A chemist decomposes a compound; the masses of its constituent elements are shown below. Calculate the empirical formula for the compound:

1.01 g Be

1.80 g O

7.19 g S

Please report in the form Be#O#S#

Question

A chemist decomposes a compound; the masses of its constituent elements are shown below. Calculate the empirical formula for the compound:

1.01 g Be

1.80 g O

7.19 g S

Please report in the form Be#O#S#

BlackTom AI · Accepted Answer

The task presents a decomposition data set and asks for the empirical formula Be#O#S# based on the masses given.
First, convert each mass to moles using approximate atomic masses: Be: 1.01 g / 9.01 g·mol⁻¹ ≈ 0.112 mol; O: 1.80 g / 16.00 g·mol⁻¹ = 0.1125 mol; S: 7.19 g / 32.06 g·mol⁻¹ ≈ 0.224 mol.
Now compare the mole amounts to obtain the simplest whole-number ratio. Divide all three by the smallest number among them, which is ~0.112 mol (Be). This yields Be ≈ 1.00, O ≈ 0.1125 / 0.112 ≈ 1.00, and S ≈ 0.224 / 0.112 ≈ 2.00.
Thus, the smallest whole-number ratio of atoms is Be:O:S = 1 : 1 : 2.
Expressed as a chemical formula with these ratios, the empirical formula is BeOS2 (BeO1S2). 
Note on interpretation: when forming the empirical formula, the sequence of elements can be arranged in any conventional order (e.g., BeOS2, BeSO2, or BeOS2 depending on ordering), but the key point is the relative counts: one Be, one O, and two S per formula unit.
If you compare to a provided key or answer option that states BeSO4, that would not align with the calculated mole ratios from the given masses, since that would require O to be four times the amount of Be, not equal to it, and S to be equivalent to Be, not twice Be.
In this calculation, the mole ratio leads to a single Be atom, a single O atom, and two S atoms per empirical unit, giving BeOS2 as the empirical formula."

A chemist decomposes a compound; the masses of its constituent elements are shown below. Calculate the empirical formula for the compound: 1.01 g Be 1.80 g O 7.19 g S Please report in the form Be#O#S#

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