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Arrange the following substances in order of increasing mass percent of carbon. a. caffeine, \(\mathrm{C}_{\mathrm{s}} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\) b. sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) c. ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

Short Answer

Expert verified
The substances arranged in increasing order of mass percent of carbon are: Sucrose (42.11%), Caffeine (49.48%), and Ethanol (53.31%).

Step by step solution

01

Calculate the molar mass of C, H, N, and O

First, we need to find the molar mass of each element present in the given substances. Carbon (C): 12.01 g/mol Hydrogen (H): 1.008 g/mol Nitrogen (N): 14.007 g/mol Oxygen (O): 16.00 g/mol
02

Calculate the molar mass of each substance

Now, let's find the molar mass of each substance using the molar masses of their constituent elements. a. Caffeine, C8H10N4O2: Molar mass = (8 × 12.01) + (10 × 1.008) + (4 × 14.007) + (2 × 16.00) = 96.08 + 10.08 + 56.028 + 32 = 194.188 g/mol b. Sucrose, C12H22O11: Molar mass = (12 × 12.01) + (22 × 1.008) + (11 × 16.00) = 144.12 + 22.176 + 176 = 342.296 g/mol c. Ethanol, C2H5OH: Molar mass = (2 × 12.01) + (5 × 1.008) + 16.00 = 24.02 + 5.04 + 16 = 45.06 g/mol
03

Calculate the mass percent of carbon in each substance

Next, we will find the mass percent of carbon in each substance. a. Caffeine: Mass percent of carbon = (Total mass of carbon / Molar mass of caffeine) × 100 = (96.08 / 194.188) × 100 = 49.48% b. Sucrose: Mass percent of carbon = (Total mass of carbon / Molar mass of sucrose) × 100 = (144.12 / 342.296) × 100 = 42.11% c. Ethanol: Mass percent of carbon = (Total mass of carbon / Molar mass of ethanol) × 100 = (24.02 / 45.06) × 100 = 53.31%
04

Arrange the substances in increasing order of mass percent of carbon

Now we will arrange the substances in increasing order of their mass percent of carbon. Sucrose (42.11%) < Caffeine (49.48%) < Ethanol (53.31%) So, the order of substances in increasing mass percent of carbon is Sucrose, Caffeine, and Ethanol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
When dealing with chemical substances, understanding molar mass is crucial. Molar mass is the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). To calculate it for a compound, you need to sum up the products of the atomic masses of each constituent element and their respective quantities in the formula.
For instance, let's consider our example caffeine (\( \text{C}_8\text{H}_{10}\text{N}_4\text{O}_2 \)).
  • First, identify the elements and their quantities: Carbon (C) with 8 atoms, Hydrogen (H) with 10 atoms, Nitrogen (N) with 4 atoms, and Oxygen (O) with 2 atoms.
  • Next, use the atomic masses: Carbon is 12.01 g/mol, Hydrogen is 1.008 g/mol, Nitrogen is 14.007 g/mol, and Oxygen is 16.00 g/mol.
Using these numbers, you can easily compute the molar mass of caffeine by multiplying each atomic mass by the number of atoms and summing the results, as shown in the original exercise. Understanding molar mass aids in comparing different substances and is foundational for further chemistry calculations.
Elemental Composition
Elemental composition refers to the proportion of each element within a chemical compound. This proportion is generally expressed as a percentage by mass. Knowing the elemental composition helps to understand the specific makeup of a substance.
To determine the mass percent of an element, such as carbon in sucrose (\( \\text{C}_{12}\text{H}_{22}\text{O}_{11} \)), follow these steps:
  • Calculate the total mass of carbon: 12 atoms of carbon times 12.01 g/mol each equals 144.12 g/mol.
  • Find the molar mass of sucrose: This we've already learned to calculate as 342.296 g/mol.
  • Finally, compute the mass percent: The formula is \( \frac{\text{mass of carbon}}{\text{molar mass of sucrose}} \times 100 \), resulting in 42.11% carbon by mass.
These steps can be replicated for any compound to find the proportion of various elements within. A clear grasp of elemental composition helps chemists adjust formulations, compare substances, and improve material properties.
Chemical Substance Comparison
Comparing chemical substances involves examining their composition, structure, and properties. One key aspect of this comparison is assessing the mass percent of specific elements, like carbon. This percentage indicates how much of a compound is made up of a particular element. By comparing mass percents, chemists can infer certain qualities about substances, such as their reactivity or energy content.
Let's consider ethanol (\( \\text{C}_2\text{H}_5\text{OH} \)) and caffeine.
  • Ethanol has a higher mass percent of carbon (53.31%) compared to caffeine (49.48%), indicating that, proportionally, ethanol contains more carbon per gram of substance.
  • By understanding these values, you can predict that ethanol might release more energy per gram when burned compared to caffeine, based on its higher carbon content.
  • Such comparisons of elemental composition provide insight into potential chemical behaviors and applications.
Through these analyses, chemists can prioritize the use of one substance over another in products or processes based on targeted properties or characteristics.

