You have been given confiscated compounds from a makeshift laboratory and you ha

Question

You have been given confiscated compounds from a makeshift laboratory and you have to characterizae these compounds. In an effort to characterize this compound, you performed the following experiment. 1.00g of compound A (liquid) was burned in an adiabatic calorimeter at constant pressure. A temperature change of +12.22 C was observed in the water bath surrounding the vessel. Analytical analysis indicated that the products of combustion were: 3.14 g CO2 (g) and 1.29 g H20 (g). The calorimeter has been calibrated with the combustion of a standard compound. The complete combustion of 1.00 g of n-octane (C8H18, (liquid)) resulted in a temperature change of +12.00 C in the water bath of the calorimeter. Analytical analysis reveals that the compound A contains 4 carbon atoms, with H as the only other element present. What is the enthaly of formation of compound A? Enthalpy of Formation of CO2 (g) = -393.5 kJ/mol Enthalpy of Formation of H20 (g) = -241.82 kJ/mol Enthalpy of Formation of C8H18 (l) = -250.10 kJ/mol Please explain all the steps and type out your answer (its easier to follow). This is all the information that the question gives you. You are supposed to be able to answer it using q = mc(t2-t1), find the calorimeter constant and use that. I dont know how to do that

Explanation / Answer

1. As per the given data combustion of compound A gives-

C 4Hx + O 2 ——> CO 2 + H 2O

1.00g. 3.14g 1.29g

Molar mass of CO 2 = 44.008g/mol

Molar mass of H 2O = 18.015 g/mol

So moles of CO 2 = 3.14g/ 44.008gmol-1 = 0.0713mol

0.0713 mol of CO 2 will produce 0.0713 mol of Carbon.

Moles of H 2O = 1.29g/ 18.015 gmol -1 = 0.0644 mol

0.0644 mol of H 2O will produce 0.0644×2 = 0.1288 mol of hydrogen.

To find the empirical formula , we have to divide 0.0713 to both the values.

This gives moles of C = 0.0713/0.0713 = 1mol

No. Of moles of H = 0.1288 / 0.0713 = 1.8 2 mol

So the empirical formula is CH 2

Molecular formula = n × empirical formula

Here n=4, given

So molecular formula = CH 2 ×4 = C 4 H 8 is compound A.

We are given,

C 8H18 (l) + 25/2 O 2 ———> 8 CO 2 (g) +9H 2O

H°f =  -250.10kJ/mol . -393.5kJ/mol -241.82kJ/mol

rH = 8× -393.5kJ/mol + 9× -241.82kJ/mol - 1× -250.10kJ/mol

= -5074.28 kJ/mol for C 8H 18

Molar mass of C 8H 18 = 114 g/mol

So the enthalpy of reaction per gram = -5074.28/114

= -44.511 kJ/g

Now we have to find out the caloriemeter constant that is C cal

Given T = 12°C

q= mC cal T and q = -rH

So C cal = q/mT = 44.511÷ 12× 1.00 = 3.709 kJ/°C

This remains constant for compound A also.

So q for A can be calculated as -

q = C cal m T = 3.709kJ/°C × 1.00g × 12.22°C

= 45.323kJ/g

Molar mass of compound A i.e., C 4H 8  = 56g/mol

So molar enthalpy of combustion of A = -45.323kJ/g × 56g/mol

= -2538.09 kJ/mol

Combustion of compound A is given as -

  C 4H 8 + 6 O 2 ——> 4 CO 2 + 4 H 2 O

H°f of C 4H8 = H°f (product) - rH

= 4× -393.5 kJ/mol +4× -241.82kJ/mol - (-2538.09kJ/mol)

= -3.19 kJ/mol. (Answer)

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