Question

Calculate the percent composition by mass of the following compounds that are important starting materials for synthetic polymers.

a. ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{2}}$(acrylic acid, from which acrylic plastic are made)

b.${\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}$ (methyl acrylate, from which Plexiglas is made)

c. ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{3}}\mathbf{N}$ (acrylonitrile, from which Orlon is made)

Short answer

The mass percentage compositions are;

$$\begin{gathered} \mathrm{a}.50.00\%\mathrm{C},5.59\%\mathrm{H},44.41\%\mathrm{O} \\ \mathrm{b}.55.80\%\mathrm{C},7.03\%\mathrm{H},37.17\%\mathrm{O} \\ \mathrm{c}.67.89\%\mathrm{C},71\%\mathrm{H},26.40\%\mathrm{N} \end{gathered}$$

Step-by-step solution

Definition of synthetic polymers

The polymer is said aspolyethylene used in plastic bags andfilm wraps. The synthetic polymer Polystyrene is used in the cabinets and packages.

a. Calculation of acrylic acid molar mass

The acrylic acid molar mass as,

$$\begin{gathered} 3\times 12.01+4\times 1.01+2\times 16.00=72.07\mathrm{g} \\ \mathrm{Calculating}: \\ \frac{36.03\mathrm{gC}}{72.07\mathrm{g}}=0.4999 \\ \frac{36.03\mathrm{gC}}{72.07\mathrm{g}}\gg 0.5 \\ \frac{36.03\mathrm{gC}}{72.07\mathrm{g}}=0.5\times 100 \\ \frac{36.03\mathrm{gC}}{72.07\mathrm{g}}=50.00\%\mathrm{C} \\ \mathrm{Similarly}, \\ \frac{4.04\mathrm{gH}}{72.07\mathrm{g}}=0.05595 \\ \frac{4.04\mathrm{gH}}{72.07\mathrm{g}}=0.05595\times 100 \\ \frac{4.04\mathrm{gH}}{72.07\mathrm{g}}=5.59\%\mathrm{H} \\ \mathrm{Now}, \\ \frac{32.00\mathrm{gO}}{72.07\mathrm{g}}=0.4441 \\ \frac{32.00\mathrm{gO}}{72.07\mathrm{g}}=0.4441\times 100 \\ \frac{32.00\mathrm{gO}}{72.07\mathrm{g}}=44.41\%\mathrm{O} \\ \mathrm{Hence},\mathrm{the}\mathbf{}\mathbf{acrylic}\mathbf{}\mathbf{acid}\mathbf{}\mathbf{molar}\mathbf{}\mathbf{mass}\mathrm{is}50.00\%\mathrm{C},5.59\%\mathrm{H},44.41\%\mathrm{O} \end{gathered}$$

b) Step 3: b. Calculation of methyl acrylate molar mass

The methyl acrylate molar mass is:

$$\begin{gathered} 4\times 12.01+6\times 1.01+2\times 16.00=86.10\mathrm{g} \\ \mathrm{Calculating}: \\ \frac{48.04\mathrm{gC}}{86.10\mathrm{g}}=0.5580 \\ \frac{48.04\mathrm{gC}}{86.10\mathrm{g}}=0.5580\times 100 \\ \frac{48.04\mathrm{gC}}{86.10\mathrm{g}}=55.80\%\mathrm{C} \\ \mathrm{Similarly}, \\ \frac{6.06\mathrm{gH}}{86.10\mathrm{g}}=0.0703 \\ \frac{6.06\mathrm{gH}}{86.10\mathrm{g}}=0.0703\times 100 \\ \frac{6.06\mathrm{gH}}{86.10\mathrm{g}}=7.03\%\mathrm{H} \\ \mathrm{Now} \\ \frac{32.00\mathrm{gO}}{86.10\mathrm{g}}=0.3717 \\ \frac{32.00\mathrm{gO}}{86.10\mathrm{g}}=0.3717\times 100 \\ \frac{32.00\mathrm{gO}}{86.10\mathrm{g}}=37.17\%\mathrm{O} \\ \mathrm{Hence},\mathrm{the}\mathbf{}\mathbf{methyl}\mathbf{}\mathbf{acrylate}\mathbf{}\mathbf{molar}\mathbf{}\mathbf{mass}\mathrm{is}55.80\%\mathrm{C},7.03\%\mathrm{H},37.17\%\mathrm{O} \end{gathered}$$

c. Calculation of acrylonitrile molar mass

The acrylonitrile molar mass is:

$$\begin{gathered} 3\times 12.01+3\times 1.01+1\times 14.01=53.07\mathrm{g} \\ \mathrm{Calculating}, \\ \frac{36.03\mathrm{gC}}{53.07\mathrm{g}}=0.6789 \\ \frac{36.03\mathrm{gC}}{53.07\mathrm{g}}=0.6789\times 100 \\ \frac{36.03\mathrm{gC}}{53.07\mathrm{g}}=67.89\%\mathrm{C} \\ \mathrm{Similarly}, \\ \frac{3.03\mathrm{gH}}{53.07\mathrm{g}}=0.0571 \\ \frac{3.03\mathrm{gH}}{53.07\mathrm{g}}=0.0571\times 100 \\ \frac{3.03\mathrm{gH}}{53.07\mathrm{g}}=5.71\%\mathrm{H} \\ \mathrm{Now}, \\ \frac{14.01\mathrm{gN}}{53.07\mathrm{g}}=0.2640 \\ \frac{14.01\mathrm{gN}}{53.07\mathrm{g}}=0.2640\times 100 \\ \frac{14.01\mathrm{gN}}{53.07\mathrm{g}}=26.40\%\mathrm{N} \\ \mathrm{Hence},\mathrm{the}\mathbf{acrylonitrile}\mathbf{}\mathbf{molar}\mathbf{}\mathbf{mass}\mathrm{is}67.89\%\mathrm{C},5.71\%\mathrm{H},26.40\%\mathrm{N} \end{gathered}$$

$$\begin{gathered} 3\times 12.01+3\times 1.01+1\times 14.01=53.07\mathrm{g} \\ \mathrm{Calculating}. \\ \frac{36.03\mathrm{gC}}{53.07\mathrm{g}}=0.6789 \\ \frac{36.03\mathrm{gC}}{53.07\mathrm{g}}=0.6789\times 100 \\ \frac{36.03\mathrm{gC}}{53.07\mathrm{g}}=67.89\%\mathrm{C} \\ \mathrm{Similarly}, \\ \frac{3.03\mathrm{gH}}{53.07\mathrm{g}}=0.0571 \\ \frac{3.03\mathrm{gH}}{53.07\mathrm{g}}=0.0571\times 100 \\ \frac{3.03\mathrm{gH}}{53.07\mathrm{g}}=5.71\%\mathrm{H} \\ \mathrm{Now}, \\ \frac{14.01\mathrm{gN}}{53.07\mathrm{g}}=0.2640 \\ \frac{14.01\mathrm{gN}}{53.07\mathrm{g}}=0.2640\times 100 \\ \frac{14.01\mathrm{gN}}{53.07\mathrm{g}}=26.40\%\mathrm{N} \\ \mathrm{Hence},\mathrm{the}\mathrm{acrylonitrile}\mathrm{molar}\mathrm{mass}\mathrm{is}67.89\%\mathrm{C},5.71\%\mathrm{H},26.40\%\mathrm{N} \end{gathered}$$