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Chapter 3: Problem 20

How many of the following are found in 15.0 kmol of xylene \(\left(\mathrm{C}_{8} \mathrm{H}_{10}\right) ?\) (a) \(\mathrm{kg} \mathrm{C}_{8} \mathrm{H}_{10}\); (b) mol \(\mathrm{C}_{8} \mathrm{H}_{10}\); (c) lb-mole \(\mathrm{C}_{8} \mathrm{H}_{10} ;\) (d) mol (g-atom) \(\mathrm{C} ;\) (e) mol \(\mathrm{H} ;\) (f) \(\mathrm{g} \mathrm{C} ;\) (g) \(\mathrm{g} \mathrm{H} ;\) (h) molecules of \(\mathrm{C}_{8} \mathrm{H}_{10}\).

Short Answer

Expert verified
(a) 1592.4 kg of xylene, (b) 15000 mol of xylene, (c) 33.07 lb-mole of xylene, (d) 120000 mol (g-atom) of Carbon, (e) 150000 mol of Hydrogen, (f) 1441200 g of Carbon, (g) 151200 g of Hydrogen, (h) 9.033 x 10^28 molecules of xylene.

Step by step solution

01

Calculate Molecular Weight of Xylene.

First, we need to calculate the molecular weight of xylene (C8H10). For this, we use the atomic weights of Carbon (C) and Hydrogen (H). Carbon has an atomic weight of about 12.01 g/mol, and Hydrogen has an atomic weight of about 1.008 g/mol. So the total molecular weight of xylene is \(8 \times 12.01 + 10 \times 1.008 = 106.16 \, g/mol\).
02

Convert kmol to mol.

Now, we need to convert the given amount of xylene from kmol to mol. We know that 1 kmol equals 1000 mol. So in this problem, 15.0 kmol of xylene is \(15.0 \times 1000 = 15000 \, mol\).
03

Calculate mass in kg.

To calculate the mass in kg, we use the molecular weight of xylene and the amount in moles. The formula is \(Mass = moles \times molecular \, weight\). Thus, the mass of xylene would be \(15000 \, mol \times 106.16 \, g/mol \times 1 kg/1000 g = 1592.4 \, kg\).
04

Calculate amount in lb-mole.

To convert moles to lb-mole, we divide by a factor of 453.592 (since 1 lb-mole is equivalent to 453.592 moles). Thus, the amount of xylene is \(15000 \, mol / 453.592 = 33.07 \, lb-mole\).
05

Calculate amount in mol (g-atom) Carbon.

Xylene has 8 Carbon atoms per molecule. Therefore, the amount in mol (g-atom) of Carbon is \(15000 \, mol \times 8 = 120000 \, mol\).
06

Calculate amount in mol Hydrogen.

Similar to step 5, xylene has 10 Hydrogen atoms per molecule. Therefore, the amount in mol of Hydrogen is \(15000 \, mol \times 10 = 150000 \, mol\).
07

Calculate mass in g of Carbon and Hydrogen.

The mass in grams of Carbon can be calculated by \(120000 \, mol \times 12.01 \, g/mol = 1441200 \, g\). And the mass in grams of Hydrogen can be calculated by \(150000 \, mol \times 1.008 \, g/mol = 151200 \, g\).
08

Calculate number of molecules.

Using Avogadro's number (6.022 x 10^23 molecules/mol), the amount of xylene molecules is \(15000 \, mol \times 6.022 \times 10^{23} molecules/mol = 9.033 \times 10^{28} molecules\).

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Key Concepts

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

Molecular Weight Calculation
Understanding molecular weight calculation is essential for delving into stoichiometry problems. Molecular weight, also known as molecular mass, is the sum of the atomic weights of all atoms in a molecule. It's calculated by multiplying the atomic weight of each element by the number of atoms of that element in the molecule and then adding these values together.

For example, xylene has a chemical formula of \( C_8H_{10} \). The atomic weight of carbon (C) is approximately 12.01 g/mol, and hydrogen (H) is about 1.008 g/mol. The molecular weight of xylene is computed as follows: \((8 \times 12.01) + (10 \times 1.008) = 106.16 \, g/mol\).

Knowing the precise molecular weight is critical because it relates the physical mass of a substance to its amount in moles, a standard way of expressing quantities in chemistry. This is helpful when you're scaling reactions up or down or when converting between mass and moles, which is common in stoichiometry calculations.
Mole Concept
The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we experience every day. A mole is simply a counting unit similar to a dozen, except instead of 12, it represents approximately 6.022 \(\times\) 10^23 items, which is Avogadro's number.

This huge number corresponds to the number of carbon atoms in exactly 12 grams of pure carbon-12, and it's used because atoms and molecules are exceedingly tiny and numerous. When we say we have a mole of a substance, we mean we have Avogadro's number of molecules or atoms of that substance. The mole concept allows chemists to work with the submicroscopic particles in more manageable quantities of grams and kilograms, which can be weighed and handled in the laboratory.
Mass-to-Mole Conversion
Mass-to-mole conversion is at the heart of stoichiometry in chemistry. This conversion uses the molecular weight of a substance and the mole concept to interconvert between the mass of a substance and the number of moles.

The formula used is: \[\text{Number of moles} = \frac{\text{Mass (g)}}{\text{Molecular weight (g/mol)}}\]. For instance, if you have 1592.4 grams of xylene, and the molecular weight of xylene is 106.16 g/mol, the number of moles is calculated as \(\frac{1592.4}{106.16} = 15\) moles. This step is indispensable because reactions in chemistry are typically written on a mole basis, meaning you need to know the moles to understand how much of each reactant is needed and how much of each product will be made.

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Most popular questions from this chapter

In the manufacture of pharmaceuticals, most active pharmaceutical ingredients (APIs) are made in solution and then recovered by separation. Acetaminophen, a pain-killing drug commercially marketed as Tylenol", is synthesized in an aqueous solution and subsequently crystallized. The slurry of crystals is sent to a centrifuge from which two effluent streams emerge: ( 1 ) a wet cake containing 90.0 wt\% solid acetaminophen \((\mathrm{MW}=\) 151 g/mol) and 10.0 wt\% water (plus some acetaminophen and other dissolved substances, which we will neglect), and (2) a highly dilute aqueous solution of acetaminophen that is discharged from the process. The wet cake is fed to a dryer where the water is completely evaporated, leaving the residual acetaminophen solids bone dry. If the evaporated water were condensed, its volumetric flow rate would be \(50.0 \mathrm{Lh}\). Following is a flowchart of the process, which runs 24 h/day, 320 days/yr. A denotes acetaminophen. (a) Calculate the yearly production rate of solid acetaminophen (tonne/yr), using as few dimensional equations as possible. (b) A proposal has been made to subject the liquid solution leaving the centrifuge to further processing to recover more of the dissolved acetaminophen instead of disposing of the solution. On what would the decision depend?

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The feed to an ammonia synthesis reactor contains 25 mole \(\%\) nitrogen and the balance hydrogen. The flow rate of the stream is \(3000 \mathrm{kg} / \mathrm{h}\). Calculate the rate of flow of nitrogen into the reactor in \(\mathrm{kg} / \mathrm{h}\). (Suggestion: First calculate the average molecular weight of the mixture.)

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Certain solid substances, known as hydrated compounds, have well-defined molecular ratios of water to some other species. For example, calcium sulfate dihydrate (commonly known as gypsum, \(\left.\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right),\) has 2 moles of water per mole of calcium sulfate; alternatively, it may be said that 1 mole of gypsum consists of 1 mole of calcium sulfate and 2 moles of water. The water in such substances is called water of hydration. (More information about hydrated salts is given in Chapter 6 .) In order to eliminate the discharge of sulfuric acid into the environment, a process has been developed in which the acid is reacted with aragonite \(\left(\mathrm{CaCO}_{3}\right)\) to produce calcium sulfate. The calcium sulfate then comes out of solution in a crystallizer to form a slurry (a suspension of solid particles in a liquid) of solid gypsum particles suspended in an aqueous \(\mathrm{CaSO}_{4}\) solution. The slurry flows from the crystallizer to a filter in which the particles are collected as a filter cake. The filter cake, which is 95.0 wiff solid gypsum and the remainder CaSO_solution, is fed to a dryer in which all water (including the water of hydration in the crystals) is driven off to yield anhydrous (water-free) CaSO \(_{4}\) as product. A flowchart and relevant process data are given below. Solids content of slurry leaving crystallizer: \(0.35 \mathrm{kg} \mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O} / \mathrm{L}\) slurry \(\mathrm{CaSO}_{4}\) content of slurry liquid: \(0.209 \mathrm{g} \mathrm{CaSO}_{4} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) Specific gravities: \(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}), 2.32 ;\) liquid solutions, 1.05 (a) Briefly explain in your own words the functions of the three units (crystallizer, filter, and dryer). (b) Takea basis of one liter of solution leaving the crystallizer and calculate the mass (kg) and volume (L) of solid gypsum, the mass of \(\mathrm{CaSO}_{4}\) in the gypsum, and the mass of \(\mathrm{CaSO}_{4}\) in the liquid solution. (c) Calculate the percentage recovery of \(\mathrm{CaSO}_{4}-\) that is, the percentage of the total \(\mathrm{CaSO}_{4}\) (precipitated plus dissolved) leaving the crystallizer recovered as solid anhydrous \(\mathrm{CaSO}_{4}\) (d) List five potential negative consequences of discharging \(\mathrm{H}_{2} \mathrm{SO}_{4}\) into the river passing the plant.
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