3 times 7), the empirical
formula is likely ImCl₂, where three chlorine atoms balance the +3 oxidation state of one Imaginium atom. The
correct answer is (C) ImCl3. This question is designed to be a mix of simple concepts (empirical formula
calculation, oxidation states) in a more complex, theoretical context, involving a hypothetical element.
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