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StoichiometryMedium

Percent yield

The reaction CaCO3→CaO+CO2\mathrm{CaCO_3 \to CaO + CO_2} has a theoretical yield of 50.0 g of CaO. In the lab you obtain 42.5 g. Calculate the percent yield and the mass of CaCO₃ needed to obtain 50.0 g of CaO.
M(CaCO₃) = 100.1 g/mol, M(CaO) = 56.1 g/mol.
Given data
theoretical yield = 50.0 g CaOactual yield = 42.5 g CaOM(CaCO₃) = 100.1 g/molM(CaO) = 56.1 g/mol
Review the theory: Stechiometria
Steps
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  1. Calculate the percent yield.
  2. Calculate the moles of CaO corresponding to 50.0 g.
  3. Calculate the mass of CaCO₃ required (1:1 ratio).
Full worked solution
  1. Calculate the percent yield.
    η=42.550.0×100\eta = \frac{42.5}{50.0} \times 100
    η=mactual/mtheoretical×100=42.5/50.0×100=85.0 %\eta = m_{\text{actual}}/m_{\text{theoretical}} \times 100 = 42.5/50.0 \times 100 = \mathbf{85.0\,\%}. The percent yield is obtained by dividing the actual yield by the theoretical yield and multiplying by 100.
  2. Calculate the moles of CaO corresponding to 50.0 g.
    n(CaO)=50.056.1n(CaO) = \frac{50.0}{56.1}
    n(CaO)=m/M=50.0/56.1=0.891 moln(\mathrm{CaO}) = m/M = 50.0/56.1 = \mathbf{0.891\,mol}. The moles of CaO are obtained by dividing the mass by the molar mass.
  3. Calculate the mass of CaCO₃ required (1:1 ratio).
    m(CaCO3)=0.891×100.1m(CaCO_3) = 0.891 \times 100.1
    m(CaCO3)=n×M=0.891×100.1=89.2 gm(\mathrm{CaCO_3}) = n \times M = 0.891 \times 100.1 = \mathbf{89.2\,g}. The mass of CaCO₃ is obtained by multiplying the moles by the molar mass.
Result:Percent yield = 85.0%. Need 89.2 g of CaCO₃.
TheoryStechiometriaFormula sheetChemistry

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