Question

Name the products obtained in Figure 27-6 when calcium salicylate monohydrate is heated to 550⁰C and to 1000⁰C. Using the formula masses of these products, calculate what mass is expected to remain when 0.635 6 g of calcium salicylate monohydrate is heated to 550⁰C and to 1000⁰C.

Short answer

The expected mass of 0.635 6 g of calcium salicylate monohydrate is heated to 550⁰C is 0.1914 g and to 1000⁰C is 0.1073 g.

Step-by-step solution

Calculate the mass heated at 550⁰C:

Here we will name the products obtained in Figure 27-6 when calcium salicylate monohydrate is heated to 550⁰C and 1000⁰C. Also, we will calculate the mass expected to remain when 0.6356 g of calcium salicylate monohydrate is heated to 550⁰C and 1000⁰C.

${\mathrm{CaC}}_{14}{\mathrm{H}}_{10}{\mathrm{O}}_6\times {\mathrm{H}}_2\mathrm{O}$calcium salicylate monohydrate$\to \mathrm{M}=332.32\mathrm{g}/\mathrm{mol}$

• CaCO₃ calcium carbonate$\to \mathrm{M}=100.09\mathrm{g}/\mathrm{molat}{550}^{\circ }\mathrm{C}$
• CaO calcium oxide$\to \mathrm{M}=56.08\mathrm{g}/\mathrm{molat}{1000}^{\circ }\mathrm{C}$

${\mathrm{CaC}}_{14}{\mathrm{H}}_{10}{\mathrm{O}}_6\times {\mathrm{H}}_2\mathrm{O}$is converted to CaCO₃ and from 1 mol of${\mathrm{CaC}}_{14}{\mathrm{H}}_{10}{\mathrm{O}}_6\times {\mathrm{H}}_2\mathrm{O}$ there would be 100.09 grams of CaCO₃ produced and 56.08 grams of CaO.

Expected mass at 550⁰C would be:

role="math" localid="1663662310806" $$\begin{gathered} {\mathrm{m}}_{550}=\frac{\mathrm{M}\left({\mathrm{CaCO}}_3\right)}{\mathrm{M}({\mathrm{CaC}}_{14}{\mathrm{H}}_{10}{\mathrm{O}}_6\times {\mathrm{H}}_2\mathrm{O})}\times \mathrm{m}({\mathrm{CaC}}_{14}{\mathrm{H}}_{10}{\mathrm{O}}_6\times {\mathrm{H}}_2\mathrm{O}) \\ {\mathrm{m}}_{550}=\frac{100.09}{332.32}\times 0.6356\mathrm{g} \\ {\mathrm{m}}_{550}=0.1914\mathrm{g} \end{gathered}$$

 Calculate the mass heated at 1000⁰C:

$$\begin{gathered} {\mathrm{m}}_{1000}=\frac{\mathrm{M}\left(\mathrm{CaO}\right)}{\mathrm{M}({\mathrm{CaC}}_{14}{\mathrm{H}}_{10}{\mathrm{O}}_6\times {\mathrm{H}}_2\mathrm{O})}\times \mathrm{m}({\mathrm{CaC}}_{14}{\mathrm{H}}_{10}{\mathrm{O}}_6\times {\mathrm{H}}_2\mathrm{O}) \\ {\mathrm{m}}_{1000}=\frac{56.08}{332.32}\times 0.6356\mathrm{g} \\ {\mathrm{m}}_{1000}=0.1073\mathrm{g} \end{gathered}$$