/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Calculate the following quantiti... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the following quantities: (a) mass, in grams, of 0.105 mole of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) (b) moles of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) in \(143.50 \mathrm{~g}\) of this substance (c) number of molecules in \(1.0 \times 10^{-6} \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (d) number of \(\mathrm{N}\) atoms in \(0.410 \mathrm{~mol} \mathrm{NH}_{3}\)

Short Answer

Expert verified
(a) The mass of 0.105 mole of sucrose (C12H22O11) is approximately \(35.942g\). (b) There are 0.757 moles of Zn(NO3)2 in 143.50 g of this substance. (c) There are approximately \(6.022 \times 10^{17}\) molecules of CH3CH2OH in 1.0 x 10^{-6} mol. (d) There are approximately \(2.469 \times 10^{23}\) N atoms in 0.410 mol NH3.

Step by step solution

01

(a) - Calculate the molar mass of sucrose

To find the mass of 0.105 mole of sucrose, we first need to calculate the molar mass of sucrose. Sucrose (C12H22O11) has: - 12 Carbon atoms (C) - 22 Hydrogen atoms (H) - 11 Oxygen atoms (O) The molar masses of these elements are: - Carbon: 12.01 g/mol - Hydrogen: 1.01 g/mol - Oxygen: 16.00 g/mol So, the molar mass of sucrose is: \( Molar\_mass = 12*12.01 + 22*1.01 + 11*16.00 = 342.30 \,g/mol \)
02

(a) - Calculate the mass of 0.105 mole of sucrose

Now we can calculate the mass of 0.105 mole of sucrose using its molar mass: \( Mass = moles * molar\_mass \) \( Mass = 0.105 * 342.30 \,g/mol = 35.942 \,g\) So the mass of 0.105 mole of sucrose is approximately \(35.942g\).
03

(b) - Calculate the molar mass of Zn(NO3)2

To calculate the molar mass of Zn(NO3)2, we have: - 1 Zinc atom (Zn) - 2 Nitrogen atoms (N) in each NO3 group, total 4 Nitrogen atoms - 3 Oxygen atoms (O) in each NO3 group, total 6 Oxygen atoms The molar masses of these elements are: - Zinc: 65.38 g/mol - Nitrogen: 14.01 g/mol - Oxygen: 16.00 g/mol So, the molar mass of Zn(NO3)2 is: \( Molar\_mass = 65.38 + 4*14.01 + 6*16.00 = 189.39 \,g/mol\)
04

(b) - Calculate the moles of Zn(NO3)2

Now we can calculate the moles of Zn(NO3)2 in 143.50 g of this substance: \( Moles = \frac{mass}{molar\_mass} \) \( Moles = \frac{143.50 \,g}{189.39 \,g/mol} = 0.757 \,mol\) So, there are 0.757 moles of Zn(NO3)2 in 143.50 g of this substance.
05

(c) - Calculate the number of molecules of CH3CH2OH

To find the number of molecules in 1.0 x 10^{-6} mol CH3CH2OH, we will use Avogadro's number (6.022 x 10^{23}). \( Number\_of\_molecules = moles * Avogadro's\_number \) \( Number\_of\_molecules = 1.0 \times 10^{-6} \,mol * 6.022 \times 10^{23} \,molecules/mol \) \( Number\_of\_molecules = 6.022 \times 10^{17} \,molecules\) So there are approximately \(6.022 \times 10^{17}\) molecules of CH3CH2OH in 1.0 x 10^{-6} mol.
06

(d) Calculate the number of N atoms in 0.410 mol NH3

To find the number of N atoms in 0.410 mol NH3, first, we calculate the number of NH3 molecules: \( Number\_of\_NH3\_molecules = moles * Avogadro's\_number \) \( Number\_of\_NH3\_molecules = 0.410 \,mol * 6.022 \times 10^{23} \,molecules/mol \) \( Number\_of\_NH3\_molecules = 2.469 \times 10^{23} \,molecules\) Since there is 1 N atom in each NH3 molecule: \( Number\_of\_N\_atoms = Number\_of\_NH3\_molecules = 2.469 \times 10^{23} \,atoms\) So there are approximately \(2.469 \times 10^{23}\) N atoms in 0.410 mol NH3.

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Key Concepts

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

Molar Mass Calculation
Understanding how to calculate molar mass is fundamental in chemistry. It is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). To calculate the molar mass, you need to know the chemical formula of the substance and the atomic masses of its constituent elements, which can be found on the periodic table.

