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mass loss question |
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| Oct30-04, 11:43 PM | #1 |
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mass loss question
I am beginner chem student and I am lost with mass loss. We did an experiment in lab where we are trying to identify a mystery metal. let me describe. We weighed our flask, then weighed our flask with about 1.5g of the mystery metal. Then we added 20ml of 6M HCl to a graduated cylinder and added that to the flask of metal. Then we weighed the empty graduated cylinder. and the flask with the final solution. Now for the question. I need to determine the mass and moles of CO2 released. where do I start? I was thinking that I would figure out the moles of HCl but then i realized that I don't think that is right. I think it has something to do with the mass of the graduated cylinder when full and empty but that is is grams of HCl. HELP !!
Ellie
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| Oct30-04, 11:47 PM | #2 |
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Let's start out with the fully balanced equation for the reaction you are considering.
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| Oct30-04, 11:48 PM | #3 |
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how do I get a fully balanced equation if I don't know one of the elements? as a matter of fact the last question is "Using your identification of the metal write a balanced equation for the reaction that occurs ...."
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| Oct30-04, 11:52 PM | #4 |
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mass loss question
okay here are the questions so you see what I need:
1. Clearly show the following calculation of the experimental molar mass of the alkali metal from the recorded data of one of your trials. Identify all numbers and units. a. Find the mass of CO2 released. b. Find the number of moles of CO2 released. (This is the same as the number of moles of M2CO3 needed to produce the amount of CO2.) c. From the mass and moles of M2CO3 find the experimental molar mass of M2CO3. d. Using the molar masses of carbon and oxygen from the periodic table, find the experimental molar mass of the metal. Ellie |
| Oct31-04, 02:57 AM | #5 |
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Hello, you may try multiplying the molarity and volume of [itex]\displaystyle HCl[/itex] to learn how many moles (or millimoles) are there. Then write the balanced equation between [itex]\displaystyle M_2CO_3[/itex] and [itex]\displaystyle HCl[/itex], to learn how much [itex]\displaystyle CO_2[/itex] is released. You may use this info:
[itex]\displaystyle metal~carbonates~give~CO_2,~H_2O,~and~metal~chloride~with~HCl[/itex] If you can learn how many moles of [itex]\displaystyle CO_2[/itex] is produced, than you can convert it to the moles of [itex]\displaystyle M_2CO_3[/itex], from here, the molar mass by calculating some proportions (1 mole of compound would be x grams, but I have n moles of compound, so I have y grams of that compound, etc.). The only task after doing these will be subtracting the molar masses of carbon and oxygen from the compound. |
| Oct31-04, 07:44 AM | #6 |
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okay so if I do that then I it is (6M/1L)(20ml/1000L)= .12 moles
What I was thinking though is then why do they have us weigh the full cylinder and then the empty cylinder? because then I get 20.586 g of HCl used and I am not really sure where that is coming into play? My other thought was thst maybe it would be the weight of the flask with the metal + the amount in grams of HCl used= 122.56g but the final weight of the flask was 122.184. Which would mean taht .376g were lost as CO2??maybe. |
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