A sample of phosphorus of mass 0.808 g reacts with oxygen to form 1.852 g of a phosphorus oxide. Determine the empirical formula of the oxide.

Question

BlackTom AI · Accepted Answer

We start by identifying the masses involved and compute the mass of oxygen that reacted. The oxide has mass 1.852 g and the phosphorus sample mass is 0.808 g, so oxygen mass = 1.852 g − 0.808 g = 1.044 g.

Next, convert masses to moles using atomic masses: phosphorus (P) has an approximate atomic mass of 30.97 g/mol, and oxygen (O) has 16.00 g/mol.
- Moles of P = 0.808 g / 30.97 g/mol ≈ 0.0261 mol.
- Moles of O = 1.044 g / 16.00 g/mol ≈ 0.0653 mol.

Now determine the simplest whole-number ratio by dividing each amount by the smallest number of moles, which is 0.0261:
- P: 0.0261 / 0.0261 ≈ 1.00
- O: 0.0653 / 0.0261 ≈ 2.50
The oxide thus has a P:O ratio of approximately 1 : 2.50. To convert this to whole numbers, multiply both by 2, giving P2O5 as the empirical formula. This scaling step is standard when you obtain a non-integer ratio like 2.50.

Now assess each answer option:
- P4O5: This would correspond to a P:O ratio of 4:5, or 0.8, which does not match the calculated ~1:2.50 (or 2:5 after scaling). Therefore this option does not fit the data.
- P2O5: This matches the derived ratio 1:2.50 (scaled to 2:5), yielding the empirical formula P2O5. It aligns with the calculation exactly.
- PO4: This implies a 1:4 ratio, or P:O = 1:4, which is far from the observed 1:2.50, so this is incorrect.
- PO3: This implies a 1:3 ratio, which again does not match the obtained 1:2.50 ratio, so it is incorrect.
- P2O: This would give a 2:1 ratio, or P:O = 2:1, which is not consistent with the data either, so this option is incorrect.

In summary, the calculated mole ratio leads to the empirical formula that corresponds to a P:O ratio of about 1:2.5, which scales to P2O5. The other options fail to fit that ratio when compared to the experimental data.

A sample of phosphorus of mass 0.808 g reacts with oxygen to form 1.852 g of a phosphorus oxide. Determine the empirical formula of the oxide.

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