Given a 10.0g sample of MgCO3*5H2O, what weight of anhydrous magnesium carbonate can be obtained after driving off the water?
84.3g?
3 answers
wait.. that can't be right.
no i don't think the weight will be increasing after getting rid of the water.
work out the molar mass of hydrated magnesium carbonate, then find the ratio of water to the whole thing in regard to molar mass
eg
M(MgCO3*5H20)= 144
M(5H2O) = 70
ratio = 70/144 = 0.486
this means that 48.6% of that 10grams is water
so by multiplying the 10grams by .486 we get the mass of water, then subtract from 10grams to get the mass of magnesium carbonate once the water has been removed
the ratio bit may be a bit vague, if it doesn't make sense just say so and I'll try and clear it up
work out the molar mass of hydrated magnesium carbonate, then find the ratio of water to the whole thing in regard to molar mass
eg
M(MgCO3*5H20)= 144
M(5H2O) = 70
ratio = 70/144 = 0.486
this means that 48.6% of that 10grams is water
so by multiplying the 10grams by .486 we get the mass of water, then subtract from 10grams to get the mass of magnesium carbonate once the water has been removed
the ratio bit may be a bit vague, if it doesn't make sense just say so and I'll try and clear it up
apologies M(MgCO3*5H20) = 154
ratio = 70/154 = 0.4545
so 45.5% is water, same process as above just use this value instead
ratio = 70/154 = 0.4545
so 45.5% is water, same process as above just use this value instead