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### Topic: Calorimetry- How to calculate the amount of energy released by a fuel per mole?  (Read 14488 times)

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#### MattA147

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##### Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« on: March 09, 2013, 09:35:34 AM »
Hi guys essentially we have been covering calorimetry in class at the moment and I am completely Ok with finding the amount of energy being released by a fuel and then the amount of energy released per gram but not the amount of energy released per mole? Can someone give me some guidance? Thanks.

#### Borek

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #1 on: March 09, 2013, 09:52:12 AM »
But you know how to convert mass to moles?

Note: in the case of fuels that are mixtures, that's not necessarily possible/simple.
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#### MattA147

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #2 on: March 09, 2013, 12:30:56 PM »
I know that Mass=Mr x Moles if that's what you mean?

#### Borek

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #3 on: March 09, 2013, 12:55:53 PM »
Right.

So converting between molar heat and heat per mass can't be more difficult that that.
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#### MattA147

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #4 on: March 09, 2013, 01:32:07 PM »
Can you guide me through how to do exactly that? I am still not understanding.

#### Borek

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #5 on: March 09, 2013, 02:30:23 PM »
Give an example of a problem you can't solve.
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#### MattA147

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #6 on: March 10, 2013, 07:43:21 AM »
1) 0.45g of fuel A was burned and heated 150g of water in a calorimeter. The temperature of the water changed from 19°C to 45°C how much was released by 1.0g of fuel?

I answered this question using Q=mcΔT and I came to 36.4KJ/g. Which I checked and this was said to be correct.

2) In a similar question to that in Question 1) fuel B released 35.6 KJ/g. The relative formula mass of fuel A is 72 and Fuel B is 114. Which fuel releases more energy per mole?

Which is the part of the question which has got me stuck. Any help would be appreciated.

#### sjb

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #7 on: March 10, 2013, 08:27:15 AM »
How many grams are there in a mole of A? of B?

#### MattA147

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #8 on: March 10, 2013, 11:02:22 AM »
The Mass of Fuel A would be 72g and the Mass of Fuel B would be 114g. What is the next stage?

#### Borek

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #9 on: March 10, 2013, 12:24:04 PM »
Can you set up proportion? If n grams produce ΔE energy, 72g (molar mass) produces x energy?
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#### MattA147

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #10 on: March 16, 2013, 02:28:59 PM »
I'm sorry but I'm afraid that doesn't make any sense to me. Can you explain it to me in another way please?

#### Borek

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #11 on: March 16, 2013, 02:38:52 PM »

1. If 1 mole of A produces 1 kJ, how many kJ are produced by 100 moles?

2. If 0.01 mole of B produces 1 kJ, how many kJ are produced by 1 mole?

3. If molar mass of C is 50 g/mol, how many moles in 10 g?

4. If 10 g of C (molar mass 50 g/mol) produce 1 kJ, how many kJ are produced by 1 mole?
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#### consul

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##### Re: Calorimetry- How to calculate the amount of energy released by a fuel per mole?
« Reply #12 on: March 16, 2013, 09:31:10 PM »

36.4 KJ  A          ? g  A              ? kJ  A
------------  *    -------- =   --------------
g  A                 mol  A              mol A

Fill up the missing details. Notice how the units cancel?
To cancel the unit g A in the first term, you need to have g A
in the numerator of your conversion factor.

The conversion factor is the relative mass per mole of A.
It can be stated as either:

72 g A               1 mol A
------   or          -------
mol A                 72 g A

Since 72 g A = 1 mol A, the two ratios are virtually equal to 1.
Thus, it does not change the value of the original term that it
is multiplied to, but it only changes the units. This is called
dimensional analysis.

To give you another example.

To convert say 1250 inches to miles:

Is there a direct conversion from inches to miles that you know of?
If none, then how about feet to mile? There is! It is 5,280 feet per mile.
Now you work your way by converting inches to ____ first then ____ to miles. Here it is:

? feet            1 ?
1250 inches *  ----------  *  --------- = 0.0197 or 1.97 x 10^-2 mile
? inches        ? feet

Practice and more practice. What is important is if you want to cancel the units in the denominator, then you should multiply by a conversion factor with the same unit in the numerator.

God bless.

Allan S. Hugo, Ch.E.
Editor, ChemBookStore
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