Respuesta :
Answer:
[tex]\large\boxed{\large\boxed{C_5H_3O_2N_2}}[/tex]
Explanation:
1. Build a table with the analysis and the atomic masses
Element Massgrams Atomic mass
(g) g/mol
C 97.56 12.011
H 4.878 1.008
O 52.03 15.999
N 45.53 14.007
2. Divide the mass of each element by its atomic mass to obtain the amount of each element in moles
Element Massgrams Atomic mass moles
(g) g/mol
C 97.56 12.011 8.1226
H 4.878 1.008 4.8393
O 52.03 15.999 3.2521
N 45.53 14.007 3.2505
3. Divide each amount in moles by the least number of moles: 3.2505
Element moles ratio
C 8.1226 2.5
H 4.8393 1.5
O 3.2521 1.0
N 3.2505 6.0
4. Mutiply each ratio by 2 to obtain whole numbers
Element ratio ratio
C 2.5 5
H 1.5 3
O 1.0 2
N 1.0 2
5. Write the empirical formula
The mole ratios are represented with suberscripts in the chemical formula:
[tex]C_5H_3O_2N_2[/tex].
The empirical formula will be: [tex]C_5H_3O_2N_2[/tex]
What information we have:
Element Mass Atomic mass
C 97.56 12.011
H 4.878 1.008
O 52.03 15.999
N 45.53 14.007
Calculation for moles:
C 97.56 g / 12.011 g/mol = 8.1226 moles
H 4.878 g / 1.008 g/mol = 4.8393 moles
O 52.03 g / 15.999 g/mol = 3.2521 moles
N 45.53 g / 14.007 g/mol =3.2505 moles
Dividing each mole by the least number of moles: 3.2505
C = 2.5
H = 1.5
O = 1.0
N= 6.0
On further simplifying:
C = 5
H = 3
O = 2
N = 2
Determination of empirical formula:
The mole ratios are represented with subscripts in the chemical formula:
[tex]C_5H_3O_2N_2[/tex].
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