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.

View Explanation

Log in for full answers

Similar Questions

Surgical grade titanium has the mass percent composition of 64.39% titanium, 11.42% vanadium, and 24.19% aluminum. What is the empirical formula of this compound?

A compound that contains only xenon and oxygen is 73.2% Xe by mass. What is the empirical formula for this compound?

A compound has an elemental analysis of 48.2% C, 8.2% H, and 43.2% O by mass. What is the empirical formula?

A compound has an empirical formula of CH and a molecular ion at m/z = 26. The compound could be

Question at position 2 A sample of an oxide of vanadium is separated into its elements, resulting in 1.02 g V and 0.48 g O. Which of the following is the formula of the oxide?V2OV2O5V2O3VO2

A common anti-inflammatory drug is 75.69% C, 8.80% H, and 15.51% O. What is its empirical formula?

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#

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#

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler