Urea is a very important nitrogenous fertilizer. Its formula is $C O N_{2} H_{4} .$ Calculate the percentage of nitrogen in urea.

Open in App

Solution

Step 1:- First calculating the molecular mass of $C O N_{2} H_{4 .}$ (Urea)
$A t o m i c m a s s o f C (C a r b o n) = 12$
$A t o m i c m a s s o f N (N i t r o g e n) = 14$
$A t o m i c m a s s o f H (H y d r o g e n) = 1$
So, the molecular mass of $C O N_{2} H_{4} = 12 + 16 + 28 + 4$
$= 60$
Step 2:- Then ,calculating the percentage of nitrogen in urea
$= \frac{M a s s o f n i t r o g e n i n t h e c o m p o u n d}{M o l e c u l a r M a s s} \times 100$
$= \frac{28}{60} \times 100 = 46.67 %$

Suggest Corrections

Similar questions

Urea is a very important nitrogenous fertilizer. Its formula is $C O N_{2} H_{4}$ Calculate the percentage of nitrogen in urea (C = 12, O = 16, N = 14 and H = 1).

Urea is a very important nitrogenous fertilizer. Its formula is $C O N_{2} H_{4}$ . Calculate the percentage of carbon in urea.

(C = 12, O = 16, N = 14 and H = 1)

Urea $[C O (N H_{2})_{2}]$ is an important nitrogenous fertilizer and is sold in 50 kg sacks. What is the mass of nitrogen in one sack of urea?

Calculate the percentage by weight of nitrogen found in urea.

Q. Urea is a very important fertiliser. Its formula is CON₂H₄. Calculate the % of nitrogen in the fertilizer.

Join BYJU'S Learning Program

Urea is a very important nitrogenous fertilizer. Its formula is CON2H4. Calculate the percentage of nitrogen in urea.

Urea is a very important nitrogenous fertilizer. Its formula is $C O N_{2} H_{4} .$ Calculate the percentage of nitrogen in urea.