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

1.21 g Be

2.15 g O

8.63 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:

1.21 g Be

2.15 g O

8.63 g S

Please report in the form Be#O#S#

BlackTom AI · Accepted Answer

To determine the empirical formula from mass data, I will convert each mass to moles using the respective atomic weights.

First, for Be: mass = 1.21 g and atomic weight ≈ 9.01 g/mol. Moles of Be = 1.21 / 9.01 ≈ 0.1343 mol.

Next, for O: mass = 2.15 g and atomic weight ≈ 16.00 g/mol. Moles of O = 2.15 / 16.00 ≈ 0.1344 mol.

Then, for S: mass = 8.63 g and atomic weight ≈ 32.07 g/mol. Moles of S = 8.63 / 32.07 ≈ 0.2691 mol.

Now compare the mole amounts: Be ≈ 0.1343, O ≈ 0.1344, S ≈ 0.2691. The smallest number is about 0.1343, so I divide each quantity by that value to get the simplest whole-number ratio.

Be: 0.1343 / 0.1343 ≈ 1.00
O: 0.1344 / 0.1343 ≈ 1.00
S: 0.2691 / 0.1343 ≈ 2.00

Thus, the simplest whole-number ratio is Be:O:S = 1:1:2, which corresponds to the empirical formula BeO S with the appropriate subscripts.

Note: When working with empirical formulas, small rounding differences can occur due to atomic weights used or measurement precision. Rechecking the calculations shows the ratios remain very close to integers, reinforcing the 1:1:2 relationship.

A chemist decomposes a compound; the masses of its constituent elements are shown below. Calculate the empirical formula for the compound: 1.21 g Be 2.15 g O 8.63 g S Please report in the form Be#O#S#

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