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kow [346]
3 years ago
15

Se someten a combustion 0,452g de un compuesto de C,H y N de masa molecular 80. Al recoger el CO2 y el H2O producidas obtenemos

0,994 g y 0,203 respectivamente. Determine formula empirica y formula molecular
Chemistry
1 answer:
NARA [144]3 years ago
3 0

Answer:

Fórmula empírica: C₂H₂N

Fórmula molecular: C₄H₄N₂

Explanation:

Un compuesto que contiene carbono hidrógeno y nitrógeno con fórmula CₐHₓNₙ es sometido a combustion produciendo:

CₐHₓNₙ + O₂ → aCO₂ + x/2 H₂O + nNO₂

Con la masa de dióxido de carbono y agua podemos encontrar las moles de carbono e hidrógeno y su aporte a los 0.452g de muestra que fueron puestos en combustión, así:

<em>Moles C:</em>

Moles C = Moles CO₂ = 0.994g CO₂ ₓ (1mol / 44g) = 0.0226 moles C

Masa C: 0.0226 moles C ₓ (12.01g / mol) = 0.271g Carbono hay en la muestra

<em>Moles H:</em>

Moles H = 2 Moles H₂O = 0.203g H₂O ₓ (1mol / 18g) = 0.0113 moles H₂O = 0.0226 moles H

Masa H: 0.0226 moles H ₓ (1.01g / mol) = 0.023g Hidrógeno hay en la muestra

Así, la masa de nitrógeno en la muestra y sus moles son:

Masa N = 0.452g - 0.271g C - 0.023g H

Masa N = 0.158g Nitrógeno

Y su moles son:

0.158g ₓ (1 mol / 14.01g) = 0.0113 moles N

Con las moles de C, H y N podemos determinar la formula empírica que se define como: "La relación de números enteros más simple entre la cantidad de átomos presentes en una mólecula. Si usamos como base las moles de nitrógeno (Valor menor):

Relación H/N: 0.0226 mol / 0.0113 mol = 2

Relación C/N: 0.0226 mol / 0.0113 mol = 2

Relación N/N: 0.0113 mol / 0.0113 mol = 1

Así, la <em>fórmula empírica es:</em>

<h3>C₂H₂N</h3>

Esta fórmula empírica tiene una masa molar de:

2C = 2*12 g/mol = 24g/mol

2H = 2*1g/mol = 2g/mol

N = 14g/mol

24+14+2 = 40g/mol

Como la masa molecular del compuesto es 80g/mol (Dos veces la de la fórmula empírica, la <em>fórmula molecular es 2 veces la fórmula empírica:</em>

<h3>C₄H₄N₂</h3>
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Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).
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a) before addition of any KOH : 

when we use the Ka equation & Ka = 4 x 10^-8 : 

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

           = 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

       = 9.2 x 10^-5 M

when PH = -㏒[H+]

   PH = -㏒(9.2 x 10^-5)

        = 4  

b)After addition of 25 mL of KOH: this produces a buffer solution 

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]


first, we have to get moles of HCIO= molarity * volume

                                                           =0.21M * 0.05L

                                                           = 0.0105 moles

then, moles of KOH = molarity * volume 

                                  = 0.21 * 0.025

                                  =0.00525 moles 

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L =  0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

                                        = 0.00525 / 0.075

                                        =0.07 M

and molarity of KCIO = moles KCIO / total volume

                                    = 0.00525 / 0.075

                                    = 0.07 M

and when Ka = 4 x 10^-8 

∴Pka =-㏒Ka

         = -㏒(4 x 10^-8)

         = 7.4 

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4 

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume 

                                                        = 0.21 M * 0.05L

                                                        = 0.0105 moles

then moles KOH = molarity * volume
                            =  0.22 M* 0.035 L 

                            =0.0077 moles 

∴ moles of HCIO remaining = 0.0105 - 0.0077=  8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume 

                                      = 8 x 10^-5 / 0.085

                                      = 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

                                          = 0.0077M / 0.085L

                                          = 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

                                   = 0.0105mol / 0.1 L

                                   = 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

       = 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO] 

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

           = 3.8
∴PH = 14- POH

        =14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

 when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M 

V2 is the excess volume added  of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

           = -㏒0.02 

           = 1.7

∴PH = 14- POH

       = 14- 1.7 

      = 12.3 
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