Question

What is the empirical formula for a compound containing 38.8% carbon, 16.2 % hydrogen, and 45.1% nitrogen?

Answer 1

CH51N10

Step-by-step explanation:

Start with the number of grams of each element, given in the problem.

If percentages are given, assume that the total mass is 100 grams so that

the mass of each element = the percent given.

Given

C= 38.0g

H =16.2g

N= 45.1g

Convert the mass of each element to moles using the molar mass from the periodic table.

C= 38*1/120=0.32mol

H=16.2*1/1=16.2mol

N=45.1*1/14=3.22mol

Divide each mole value by the smallest number of moles calculated. Round to the nearest whole number

C=0.32/0.32=1

H=16.2/0.32=50.6 =51

N=3.22/0.3 =10

This is the mole ratio of the elements and is represented by subscripts in the empirical formula.

CH51N10

Answer 2

`CH_5N`

Step-by-step explanation:

Hello,

Percent composition is defined as:

`Percent_i=(n*m_i)/(M) *100`%

Whereas
`n` is the subscript in the formula,
`m_i` the atomic mass of the ith element and
`M` the compound's molecular mass, thus, solving for
`n`:

`n_i=(percent*M)/(100*m_i) \\n_C=(38.8*M)/(100*12) =0.0323M\\n_H=(16.2*M)/(100*1) =0.162M\\n_N=(45.1*M)/(100*14) =0.0322M\\`

Now, divining each
`n_i` over the least one (
`n_N`) we get:

`n_C=(0.0323)/(0.0322)=1\\\\n_H=(0.162)/(0.0322)=5\\\\n_N=(0.0322)/(0.0322)=1`

Finally the empirical formula turns out into:

`CH_5N`

Best regards.