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Stels [109]
2 years ago
13

An unknown compound contains only carbon, hydrogen, nitrogen, and oxygen. Complete combustion of a 2.77 g sample of the compound

gives 6.64 g CO2 and 1.67 g H2O. Another sample of mass 1.69 g was found to contain 0.143 g N. The molar mass of the compound was found to be 165 g mol−1. Find the molecular formula of the unknown compound
Chemistry
1 answer:
alisha [4.7K]2 years ago
4 0

The molecular formula of the unknown compound is determined as  C₉H₁₁NO₂.

<h3>Molecular formula of the compound</h3>

The molecular formula is calculated as follows;

CHNO  +  O₂ ------------> CO₂  + H₂O

Mass of carbon, C:  = (6.64 x 12)/44 = 1.81 g in 2.77 g sample

Mass of hydrogen, H: = (1.67 x 2)/18 = 0.186 g in 2.77 g sample

Mass of Nitrogen, N: = (2.77 x 0.143)/1.69 = 0.234 g

Mass of oxygen, O:  = 2.77 g - 1.81 g - 0.186 g - 0.234 g = 0.54 g

<h3>molar ratio of the elements: </h3>

C = 1.81 g = 0.15 mol

H = 0.186 g = 0.186 mol

N = 0.234 g = 0.017 mol

O = 0.54 g = 0.0337 mol

divide through with the smallest number of moles (0.017 mol);

C = 9

H = 11

N = 1

O = 2

Molecular formula = C₉H₁₁NO₂

Check the molar mass of the compound = (9 x 12) + (11 x 1) + (14) + (2 x 16) = 165 g/mol

Thus, the molecular formula of the unknown compound is determined as  C₉H₁₁NO₂.

Learn more about molecular mass here: brainly.com/question/21334167

#SPJ1

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(b) 2.82

(c) 1.95

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The neutralization reaction in this question is

(CH3CH2)3N + HCl   ⇒ (CH3CH2)3NH⁺ + Cl⁻

The problem  can be  solved by calculating the number of moles of triethylamine after  addition of the portions of HCl. Since it is a weak base if it is not consumed completely, that is in excess we will have a buffer of a waek base. If its consumed completely the pH will be determined by the strong acid HCl.

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mol (CH3CH2)3NH⁺ produced = 0.0011 mol

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This is so  because the acid will consume completely the 0.0020 mol of the weak base  we had originally present.

Now the problem circumscribes to that of calculating the pH of the unreacted HCl

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[HCl] = 0.0005 mol / 0.045 L = 0.111

pH = - log ( 0.111) = 1.95                                            

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