Quality control of fertilizers – determining urea content from nitrogen assay A urea sample shows an N (nitrogen) content of 42% by weight. Given molecular weight of urea = 60, what is the actual urea content of the sample (mass %)?

Introduction / Context:
Fertilizer grade verification often uses nitrogen analysis. For urea, knowing the theoretical nitrogen fraction allows back-calculating the mass fraction of urea in a mixture or degraded sample. This is a routine quality-control calculation in fertilizer plants and agro-retail labs.

Given Data / Assumptions:

• Molecular weight of urea (NH2CONH2) = 60 g/mol.
• Nitrogen mass per mole of urea = 2 * 14 = 28 g.
• Measured nitrogen percentage in the sample = 42% by mass.

Concept / Approach:
Theoretical nitrogen in pure urea is 28/60 by mass = 0.4667 → 46.67%. If a sample analyzes at 42% N, the mass fraction of urea is the ratio of measured N% to theoretical N%, assuming all nitrogen arises from urea.

Step-by-Step Solution:

Verification / Alternative check:
Take 100 g sample → contains 42 g N → equivalent urea mass = 42 / 0.4667 ≈ 90.0 g, confirming 90%.

Why Other Options Are Wrong:

• 80%, 95%, 98%: Do not match the stoichiometric ratio when using the correct theoretical nitrogen content of urea.

Common Pitfalls:
Using 50% as rough N content for urea (incorrect); ignoring that all nitrogen detected may not come from urea in adulterated samples (an assumption stated here).

Final Answer:
90%