GChem - Molecular Formula

[skyisblue]

How do you guys even set this problem up?

Unknown compound with MM 114g/mol. 1.55g sample of this compound was burned in excess Oxygen and 2.21g of H2O and 4.80g of CO2 were produced. What's the molecular formula?

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[skyisblue]

I just figured out that there are 2 ways to solve this problem.

But you guys are more than welcome to try it.

 

[Streetwolf]

Okay I'll bite. Since it's a combustion problem we are working with an alkane. Alkanes have the formula CH(2n+2). This is easy since we are given the MM = 114g/mol. Since C = 12g/mol and H = 1g/mol we have 12n + 1(2n+2) = 114, so 14n + 2 = 114, so 14n = 112, and n = 8.

Therefore we have C8H18 (octane).

 

[skyisblue]

Okay I'll bite. Since it's a combustion problem we are working with an alkane. Alkanes have the formula CH(2n+2). This is easy since we are given the MM = 114g/mol. Since C = 12g/mol and H = 1g/mol we have 12n + 1(2n+2) = 114, so 14n + 2 = 114, so 14n = 112, and n = 8.

Therefore we have C8H18 (octane).

That's correct.

Streetwolf! that is such a cool way of solving this problem. I did it the long way and had another way of doing it, which was even longer.

Thank you! Makes me want to go to Stony Brook just to meet up with you. That's Hot! Mad Props to you.

 

[Streetwolf]

Alrighty, maybe you did this?

Again we know it's a combustion problem so we have an alkane. The molar ratio of the alkane to CO2 is 1:n, in other words there is 1 mole of alkane for every n moles of CO2, n being the # of carbons in the alkane.

We have MM = 114 g/mol with 1.55 grams meaning we have roughly 0.0136 mol of alkane. So we have (0.0136)n moles of CO2. We have 44g/mol = CO2 MM meaning we have 44*(0.0136)n grams of CO2 = (0.6)n grams CO2. Thus (0.6)n = 4.80 and n = 8.

 

[skyisblue]

Alrighty, maybe you did this?

Again we know it's a combustion problem so we have an alkane. The molar ratio of the alkane to CO2 is 1:n, in other words there is 1 mole of alkane for every n moles of CO2, n being the # of carbons in the alkane.

We have MM = 114 g/mol with 1.55 grams meaning we have roughly 0.0136 mol of alkane. So we have (0.0136)n moles of CO2. We have 44g/mol = CO2 MM meaning we have 44*(0.0136)n grams of CO2 = (0.6)n grams CO2. Thus (0.6)n = 4.80 and n = 8.

I didn't even do this. What I did was figure out the amounts of C and H in the sample compound using the amounts of CO2 and H2O given. Then converted it to moles. Finally, figured out empirical formula from there.

I get how you got 0.0136 moles of alkane, but how did you go from there to
(0.0136)n moles of CO2? Then multiply by 44?

 

[Streetwolf]

I didn't even do this. What I did was figure out the amounts of C and H in the sample compound using the amounts of CO2 and H2O given. Then converted it to moles. Finally, figured out empirical formula from there.

I get how you got 0.0136 moles of alkane, but how did you go from there to
(0.0136)n moles of CO2? Then multiply by 44?

In any combustion equation you have (alkane) + O2 --> H2O + CO2. The only compound on the left with a C is the alkane, and the only one on the right is CO2. So if the alkane is C9H20, you will have 9CO2. If it's C2H6, you have 2CO2. Point is, with CH(2n+2), you have CO2. Knowing this, the ratio of moles CO2 to moles alkane is n:1.

If you have 0.0136 moles of alkane, multiply that by moles CO2 per every 1 mole alkane, and you get *0.0136 moles CO2 (try it with different values of n if you want a 'real' example). Since the MM of CO2 is 44g/mol, you take 0.0136*n moles CO2 * 44g/mol = 0.6*n grams of CO2. In your problem you said you end up with 4.80g of CO2. So you want the value of n such that 0.6*n = 4.80. Some quick math gives you n = 8.

 

[skyisblue]

In any combustion equation you have (alkane) + O2 --> H2O + CO2. The only compound on the left with a C is the alkane, and the only one on the right is CO2. So if the alkane is C9H20, you will have 9CO2. If it's C2H6, you have 2CO2. Point is, with CH(2n+2), you have CO2. Knowing this, the ratio of moles CO2 to moles alkane is n:1.

If you have 0.0136 moles of alkane, multiply that by moles CO2 per every 1 mole alkane, and you get *0.0136 moles CO2 (try it with different values of n if you want a 'real' example). Since the MM of CO2 is 44g/mol, you take 0.0136*n moles CO2 * 44g/mol = 0.6*n grams of CO2. In your problem you said you end up with 4.80g of CO2. So you want the value of n such that 0.6*n = 4.80. Some quick math gives you n = 8.

COOL thanks streetwolf

 

[scottyhoop]

Streetwolf....been thinking about your method. Just curious as to how you can assume it's an alkane just by knowing it's a combustion reaction. C2H2 and C6 H12O6 will both burn and neither are alkanes. I'm not questioning your method as it is awesome if you do know it's an alkane. Let me know what you think....
Scott

 

[scottyhoop]

By the way..skyisblue here's the method I remember from gchem. Take mass CO2 goto moles of CO2 goto moles C goto mass C. Do the same for H2O to get mass H. Add the 2 masses then subtract from original sample mass. This gives you mass of O which you can turn into moles. Using the moles of each you can get emperical formula. Find emperical weight and get ratio b/w emperical weight and sample weight. Multiply mole subscripts by this ratio. Prolly a shorter way, but that's how i did it and it worked out. I'm starting to brush up the gen. chem so any tricks you know would be great! Street seems to know his stuff!!
Scott

 

[Streetwolf]

Streetwolf....been thinking about your method. Just curious as to how you can assume it's an alkane just by knowing it's a combustion reaction. C2H2 and C6 H12O6 will both burn and neither are alkanes. I'm not questioning your method as it is awesome if you do know it's an alkane. Let me know what you think....
Scott

Well once I got n=8 I went back and made sure that was the correct answer. I just assumed (don't assume haha) that they'd have an alkane since it's the DAT. I suppose the empirical formula thing would be best. Since I got an exact number I figured I was right.

Though honestly it's most likely that an alkane is the reactant in these sorts of problems.

 

[Lonely Sol]

Well once I got n=8 I went back and made sure that was the correct answer. I just assumed (don't assume haha) that they'd have an alkane since it's the DAT. I suppose the empirical formula thing would be best. Since I got an exact number I figured I was right.

Though honestly it's most likely that an alkane is the reactant in these sorts of problems.

Lets say if it were to be a CHO group, could you also assume that it is CH(2n)O, and do it that way too, or does it only work for alkanes?

 

[Streetwolf]

Lets say if it were to be a CHO group, could you also assume that it is CH(2n)O, and do it that way too, or does it only work for alkanes?

Well you'd have 12n + 2n + 16n = 114
30n = 114
n = not a natural number

So that wouldn't work.

If MM was 120 it might.

 

[scottyhoop]

Ok...I see what you mean. It's a useful tool thats for sure. Thanks for clearing up the confusion!
Scott