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A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157g of the compound produced 0.213g of CO2 and 0.0310g of H2O. In another experiment, it is found that 0.103g of the compound produces 0.0230g of NH3. What is the empirical formula of the compound? (hint : combustion involves reacting with excess O2. Assume that all the carbon ends up in CO2 and all the hydrogen ends up in H2O. Also assume that all the nitrogen ends up in the NH3 in the second experiment.)

Respuesta :

Answer:

C7H5N3O6

Explanation:

C = 12 g/mol

O = 16 g/mol

H = 1 g/mol

N = 14 g/mol

CO2 = 44 g/mol

H2O = 18 g/mol

NH3 = 17 g/mol

* 44 g CO2 is produced from 12 g C

0.213 g CO2 is produced from = (12)(0.213) / 44 = 0.0581 g C present in 0.157 g compound.

* 18 g H2O is produced from 2 g H

0.0310 g H2O is produced from = (2)(0.0310) / 18 = 0.0034 g H present in 0.157 g compound.

* 17 g NH3 is produced from 14 g N

0.0230 g NH3 is produced from = (14)(0.0230) / 17 = 0.0189 g N present in 0.103 g compound.(this is different)

If in 0.103 g compound there is 0.0189 g N

in 0.157 g compound there is ;

(0.157)(0.0189) / 0.103 = 0.0288 g N

* Mass of C, H and N = 0.0581 + 0.0034 + 0.0288 = 0.0903 g

The rest is mass of O = 0.157 - 0.0903 = 0.0667 g

Moles of elements in the compound sample:

C = 0.0581 g / 12 g/mol = 0.00484 mol

H = 0.0034 g / 1 g/mol = 0.0034 mol

N = 0.0288 g / 14 g/mol = 0.00205 mol

O = 0.0667 g / 16 g/mol = 0.00417 mol

Smallest mole ratio:

N = 0.00205 / 0.00205 = 1.00

C = 0.00484 / 0.00205 = 2.36

H = 0.0034 / 0.00205 = 1.66

O = 0.00417 / 0.00205 = 2.00

Multiplying by 3 we get;

N = 3

C = 7

H = 5

O = 6

Simplest formula:

C7H5O6N3

This is probanly TNT (trinitrotoluen)

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