18 Chemistry 103 Finals NAME DATE SHOW SET UP AND SOLUTION 1. Determine the mole

Question

18 Chemistry 103 Finals NAME DATE SHOW SET UP AND SOLUTION 1. Determine the molecular formula of a compound that contains 66 e% c. 11 and has a molecular weight of 72 2. A compound contains only carbon, hydrogen, and nitrogen. After combustion of a 0 500 ssmple, the mass of the first trap (collecting H/0) Increases by 0.698 g and the mass of the ond trap (collecting COl Increases by 0 977 E: What is the emplrical formuia of this compourd? ANS 3. What is the molecular formula of a substance that contains only 0.801 g of carbon and 0.10 of hydrogen and has a molecular weight of 54? 4. Given the following equation, calculate the mass of O: needed to react completelyyw NO. 2NO + O2 2NO2

Explanation / Answer

1) we have given molecular weight of unkonown compound is 72

there are percentage given so,

Amount of C : Amount of H : Amount of O = 66.6 : 11.2 : 22.2
Atoms of C : Atoms of H : Atoms of O

atoms of C = 66.6/12 = 5.55

atoms of H = 11.2/1.008 = 11.11

atoms of O = 22.2/16 = 1.3875

now we convert in simple ration = 4 : 8 : 1

molecular formula will be = C4H8O

2) write the chemical equation

CxHyNz + O2 --> CO2 + H2O + NH3

With x , y and z is unknown

C = 12 g/mol
O2 = 32 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

0.977 g CO2 is produced from = (12)(0.977) / 44 = 0.266 g C present in 0.5 g compound.

18 g H2O is produced from 2 g H

0.698 g H2O is produced from = (2)(0.698) / 18 = 0.0775 g H present in 0.5 g compound.

now remainig will be N = 0.5 - (0.266 +0.0775) = 0.1564 g

no. of moles of C = (0.266 g / 12 g/mol ) = 0.02216

no. of moles of H = (0.0775 g / 1.0 g/mol ) = 0.0775

no. of moles of N = (0.1564 g / 14 g/mol) = 0.01117

ratio of C : H : N = 2 : 7 : 1

molecular formula will be = C2H7N

x = 2 , y = 7 and z = 1

:- { other 3 and 4 question are not shown }

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.