I\'m doing a chemistry lab on gravimetric analysis of a group 1 metal carbonate

Question

I'm doing a chemistry lab on gravimetric analysis of a group 1 metal carbonate compound... So basically, we heated the unknown metal carbonate, and then mixed it with calcium chloride solution. Then, we dried and weighed these products and recorded their masses. I'm having trouble finding the final mass & identity of the precipitate, alongside its molar mass.

My balanced chemical equation is: CaCl2 + X2CO3 (where X = unknown group 1 metal) --> CaCO3 + 2XCl

The mass of the heated compound was 2.054-1.001 (where 1.001 = mass of filter paper) in the beginning and 2.242-1.001 at the end. The mass of the non-heated compound was 2.008 in the beginning and 2.325 in the end.

Thanks in advance!

Explanation / Answer

Answer:

Mass of CaCO3 + Filter paper = 2.375g

Mass of CaCO3 = 2.325-1.001 = 1.324

Then Moles of CaCO3 = 1.324 / 100.0875 (Mass / Molecular Mass)

= 0.0132mol

(0.0132mol of CaCO3) x( 1 mol of x2CO3 /1 mol of CaCO3) = 0.0312 mol of x2CO3

Molar mass of  x2CO3 = 2.054 / 0.0132 = 155.06g/mol

(155.06gX2CO3/mol) - (60.00907gCO3 /mol) = 95.05gX2/ mol = 47.52 g x/mol

Identify of X2CO3 : K2CO3

Percent error:(155.06g/mol -138.205g K2CO3/mol) / 138.205 g K2CO3/mol = 0.121%

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