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### Topic: molecular formula of compound  (Read 2055 times)

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#### auksas

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« on: September 26, 2022, 11:23:10 AM »
Kerosene contains 86% of carbon and 14% of hydrogen. Calculate the volume of air for burning
80 g of kerosene

I tried to find molecular formula but after calculation of empirical formula I'm stuck...I think that the next step is to use chemical equation to solve the problem, but I can't find how.

#### sjb

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« Reply #1 on: September 26, 2022, 12:59:00 PM »
For a generic compound CxHy what is the equation for this burning?

#### auksas

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« Reply #2 on: September 26, 2022, 01:45:42 PM »
For a generic compound CxHy what is the equation for this burning?

CxHy+O2 ...  But I can't find the solution without molecular formula.

#### Corribus

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« Reply #3 on: September 26, 2022, 03:41:55 PM »
What is the mass % of carbon in 1 mole of hydrocarbon CxHy, in terms of x?
Hint: think about how you calculate the molecular weight of a hydrocarbon where you know what the formula is. can you think of a way to do that in reverse?

Although if you think carefully about it, you'll realize that for the series of alkanes (for example), after you get past the first few, the mass % of carbon in all of them is about the same. This should make sense to you. As the chain gets longer and longer, the ratio of carbons to hydrogens gets closer and closer to 2:1, which is the ideal ratio in the infinitely long chain. In finite chains, the only difference is the terminal CH3 groups... and the longer your chaain gets, the less important those become.  A molecule with only carbon and hydrogen with a 2:1 hydrogen to carbon ratio has a % carbon by mass of ~12/14 or... ~85.7%

So actually you can just about pick any long chain alkane here if it helps you and you'll probably get pretty close to the right answer. In reality kerosine is a mixture of high molecular weight hydrocarbons, so there's no exact formula.

What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### Borek ##### Re: molecular formula of compound
« Reply #4 on: September 26, 2022, 07:02:37 PM »
Actually it is quite simple and the composition of the mixture nor the formula doesn't matter at all.

You have 80g of something of which 86% is carbon. What is the carbon mass? Number of moles?

What about the remaining 14% of hydrogen?
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#### auksas

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« Reply #5 on: September 27, 2022, 05:27:54 AM »
Actually it is quite simple and the composition of the mixture nor the formula doesn't matter at all.

You have 80g of something of which 86% is carbon. What is the carbon mass? Number of moles?

What about the remaining 14% of hydrogen?

I've found how to calculate empirical formula-CH2 but with this equation is not correct. And textbook is without any answer for the task and I can't check the solution.

#### Aldebaran

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« Reply #6 on: September 27, 2022, 10:54:59 AM »
Keep in mind the question asks for volume of air so once you’ve got moles of oxygen you will need to allow for its proportion in air and make some reasonable assumption about temperature and pressure for the volume.

#### Borek ##### Re: molecular formula of compound
« Reply #7 on: September 27, 2022, 11:09:45 AM »
I've found how to calculate empirical formula-CH2 but with this equation is not correct. And textbook is without any answer for the task and I can't check the solution.

You are not asked to find any formula, just about air volume. You can easily calculate mass of carbon and mass of hydrogen. Don't make the question harder than it is Imagine the question is worded "what is the volume of air required to burn mixture of 10g of carbon and 10g of hydrogen?" - is it still difficult? Does it require using any "formula" of the mixture, or is it it enough to write two equations for separate burning of carbon and hydrogen?
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