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### Topic: Need help on getting the Emperical formula?  (Read 6481 times)

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#### Yusuf

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##### Need help on getting the Emperical formula?
« on: August 29, 2012, 05:54:54 PM »
Here is the problem

3) A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compound yields 16.01 mg CO2 and 4.37 mg H2O. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound.

And here is the solution

Quote
just to give another way of calculation that leads to the same answer C6H8O6:

16.01 mg CO2 = 0.364 mmol CO2 --> contains 0.364 mmol C: mass = 4.369 mg

4.37 mg H2O = 0,243 mmol H2O --> contains 0.486 mmol H: mass = 0.489 mg.

--> mass O: 10.68 - (4.369 + 0.489) = 5.822 mg --> thus 0.364 mmol O

Ratio C : H : O = 0.364 : 0.486 : 0.364 = 1 : 1.34 : 1 = 3 : 4 : 3

Empirical formula: C3H4O3 --> molar mass = 88.062 g/mol

--> molar mass is 176.1, thus that is 176.1 / 88.062 = 2 times the empirical formula.

--> molecular formula: C6H8O6.

I don't understand how first of all the compound went from having a total mass of 10.68 mg? to gaining a mass of 20.38 mg? Also, how do you get the mole  number for carbon in "16.01 mg CO2 = 0.364 mmol CO2 --> contains 0.364 mmol C: mass = 4.369 mg", are you not taking the amu of oxygen into account in order to get .364? It just makes no sense, so can someone please help me understand this, thank you!

#### Arkcon

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##### Re: Need help on getting the Emperical formula?
« Reply #1 on: August 29, 2012, 06:12:54 PM »

I don't understand how first of all the compound went from having a total mass of 10.68 mg? to gaining a mass of 20.38 mg?

The products of a combustion reaction will have more mass than the fuel, because oxygen is also a reactant.  If you try to write a balanced chemical reaction, you'll noticed it.  You don't really need that because of the way this problem is set up, but its a good habit to get into.

Quote
Also, how do you get the mole  number for carbon in "16.01 mg CO2 = 0.364 mmol CO2 --> contains 0.364 mmol C: mass = 4.369 mg", are you not taking the amu of oxygen into account in order to get .364? It just makes no sense, so can someone please help me understand this, thank you!

Now, you want to forget the added oxygen, to compute the amount of carbon in the sample.  Hard for me to see your problem, so maybe you should try to see where these calculations come from?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

#### Dan

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##### Re: Need help on getting the Emperical formula?
« Reply #2 on: August 29, 2012, 06:25:04 PM »
I don't understand how first of all the compound went from having a total mass of 10.68 mg? to gaining a mass of 20.38 mg?

The compound (10.68 mg) was burned - it combined with oxygen (9.70 mg) to produce products with a total mass of 20.38 mg.

Quote
Also, how do you get the mole  number for carbon in "16.01 mg CO2 = 0.364 mmol CO2 --> contains 0.364 mmol C: mass = 4.369 mg", are you not taking the amu of oxygen into account in order to get .364?

The molar mass of CO2 is 44 g/mol. Therefore mmol CO2 = 16.01/44 = 0.364 mmol

For every 1 CO2 molecule, there is 1 C atom
For every 1 mol of CO2 molecules, there is 1 mol of C atoms
For every 0.364 mmol of CO2 molecules, there is 0.364 mmol of C atoms
Calculate the mass of 0.364 mmol of C atoms.

An important thing to understand is that all of the C in the products originated in the starting material, and the same is true for the H. Some of the O came from the starting material, and some came from oxygen gas:

CxHyOz + nO2  xCO2 + (y/2)H2O

My research: Google Scholar and Researchgate

#### Yusuf

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##### Re: Need help on getting the Emperical formula?
« Reply #3 on: August 29, 2012, 06:31:10 PM »

I don't understand how first of all the compound went from having a total mass of 10.68 mg? to gaining a mass of 20.38 mg?

The products of a combustion reaction will have more mass than the fuel, because oxygen is also a reactant.  If you try to write a balanced chemical reaction, you'll noticed it.  You don't really need that because of the way this problem is set up, but its a good habit to get into.

Quote
Also, how do you get the mole  number for carbon in "16.01 mg CO2 = 0.364 mmol CO2 --> contains 0.364 mmol C: mass = 4.369 mg", are you not taking the amu of oxygen into account in order to get .364? It just makes no sense, so can someone please help me understand this, thank you!

Now, you want to forget the added oxygen, to compute the amount of carbon in the sample.  Hard for me to see your problem, so maybe you should try to see where these calculations come from?

Thanks man, I figured out some of the problems I was having; however I'm wondering if their is any sign to tell you that you must multiply the emperical formula by a certain number like 2 or 3. Like If I had C:.5, O:.333, and H: 1.333?

#### Hunter2

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##### Re: Need help on getting the Emperical formula?
« Reply #4 on: August 30, 2012, 01:14:28 AM »
Yes you  have to.

C 0.5, O 0.333, H 1.333  devided by 0.5

C 1, O 0.666 H 2.666  times 2

C 2 O 1.333 H 5.3333 times 3

C 6 O 4 H 16

C6H16O4

#### Borek

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##### Re: Need help on getting the Emperical formula?
« Reply #5 on: August 30, 2012, 02:27:37 AM »
Yes you  have to.

C 0.5, O 0.333, H 1.333  devided by 0.5

C 1, O 0.666 H 2.666  times 2

C 2 O 1.333 H 5.3333 times 3

C 6 O 4 H 16

C6H16O4

This is incorrect. Empirical formula should have the smallest possible integer coefficients, so you have to divide all by 2 now.

Besides, I don't see a reason to multiply by 2 in the second step. If you see something like 0.666 multiplying it by 3 gives (almost exactly) 2, so that would be the logical move. Accidentally that would also get rid of the 2.666 and it would also immediately gave the correct final answer.
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#### Hunter2

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##### Re: Need help on getting the Emperical formula?
« Reply #6 on: August 30, 2012, 06:45:01 AM »
I agree my fault.