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Chapter 3: Problem 81

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, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Short Answer

Expert verified
The mass percents of the elements in the compounds are as follows: a. Formaldehyde: 39.98% carbon, 6.71% hydrogen, and 53.31% oxygen. b. Glucose: 40.00% carbon, 6.72% hydrogen, and 53.28% oxygen. c. Acetic acid: 6.73% hydrogen, 40.02% carbon, and 53.25% oxygen.

Step by step solution

01

a. Formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\)#

To find the mass percent of each element in formaldehyde, we must first find the molar mass of the compound. To determine the molar mass of formaldehyde, we need to know the molar masses of its constituent elements, which are carbon (C), hydrogen (H), and oxygen (O). The molar masses are roughly: - Carbon (C): 12.01 g/mol - Hydrogen (H): 1.01 g/mol - Oxygen (O): 16.00 g/mol Now, we can calculate the molar mass of formaldehyde by summing up the molar masses of its elements: \[ (1 \times 12.01) + (2 \times 1.01) + (1 \times 16.00) = 30.03 \ g/mol \] Next, we will find the mass percent of each element in formaldehyde: - Mass percent of carbon: \[ \frac{1 \times 12.01}{30.03} \times 100 = 39.98\% \] - Mass percent of hydrogen: \[ \frac{(2 \times 1.01)}{30.03} \times 100 = 6.71\% \] - Mass percent of oxygen: \[ \frac{1 \times 16.00}{30.03} \times 100 = 53.31\% \] Thus, the mass percents in formaldehyde are approximately 39.98% carbon, 6.71% hydrogen, and 53.31% oxygen.
02

b. Glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)#

To find the mass percent of each element in glucose, we must first find the molar mass of the compound: \[ (6 \times 12.01) + (12 \times 1.01) + (6 \times 16.00) = 180.18 \ g/mol \] Next, we will find the mass percent of each element in glucose: - Mass percent of carbon: \[ \frac{(6 \times 12.01)}{180.18} \times 100 = 40.00\% \] - Mass percent of hydrogen: \[ \frac{(12 \times 1.01)}{180.18} \times 100 = 6.72\% \] - Mass percent of oxygen: \[ \frac{(6 \times 16.00)}{180.18} \times 100 = 53.28\% \] Thus, the mass percents in glucose are approximately 40.00% carbon, 6.72% hydrogen, and 53.28% oxygen.
03

c. Acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)#

To find the mass percent of each element in acetic acid, we must first find the molar mass of the compound: \[ (1 \times 1.01) + (2 \times 12.01) + (3 \times 1.01) + (2 \times 16.00) = 60.05 \ g/mol \] Next, we will find the mass percent of each element in acetic acid: - Mass percent of hydrogen: \[ \frac{(4 \times 1.01)}{60.05} \times 100 = 6.73\% \] - Mass percent of carbon: \[ \frac{(2 \times 12.01)}{60.05} \times 100 = 40.02\% \] - Mass percent of oxygen: \[ \frac{(2 \times 16.00)}{60.05} \times 100 = 53.25\% \] Thus, the mass percents in acetic acid are approximately 6.73% hydrogen, 40.02% carbon, and 53.25% oxygen.

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Most popular questions from this chapter

Fungal laccase, a blue protein found in wood-rotting fungi, is 0.390\(\% \mathrm{Cu}\) by mass. If a fungal laccase molecule contains four copper atoms, what is the molar mass of fungal laccase?

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{kg}\) \(\mathrm{HNO}_{3}\) by the Ostwald process? Assume 100\(\%\) yield in each reaction, and assume that the NO produced in the third step is not recycled.

The compound As \(_{2} \mathrm{L}_{4}\) is synthesized by reaction of arsenic metal with arsenic triodide. If a solid cubic block of arsenic \(\left(d=5.72 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is 3.00 \(\mathrm{cm}\) on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triodide, what mass of \(\mathrm{As}_{2} \mathrm{L}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{L}_{4}\) was 75.6\(\%\) what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?

You take 1.00 g of an aspirin tablet (a compound consisting solely of carbon, hydrogen, and oxygen), burn it in air, and collect 2.20 \(\mathrm{g} \mathrm{CO}_{2}\) and 0.400 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) You know that the molar mass of aspirin is between 170 and 190 \(\mathrm{g} / \mathrm{mol}\) . Reacting 1 \(\mathrm{mole}\) of salicylic acid with 1 mole of acetic anhydride \(\left(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) gives you 1 mole of aspirin and 1 mole of acetic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\right)\) Use this information to determine the molecular formula of salicylic acid.

In using a mass spectrometer, a chemist sees a peak at a mass of 30.0106 . Of the choices \(^{12} \mathrm{C}_{2}^{1} \mathrm{H}_{6},^{12} \mathrm{C}^{1} \mathrm{H}_{2}^{16} \mathrm{O},\) and \(^{14} \mathrm{N}^{16} \mathrm{O}\) which is responsible for this peak? Pertinent masses are \(^{1} \mathrm{H}\) \(1.007825 ; 16 \mathrm{O}, 15.994915 ;\) and \(^{14} \mathrm{N}, 14.003074\)
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