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Most popular questions from this chapter

DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction: $$ 2 \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}+\mathrm{C}_{2} \mathrm{HOCl}_{3} \longrightarrow \mathrm{C}_{14} \mathrm{H}_{4} \mathrm{Cl}_{5}+\mathrm{H}_{2} \mathrm{O} $$ \(\begin{array}{ll}\text { orobenzenc chloral } & \mathrm{D}\end{array}\) In a government lab, \(1142 \mathrm{~g}\) of chlorobenzene is reacted with \(485 \mathrm{~g}\) of chloral. a. What mass of DDT is formed? b. Which reactant is limiting? Which is in excess? c. What mass of the excess reactant is left over? d. If the actual yield of DDT is \(200.0 \mathrm{~g}\), what is the percent yield?

One of relatively few reactions that takes place directly between two solids at room temperature is $$ \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+\mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow $$ In this equation, the \(\cdot 8 \mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) indicates the pres- ence of eight water molecules. This compound is called barium hydroxide octahydrate. a. Balance the equation. b. What mass of ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right)\) must be used if it is to react completely with \(6.5 \mathrm{~g}\) barium hydroxide octahydrate?

Dimethylnitrosamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2} \mathrm{O}\), is a carcinogenic (cancercausing) substance that may be formed in foods, beverages, or gastric juices from the reaction of nitrite ion (used as a food preservative) with other substances. a. What is the molar mass of dimethylnitrosamine? b. How many moles of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2} \mathrm{O}\) molecules are present in \(250 \mathrm{mg}\) dimethylnitrosamine? c. What is the mass of \(0.050\) mol dimethylnitrosamine? d. How many atoms of hydrogen are in \(1.0 \mathrm{~mol}\) dimethylnitrosamine? e. What is the mass of \(1.0 \times 10^{6}\) molecules of dimethylnitrosamine? f. What is the mass in grams of one molecule of dimethylnitrosamine?

Commercial brass, an alloy of \(Z n\) and \(\mathrm{Cu}\), reacts with hydrochloric acid as follows: $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ (Cu does not react with HCl.) When \(0.5065 \mathrm{~g}\) of a certain brass alloy is reacted with excess \(\mathrm{HCl}, 0.0985 \mathrm{~g} \mathrm{ZnCl}_{2}\) is eventually isolated. a. What is the composition of the brass by mass? b. How could this result be checked without changing the above procedure?

Which (if any) of the following is true regarding the limiting reactant in a chemical reaction? a. The limiting reactant has the lowest coefficient in a balanced equation. b. The limiting reactant is the reactant for which you have the fewest number of moles. c. The limiting reactant has the lowest ratio of moles available/ coefficient in the balanced equation. d. The limiting reactant has the lowest ratio of coefficient in the balanced equation/moles available. Justify your choice. For those you did not choose, explain why they are incorrect.

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