Enthalpy Lab

Topics: Oxygen, Ethanol, Carbon dioxide Pages: 7 (1026 words) Published: April 22, 2013

Determine the Enthalpy change of combustion ΔHc of three different alcohols. Methanol, Ethanol and Isopropilic acid.


1. Fill the spirit micro burner with Ethanol and weight it 2. Pour 100 cm3 of water into the aluminum cup
3. Arrange the cup a short distance over the micro burner 4. Measure the temperature of water
5. When the temperature of the water has risen by 10°C, record the temperature. 6. Reweight the microburner. Record
7. Repeat steps 1 to 6 but now with Methanol
8. Repeat step 1 to 6 with Isopropilic acid.

Data and Processing
Alcohols | Initial mass of microburner fill with alcohol (g)± 0.01| Final mass of microburner fill with alcohol (g)± 0.01| Initial temperature of water(°C)± 0.1| Final temperature of water(°C)± 0.1| Volume of water in metallic calorimeter (cm3) ±0.5| Ethanol | 5.38| 5.08| 23.0| 33.0| 100.0|

Methanol | 5.33| 4.94| 24.0| 34.0| 100.0|
Isopropolic acid | 5.45| 5.20| 24.0| 34.0| 100.0|

Find the mass of water


ρ (density) H2O = 1.0 g /cm3

Calculate % Uncertainty in mass of water

As the mass of water is the same in the 3 alcohols the %uncertainty is the same for all the alcohols

Absolute uncertainty of the measuring cylindermass of water × 100
Calculating Δ mass change (alcohol’s burned mass)

(initial mass ± 0.01 g)-(final mass ± 0.01 g)

Calculating percentage uncertainty in alcohol burned mass

Absolute uncertainty of alcohol's burned massalcohol's burned mass × 100

Calculate the percentage uncertainty of alcohol burned moles

percentage uncertainty of alcohol burned mass+percentage uncertainty of alcohol`s molar mass

Calculating ΔH (enthalpy change)

ΔH=-mass of water x specific heat of water x Δ T of water mol of alcohol

* The specific heat for water is 4.18

=100x4.184x 10=4,184 J or 4,184 KJ exothermic


* H2O = 100 ml
* mH2O= 100 mg

* t1 H2O= 23°C

Mass (i) methanol= 5.38g

* tf= H2O=33°C

Mass (f) methanol= 5.08 g

ΔT= 10°C

Calculating mass change



Δm=ΔmMr=0.3032.04=0.009 mol


% uncertainity(balance)=±0.020.30x 100=6.67 %

% uncertainity(thermometer )=±110x 100=10 %

%error=-726000-(-464888.9)-726000x 100=36%

Qualitative Observations

We could see from the burn of methanol that the flame owas of color orange red, moreover therewere not dirt in the bottle.


* H2O = 100 ml
* mH2O= 100 mg

* t1 H2O= 24°C

Mass (i) ethanol= 5.33 g

* tf= H2O=34°C

Mass (f) ethanol= 4.94 g

ΔT= 10°C

Calculating mass change

Δm=mi-mf= 0.39 g

5.33-4.94= 0.39 g

ethanol 0.3946.07 g/mol=0,008 mol


% uncertainity(balance)=±0.020.39x 100=13 %

% uncertainity(thermometer )=±110x 100=10 %

%error=-1360000-(-523000.0)-1368000x 100=61.8%

Qualitative Observations

We can observe a lost of weight during the experiment, moreover the flame was orange blue but with a big strong orange , it didn’t burn complete therefore show dirt in the cup.

Isopropolic acid

* H2O = 100 ml
* mH2O= 100 mg

* t1 H2O= 24°C

Mass (i) = 5.45 g

* tf= H2O=34°C

Mass (f) ethanol= 5.20g

ΔT= TF-TI=10 c

Δm=mi-mf= 0.25 g

Isopropolic acid 0.25 60,1g/mol=0,004 mol


% uncertainitybalance=±0.020.25x 100=8%

% uncertainity(thermometer )=±110x 100=10 %

%error=-2006.9-(-1046.0)-2006.9x 100=47.9%

* At last, the alcohol used was Isopropilic acid. The flame with this alcohol was the strongest flame, it was very strong, was very yellow at the top and blue at the bottom. * We could also notice that all the 3 alcohols produced Soot. (is a general term that refers to impure carbon particles resulting from the incomplete combustion)

Conclusion =

As we...
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