The long way is to do each one. Assume 100 g sample and convert to grams CO2 which is the weight loss. For example, MgCO3, molar mass about 84.3
MgCO3 => MgO + CO2 and what you want is for CO2 to come out to be 36.1% (or 36.1 g from a 100 g sample).
mols MgCO3 = grams/molar mass = 100/84.3 = about 1.19
That will give you 1.19 mols CO2.
Convert that to grams CO2. 1.19 x molar mass CO2 = about 52.2 so MgCO3 is not the answer.
The thermal decomposition of carbonates leads to the loss of CO2. The decomposition of an unknown carbonate leads to a 35.1% weight loss. The unknown was which of the following compounds?
A. Li2CO3
B. MgCO3
C. CaCO3
D. ZnCO3
E. BaCO3
How do I solve this?
3 answers
How do you get the 52.2? 1.19 x 36.1 = 43.0?
Oh Never mind.