Given a 10.0g sample of MgCO3*5H2O, what weight of anhydrous magnesium carbonate can be obtained after driving off the water?
84.3g?
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
apologies M(MgCO3*5H20) = 154
ratio = 70/154 = 0.4545
so 45.5% is water, same process as above just use this value instead
To find the weight of anhydrous magnesium carbonate obtained after driving off the water, we need to consider the compound's molecular formula and the molar mass of the hydrate.
The formula given, MgCO3*5H2O, tells us that for every mole of the hydrate compound, there are 5 moles of water (H2O). The molar mass of the hydrate can be calculated by summing up the atomic masses of each element in the formula:
MgCO3*5H2O:
1 mole of Mg (24.31 g/mol) + 1 mole of C (12.01 g/mol) + 3 moles of O (16.00 g/mol) + 5 moles of H2O
= 24.31 g + 12.01 g + (3 * 16.00 g) + (5 * 18.02 g)
= 24.31 g + 12.01 g + 48.00 g + 90.10 g
= 174.42 g/mol
Now, we can convert the given mass of the sample (10.0 g) into moles, using the molar mass of the hydrate:
10.0 g / 174.42 g/mol ≈ 0.057 mol
Since the ratio between anhydrous magnesium carbonate (MgCO3) and the hydrate is 1:1, the moles of anhydrous magnesium carbonate obtained will be the same as the moles of the hydrate:
0.057 mol of anhydrous MgCO3
Finally, we can convert the moles of anhydrous magnesium carbonate into grams by multiplying by its molar mass:
0.057 mol * (24.31 g/mol + 12.01 g/mol + 3 * 16.00 g/mol) ≈ 2.47 g
Therefore, the weight of anhydrous magnesium carbonate that can be obtained after driving off the water is approximately 2.47 grams, not 84.3 grams.