For example, in the exercise, the molar mass of sucrose (\(C_{12}H_{22}O_{11}\)) is determined by multiplying the number of each type of atom by its atomic mass (in g/mol) and then summing these values. Here's the breakdown:
  • Carbon (\(C\)): 12 atoms \times 12.01 g/mol
  • Hydrogen (\(H\)): 22 atoms \times 1.01 g/mol
  • Oxygen (\(O\)): 11 atoms \times 16.00 g/mol
The molar mass of sucrose then becomes the sum of all these, resulting in 342.30 g/mol. Knowing the molar mass is essential as it enables you to convert between moles and grams, a vital step in stoichiometric calculations.
Avogadro's Number
Avogadro's number, approximately 6.022 x 1023, is the number of units in one mole of any substance. It can refer to atoms, ions, or molecules, depending on the context. This constant is pivotal in converting moles to discrete particles and vice versa.

For instance, in step 5 of the exercise, to find out the number of molecules in 1.0 x 10-6 moles of ethanol (\(CH_{3}CH_{2}OH\)), you multiply the moles by Avogadro's number: \[ Number\_of\_molecules = 1.0 \times 10^{-6} \,mol \times 6.022 \times 10^{23} \,molecules/mol \] yielding 6.022 x 1017 molecules. This concept is crucial for understanding how much of a substance is needed or produced in a chemical reaction.
Chemical Formula Stoichiometry
Chemical formula stoichiometry involves using the balanced equation of a chemical reaction to calculate the relative quantities of reactants and products. It is based on the principle of the conservation of mass and the mole concept. To perform these calculations, you'll need a balanced chemical equation and knowledge of the concepts of molar mass and Avogadro's number.

In the context of the given problems, once you determine the number of moles, you can calculate the number of molecules or atoms using stoichiometry. For example, in calculating the number of nitrogen atoms in 0.410 mole of ammonia (\(NH_{3}\)), you first find the number of molecules of ammonia and then use the fact that each molecule contains one nitrogen atom to determine the total number of nitrogen atoms: \[ Number\_of\_N\_atoms = Number\_of\_NH3\_molecules = Moles \times Avogadro's\_number \] Final stoichiometric calculations enable us to understand the detailed composition of substances involved in chemical reactions, crucial for laboratory preparations, and industrial processes.

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Most popular questions from this chapter

The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, principally \(\mathrm{N}_{2}(\sim 79 \%)\) and \(\mathrm{O}_{2}(\sim 20 \%) .\) In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As \(\mathrm{NO}\) is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reactions. (b) Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called " \(\mathrm{NO}_{\mathrm{x}}^{\prime \prime}\) gases. In \(2007,\) the United States emitted an estimated 22 million tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of \(\mathrm{NO}_{\mathrm{x}}\) gases is an unwanted side reaction of the main engine combustion process that turns octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), into \(\mathrm{CO}_{2}\) and water. If \(85 \%\) of the oxygen in an engine is used to combust octane and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of 500 grams of octane.

Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which means that a certain number of water molecules are included in the solid structure. Its formula can be written as \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot x \mathrm{H}_{2} \mathrm{O},\) where \(x\) is the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). When a \(2.558-\mathrm{g}\) sample of washing soda is heated at \(25^{\circ} \mathrm{C},\) all the water of hydration is lost, leaving \(0.948 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) What is the value of \(x ?\)

Section 2.9 introduced the idea of structural isomerism, with 1-propanol and 2 -propanol as examples. Determine which of these properties would distinguish these two substances: (a) boiling point; (b) combustion analysis results; (c) molecular weight; (d) density at a given temperature and pressure. You can check on the properties of these two compounds in Wolfram Alpha (http://www.wolframalpha.com/) or the CRC Handbook of Chemistry and Physics.

Very small crystals composed of 1000 to 100,000 atoms, called quantum dots, are being investigated for use in electronic devices. (a) A quantum dot was made of solid silicon in the shape of a sphere, with a diameter of \(4 \mathrm{nm} .\) Calculate the mass of the quantum dot, using the density of silicon \(\left(2.3 \mathrm{~g} / \mathrm{cm}^{3}\right)\) (b) How many silicon atoms are in the quantum dot? (c) The density of germanium is \(5.325 \mathrm{~g} / \mathrm{cm}^{3}\). If you made a 4-nm quantum dot of germanium, how many Ge atoms would it contain? Assume the dot is spherical.

The fizz produced when an Alka-Seltzer \(^{\circledast}\) tablet is dissolved in water is due to the reaction between sodium bicarbonate \(\left(\mathrm{NaHCO}_{3}\right)\) and citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) $$ \begin{aligned} 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \longrightarrow \\ & 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \end{aligned} $$ In a certain experiment \(1.00 \mathrm{~g}\) of sodium bicarbonate and \(1.00 \mathrm{~g}\) of citric acid are allowed to react. (a) Which is the limiting reactant? (b) How many grams of carbon dioxide form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed?

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