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Chapter 5: Problem 79

Express the composition of each of the following compounds as the mass percents of its elements. a. formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\) b. glucose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Short Answer

Expert verified
The mass percents of each element in the given compounds are: a. Formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\): C: 40.00%, H: 6.70%, O: 53.30% b. Glucose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\): C: 40.00%, H: 6.72%, O: 53.28% c. Acetic Acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\): C: 40.03%, H: 6.73%, O: 53.24%

Step by step solution

01

Calculating Molar Masses of Elements in Each Compound

The molar mass of the elements is as follows: Carbon (C) = 12.01 g/mol Hydrogen (H) = 1.01 g/mol Oxygen (O) = 16.00 g/mol
02

Calculating Molar Masses of Each Compound

a. Formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\): 1 C + 2 H + 1 O = \(12.01 + 2(1.01) + 16.00 = 30.03\) g/mol b. Glucose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\): 6 C + 12 H + 6 O = \(6(12.01) + 12(1.01) + 6(16.00) = 180.18\) g/mol c. Acetic Acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\): 2 C + 4 H + 2 O + 1 H = \(2(12.01) + 4(1.01) + 2(16.00) = 60.05\) g/mol
03

Calculating Mass Percents of Elements in Each Compound

a. Formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\): % C = \(\frac{12.01}{30.03} \times 100 = 40.00\% \) % H = \(\frac{2(1.01)}{30.03} \times 100 = 6.70\% \) % O = \(\frac{16.00}{30.03} \times 100 = 53.30\% \) b. Glucose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\): % C = \(\frac{6(12.01)}{180.18} \times 100 = 40.00\% \) % H = \(\frac{12(1.01)}{180.18} \times 100 = 6.72\% \) % O = \(\frac{6(16.00)}{180.18} \times 100 = 53.28\% \) c. Acetic Acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\): % H = \(\frac{4(1.01)}{60.05} \times 100 = 6.73\% \) % C = \(\frac{2(12.01)}{60.05} \times 100 = 40.03\% \) % O = \(\frac{2(16.00)}{60.05} \times 100 = 53.24\% \)
04

Presenting Final Mass Percent Composition

a. Formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\): C: 40.00%, H: 6.70%, O: 53.30% b. Glucose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\): C: 40.00%, H: 6.72%, O: 53.28% c. Acetic Acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\): C: 40.03%, H: 6.73%, O: 53.24%

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Most popular questions from this chapter

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced \(0.213 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0310 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that 0.103 g of the compound produces \(0.0230 \mathrm{g} \mathrm{NH}_{3}\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Give the balanced equation for each of the following chemical reactions: a. Glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) reacts with oxygen gas to produce gaseous carbon dioxide and water vapor. b. Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. c. Carbon disulfide liquid reacts with ammonia gas to produce hydrogen sulfide gas and solid ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right).\)

Consider samples of phosphine \(\left(\mathrm{PH}_{3}\right),\) water \(\left(\mathrm{H}_{2} \mathrm{O}\right),\) hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{S}\right),\) and hydrogen fluoride (HF), each with a mass of \(119 \mathrm{g} .\) Rank the compounds from the least to the greatest number of hydrogen atoms contained in the samples.

Natural rubidium has the average mass of 85.4678 u and is composed of isotopes \(^{85} \mathrm{Rb}\) (mass \(=84.9117 \mathrm{u}\) ) and \(^{87} \mathrm{Rb}\). The ratio of atoms \(^{85} \mathrm{Rb} /^{87} \mathrm{Rb}\) in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\).

The reusable booster rockets of the U.S. space shuttle employ a mixture of aluminum and ammonium perchlorate for fuel. A possible equation for this reaction is $$\begin{aligned}3 \mathrm{Al}(s)+3 \mathrm{NH}_{4} \mathrm{ClO}_{4}(s) & \longrightarrow \\ \mathrm{Al}_{2} \mathrm{O}_{3}(s)+& \mathrm{AlCl}_{3}(s)+3 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\end{aligned}$$ What mass of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) should be used in the fuel mixture for every kilogram of Al?
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