AP Chemistry
posted by Gabriella .
A 28.4858 g sample of impure magnesium carbonate was heated to complete decomposition according to the equation
MgCO3(s)→MgO(s) + CO2(g).After the reaction was complete, the solid
residue (consisting of MgO and the original impurities) had a mass of 17.3014 g. Assuming that only the magnesium carbonate had decomposed, how much magnesium carbonate was present in the original sample?
Answer in units of g

mass CO2 = 28.4858  17.3014 = ?
mass MgCO3 = mass CO2 x (molar mass MgCO3/molar mass CO2) = ? 
Known:
Mass of sample: s = 28.4858 g
Mass of residue: r = 17.3014 g
Unknown:
Mass of Impurities: x
Mass of MgCO3: m0 = s  x
Mass of MgO: m1 = r  x
Therefore: m1 = m0 + r  s
Lookup:
Molecular mass of MgCO3: w0 = 84.31413 ± 0.00005 g/mol
Molecular mass of MgO: w1 = 40.30449 ± 0.00002 g/mol
Stoichiometry: m0/w0 = m1/w1
Find m0:
Substituting: m1 = m0+rs
m0/w0 = (m0 + r  s)/w1
Rearranging:
m0 = (sr)w0/(w0w1)
Evaluate:
m0 = (28.4858g17.3014g)*(84.31413)/(84.3141340.30449)
= 21.4272g