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
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+r-s
m0/w0 = (m0 + r - s)/w1
Rearranging:
m0 = (s-r)w0/(w0-w1)
Evaluate:
m0 = (28.4858g-17.3014g)*(84.31413)/(84.31413-40.30449)
= 21.4272g
ysr
To determine the amount of magnesium carbonate present in the original sample, we need to calculate the mass of magnesium carbonate that decomposed based on the given information.
First, we need to find the mass of the solid residue, which consists of MgO and the original impurities. According to the problem, the mass of the solid residue is 17.3014 g.
Next, we need to calculate the mass of MgO in the solid residue. Since the magnesium carbonate decomposed according to the equation MgCO3(s) → MgO(s) + CO2(g), the mass of MgO formed is equal to the mass of the solid residue. So, the mass of MgO is also 17.3014 g.
Now, we can find the mass of magnesium carbonate in the original sample. Since only the magnesium carbonate decomposed, the mass of MgCO3 in the original sample is equal to the mass of MgO in the solid residue.
Therefore, the amount of magnesium carbonate present in the original sample is 17.3014 g.
mass CO2 = 28.4858 - 17.3014 = ?
mass MgCO3 = mass CO2 x (molar mass MgCO3/molar mass CO2) = ?