[greeneyedcritter]

I'm having trouble with an experiment I performed where I have to determine an unknown element M (metal) can be one of three: Li, Na, or K.

Here is the chemical formula:

M2CO3 + 2HCl --> CO2 + 2MCl + H20

Used:
250mL flask empty mass: 97.68g
2.00g of unknown M
20mL of 6.0M HCl

1. Added the 2.00g of M to flask, mass of flask after adding: 99.68g
2. Used 100mL beaker for HCl, empty beaker mass: 50.81g
3. Added 20mL of HCl to beaker, beaker mass w/HCl: 65.88g
4. Subtract 65.88g - 50.81g = 15.07g HCl
5. Mass of empty flask + unknown M + HCl before Rx: (99.68g + 15.07g = 114.75g)
6. Slowly poured HCl into flask with unknown M
7. Observed vigorous Rx
8. After Rx, mass of flask w/unknown M and HCl added: 120.59g

Now I need to determine the mass of M by figuring out the amount of CO2 that was released in the Rx. My question is why is the mass after the Rx greater than the mass before the Rx? If I subtract 120.59 - 114.75 = 5.84g, is that the mass of the CO2? Doesn't seem right? Any help is appreciated. Thanks

gec  

[AWK]

The loss of weight is ,of course, because of CO2 evolving. But some CO2 remains in solution because its solubility, moreover some amount of HCl still remains in a beaker.

[Borek]

Think about excess HCl.

However, something must be wrong - mass of the CO₂ evolved is almost three times larger then the mass of sample.

Have you weighted empty beaker after HCl was added, to check what was real mass of HCl solution used?

[xiankai]

u've used too much HCl.

min. possible Mr of M = 7  (Li)

Mr of M2CO3 = 7x2 + 12 + 16x3
                   = 72

mol of M2CO3 used = 2/72
                             = 0.0278 mol (3 sig. fig.)

mol of HCl reacted = 0.0278x2
                           = 0.0556 mol (3 sig. fig.)

mol of HCl used in expt. = 6 x 20/1000
                                   = 0.12 mol

try using the max possible Mr, and u'll get an even lower figure of moles. in any case, u've added 200% more HCl than u ought to.

[greeneyedcritter]

The 20mL of HCl was required by the experiment. In my opinion, the mass of the flask after the Rx should be less than the mass before the Rx since the CO2 has been released, but my results show that it's not. Hence, the experiment name "simple weight loss." Not "simple weight gain!" I'm stumped.

[Dude]

I think Borek is correct.  HCl is a dissolved gas in liquid water.  The odor you smell from HCl is actually gas escaping from the liquid.  I would suspect that you are including a whole lot of HCl and some H2O mass in your weight loss.  As Borek suggested, for the time/temperature conditions of your experiment, run a blank beaker containing only the 6 M HCl at the same time and temperature conditions and subtract it from your measured value for a crude answer.

I think a better approach would be to degas everything and weigh the remaining salt (which is non-volatile).  Run through the theoretical calculations for each salt and the one that is closest is your answer.

[Borek]

u've used too much HCl.

It shouldn't matter. Excess HCl will behave just like beaker does - ie it will be 'dead weight'. Some excess is necessary to be sure raection get to the end.

[Borek]

greeneyedcritter - I have reread your post carefully without my family hanging around. I misunderstood it at first sight.

You are absolutly right that your mass change is in the wrong direction.

I suppose you have made some error weighting the beaker with HCl solution - 20 mL of 6M HCl should weight about 22 g. If you have measured volume reasonably accurately mass of the solution should be about 21-23 g, but not 15.

[greeneyedcritter]

Well, I had a chance to repeat the experiment this week. Borek, you were right on the money. The HCl mass was off--this time I measured it at 22.92g. I can't explain why, unless the scale hadn't been zeroed, but I'm usually so careful and check twice, but there are about 30 other students using it, with crowding and chaos, so it could happen. Now my new dilemma:

Chemical equation: M2CO3 + 2HCl --> CO2 + 2MCl + H2O

Before Rx mass: 75.42g
After Rx mass:    73.67g

Amount of CO2 loss: 1.75g {in the right direction! }

Mols of CO2: 1.75g/44.01g per mol of CO2 = 3.98 x 10-2 mol of CO2

According to chemical equation: 1 mol of CO2 = 1 mol of CO3
So, amount of mols of CO3 = 3.98 x 10-2

My question:
How do I get the mass of M? Everything I've tried doesn't work. I keep getting more grams of CO3 than there is in the sample (only 2g). Which doesn't seem right.

[lemonoman]

You 'know' (approximately) how many moles of CO2 was produced.  So, you know how moles of M2CO3 you started with (how many CO2s are made from each M2CO3?)

You can get the molar mass of the M2CO3 from this 'number of moles' and the mass of sample you started with.

Each of Li2CO3, Na2CO3 and K2CO3 have different molar masses.  You can use THIS to identify it.

[greeneyedcritter]

Ok now I feel silly.

1 mol of CO2 = 1 mol of M2CO3 (not just CO3)

So, g of unknown sample M2CO3/mol of M2CO3 gives me answer.

2.00g M2CO3/3.98 x 10-2 mol M2CO3 = 50.25g/mol

LiCO3  = 66.95 g/mol (closest match)
NaCO3 = 83.00 g/mol
KCO3   = 99.11 g/mol

Taking into account the myriad of possible errors, the sample could be LiCO3.

Thanks for the help everyone. As usual, I was making the problem harder than it is.

gec

[greeneyedcritter]

Whoops! Should be:

Li2CO3  = 73.89   g/mol (still closest match)
Na2CO3 = 105.99 g/mol
K2CO3   = 138.21 g/mol

gec

[Borek]

Two things I don't like about the experiment:

1. molar masses of carbonates are in fact unknown. I don't know about Li₂CO₃ but Na₂CO₃ forms a decahydrate and K₂CO₃ 1.5 hydrate (whatever it is named in English). Especially in case of sodium carbonate amount of water in solid may vary if the solid was not kept in controlled conditions so your assumption about molar mass being 105.99 can be wrong.

2. error is too large (-32%). In case of K₂CO₃ your evaluation of molar mass will be about 95 g and in case of Na₂CO₃ about 72 g - so in both cases your identification will be wrong.

[arnyk]

When the hydrated salt dissociates, does the cation itself form the new salt or is there some water included in that mass?

[greeneyedcritter]

Hey Borek, did you use html for the subscripts? I tried them on this board but it didn't work when I previewed.

As for the experiment. Well, I didn't make it up. I'm just a grunt/mechanical engineering student trying to pass a class. I hope the professor, who I haven't met (T.A. teaches the class, very confusingly I might add, and not just because of his very thick accent) knows there is a lot of room for error.