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Chemistry help-unknown carbonate?
I'm just a little confused with this:
66.28g beaker filled with .5g of an unknown carbonate, 10mL of water, and a few drops of indicator.
Titrated with 1M HCl and then placed on a hot plate to drive off the water.
Mass of beaker after is 66.88g.
How exactly do I determine what the unknown element from family 1 is?
I just need to know the steps to take.
Thanks.
1 Answer
- HPVLv 710 years agoFavorite Answer
The titration reaction is: X2CO3 + 2HCl ==> 2XCl + CO2 + H2O . . .where X is a Group I metal. Since you are boiling off the CO2 and the H2O, the remaining residue will be XCl (NaCl, KCl, etc.).
Final mass of beaker - initial mass of beaker = 66.88 g - 66.28 g = 0.60 g = mass of XCl
Let x = the molar mass of element X. Then the molar mass of X2CO3 will be 2x + 60 (carbonate is 60). The moles of X2CO3 will be g X2CO3 / molar mass X2CO3 = 0.5 / (2x + 60).
The molar mass for XCl will be x + 35.45 (Cl is 35.45). The moles of XCl = g XCl / molar mass XCl = 0.60 / (x + 35.45).
Looking at the balanced equation, you know that 2 moles of XCl are produced for every 1 mole of X2CO3 reacted. So moles XCl = 2 times moles X2CO3, or
0.60 / (x + 35.45) = 2 (0.5 / (2x + 60))
0.60 / (x + 35.45) = 1 / (2x + 60)
x + 35.45 = (0.60)(2x + 60)
x + 35.45 = 1.2x + 36
x = -2.75 ... that's imposible! The molar mass for element X can't be -2.75.