Question

an organic compound contains 48% carbon, 8% hydrogen, 28% nitrogen and 16% oxygen. Write its empirical formula. ​

Answer 1

Answer:

:

For the available data : Percentage of C=48% , Percentage of H=8% , Percentage of N=28%

Total percentage of C, H and N=48+8+28=84%

We know that the sum of the percentages must be 100. But in this case, it is only 84%. The balance (100−84)=16% is regarded as the percentage of oxygen∗ in the compound

Step I. Calculation of simplest whole number ratios of the elements

ElementCHNOPercentage48.08.028.016.0Atomic Mass1211416Gram atoms (Moles)47.012=48.01=828.014=216.016=1Atomic ratio (Molar ratio)41=481=821=211=1Simplest whole no. ratio4821

The simplest whole number ratios of different elements are : C:H:N:O::4:8:2:1

Step II. Writing the empirical formula of the compound

The empirical formula of compound =C4H8N2O.

Explanation:

jope you like my ans pls mark as brainliest :)

Answer 2

Given: An organic compound contains 48% carbon, 8% hydrogen, 28% nitrogen and 16% oxygen.

To find: We have to find out its emperical formula.

Solution:

Let the total weight of the compound is 100g.

It contain 48% carbon, 8% hydrogen, 28% nitrogen and 16% oxygen mean 48 gram carbon, 28g nitrogen, 8g hydrogen, and 16g oxygen.

Now the atomic weight of carbon, nitrogen,hydrogen and oxygen are 12,14,1,16.

Thus it will contain 48/12=4 carbon atoms.

28/14=2 nitrogen atoms.

8/1=8 hydrogen atoms and 16/16=1 oxygen atom.

So, the emperical formula of the compound is $C_4H_8N_2O$.