Elemental Analysis (College Board AP® Chemistry): Revision Note

A 4.1 g sample of a mixture of calcium chloride (CaCl₂) and magnesium fluoride (MgF₂) are found to contain 0.30 g of Ca. What is the percent of CaCl₂ in the sample?

Answer:

• Step 1: Determine the moles of Ca in the sample

  • n = m / M

  • n =  0.30 g / 40.08 g mol^-1  

  • n = 0.00749 mol of Ca

• Step 2: Determine the moles of CaCl₂ using the moles of Ca

  • The subscripts show that there is one atom of Ca in one formula unit of CaCl₂. Therefore,
    $0.00749\mathrm{mol}\mathrm{of}\mathrm{Ca}\times \frac{1\mathrm{mol}\mathrm{of}{\mathrm{CaCl}}_2}{1\mathrm{mol}\mathrm{of}\mathrm{Ca}}=0.00749\mathrm{mol}\mathrm{of}\mathrm{Ca}$

• Step 3: Determine the mass of CaCl₂ in the sample by rearranging

  • n = m / M

  • m = n x M

  • m =  0.749 mol x 110.98 g mol^-1  

  • m = 0.83 g of CaCl₂

• Step 4: Calculate the percent of CaCl₂ in the sample

  • percent of CaCl₂ = (mass of CaCl₂ / total mass of sample) × 100

  • percent of CaCl₂ = (0.83 g of CaCl₂ / 4.1 g of sample) × 100

  • percent of CaCl₂ = 20%

5.00 g of a sodium chloride sample were analyzed and it was found that it was contaminated with sulfur. The sample is determined to contain 1.35 g of sodium, 2.00 chlorine, and the rest is sulfur. What is the purity of the sample?

Answer:

• Step 1: Identify the desired compound

  • The sample was intended to be sodium chloride so, NaCl is our desired compound

• Step 2: Calculate the mass of the desired compound

  • The statement lists the masses of sodium and chlorine. Since these are the atoms that are part of NaCl, we add both masses

  • m of desired compound = m of Na + m of Cl

  • m of desired compound = 1.35 g + 2.00 g

  • m of desired compound = 3.35 g

• Step 3: Determine the purity using formula that depends on m of desired composition and the total mass of sample

$\mathrm{Purity}=\frac{\mathrm{m}\mathrm{of}\mathrm{desired}\mathrm{compound}}{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{sample}}\times 100$

$\mathrm{Purity}=\frac{3.35\mathrm{g}\mathrm{of}\mathrm{NaCl}}{5.00\mathrm{g}}\times 100$

$\mathrm{Purity}=67.0\%$