Question

what are the percentages of the elements C6H12NO2

Answer 1

Let us consider a mole of $C_6H_{12}NO_2$

mass of one mole of the above compound is $12\times6+12\times1+14+16\times2=130 grams$

Now lets see proportion of each element in it

Carbon- 12*6=72 grams

Hydrogen- 12 grams

Nitrogen- 14 grams

Oxygen- 32 grams

now lets calculate the percentage of each element in mole of the above compound

Carbon= $\frac{72}{130}\times100$ =55.38 %

Hydrogen=$\frac{12}{130}\times100$= 9.23 %

Nitrogen= $\frac{14}{130}\times100$= 10.77 %

Oxygen= $\frac{32}{130}\times100$= 24.61 %