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Chapter 3: Problem 67

Calculate the percent composition by mass of the following compounds that are important starting materials for synthetic polymers: a. \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2}\) (acrylic acid, from which acrylic plastics are made) b. \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{2}\) (methyl acrylate, from which Plexiglas is made) c. \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}\) (acrylonitrile, from which Orlon is made)

Short Answer

Expert verified
The percent composition by mass of the given compounds are as follows: a. Acrylic acid (\(C_{3}H_{4}O_{2}\)): 50% carbon, 5.6% hydrogen, and 44.4% oxygen. b. Methyl acrylate (\(C_{4}H_{6}O_{2}\)): 55.8% carbon, 7.0% hydrogen, and 37.2% oxygen. c. Acrylonitrile (\(C_{3}H_{3}N\)): 67.9% carbon, 5.7% hydrogen, and 26.4% nitrogen.

Step by step solution

01

Determine the molar mass of the compound.

The molar mass is the sum of the atomic masses of the elements in the compound, weighted by the number of atoms of each element. The molar mass of carbon, hydrogen, and oxygen are approximately 12 g/mol, 1 g/mol, and 16 g/mol, respectively.\[Molar\:mass\:of\:C_{3}H_{4}O_{2}=(3\times12+4\times1+2\times16)g/mol=72\:g/mol\]
02

Calculate the mass of each element in the compound.

Now, we will multiply the atomic mass of each element by the number of moles of the element present in one mole of the compound. \[Mass\:of\:Carbon=3\times12=36\:g/mol\] \[Mass\:of\:Hydrogen=4\times1=4\:g/mol\] \[Mass\:of\:Oxygen=2\times16=32\:g/mol\]
03

Calculate the mass percent of each element.

We can now calculate the mass percent of each element by dividing the mass of the element by the molar mass of the compound and multiplying the result by 100. \[MassPercent_{Carbon}=\frac{36}{72}\times100=50\%\] \[MassPercent_{Hydrogen}=\frac{4}{72}\times100=5.6\%\] \[MassPercent_{Oxygen}=\frac{32}{72}\times100=44.4\%\] So, the percent composition by mass of acrylic acid is approximately 50% carbon, 5.6% hydrogen, and 44.4% oxygen. #Exercise b: C4H6O2 (methyl acrylate)#
04

Determine the molar mass of the compound.

\[Molar\:mass\:of\:C_{4}H_{6}O_{2}=(4\times12+6\times1+2\times16)g/mol=86\:g/mol\]
05

Calculate the mass of each element in the compound.

\[ Mass\:of\:Carbon=4\times12=48\:g/mol\] \[Mass\:of\:Hydrogen=6\times1=6\:g/mol\] \[Mass\:of\:Oxygen=2\times16=32\:g/mol\]
06

Calculate the mass percent of each element.

\[ MassPercent_{Carbon}=\frac{48}{86}\times100=55.8\%\] \[MassPercent_{Hydrogen}=\frac{6}{86}\times100=7.0\%\] \[MassPercent_{Oxygen}=\frac{32}{86}\times100=37.2\%\] So, the percent composition by mass of methyl acrylate is approximately 55.8% carbon, 7.0% hydrogen, and 37.2% oxygen. #Exercise c: C3H3N (acrylonitrile)#
07

Determine the molar mass of the compound.

The atomic mass of nitrogen is approximately 14 g/mol. \[Molar\:mass\:of\:C_{3}H_{3}N=(3\times12+3\times1+1\times14)g/mol=53\:g/mol\]
08

Calculate the mass of each element in the compound.

\[ Mass\:of\:Carbon=3\times12=36\:g/mol\] \[Mass\:of\:Hydrogen=3\times1=3\:g/mol\] \[Mass\:of\:Nitrogen=1\times14=14\:g/mol\]
09

Calculate the mass percent of each element.

\[ MassPercent_{Carbon}=\frac{36}{53}\times100=67.9\%\] \[MassPercent_{Hydrogen}=\frac{3}{53}\times100=5.7\%\] \[MassPercent_{Nitrogen}=\frac{14}{53}\times100=26.4\%\] So, the percent composition by mass of acrylonitrile is approximately 67.9% carbon, 5.7% hydrogen, and 26.4% nitrogen.

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Most popular questions from this chapter

Considering your answer to Exercise 73 , which type of formula, empirical or molecular, can be obtained from elemental analysis that gives percent composition?

Calculate the molar mass of the following substances. a. \(\mathrm{H}\) b. Q N \(\mathrm{N}\) c. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)

Consider the following balanced chemical equation: $$ A+5 B \longrightarrow 3 C+4 D $$ a. Equal masses of \(\mathrm{A}\) and \(\mathrm{B}\) are reacted. Complete each of the following with either "A is the limiting reactant because \(" ;{ }^{\prime *} \mathrm{~B}\) is the limiting reactant because \("\) or "we cannot determine the limiting reactant because i. If the molar mass of \(\mathrm{A}\) is greater than the molar mass of B, then ii. If the molar mass of \(\mathrm{B}\) is greater than the molar mass of A, then b. The products of the reaction are carbon dioxide (C) and water (D). Compound \(\mathrm{A}\) has the same molar mass as carbon dioxide. Compound \(\mathrm{B}\) is a diatomic molecule. Identify \(\mathrm{com}-\) pound \(\mathrm{B}\) and support your answer. c. Compound \(\mathrm{A}\) is a hydrocarbon that is \(81.71 \%\) carbon by mass. Detemine its empirical and molecular formulas.

Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \% \mathrm{H} \text { (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \text { II } & 12.58 & 87.42 \\ \text { III } & 2.34 & 97.66 \end{array} $$ When \(1.00 \mathrm{~L}\) of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$ \begin{array}{lcc} & \mathrm{H}_{2} \text { (L) } & \mathrm{N}_{2} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \text { II } & 2.00 & 1.00 \\ \text { III } & 0.50 & 1.50 \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(l)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ What masses of iron(III) oxide and aluminum must be used to produce \(15.0 \mathrm{~g}\) iron? What is the maximum mass of aluminum oxide that could be produced?
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