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Chapter 8: Problem 33

Calculate the number of moles of the indicated substance in each of the following samples. a. \(41.5 \mathrm{~g}\) of \(\mathrm{MgCl}_{2}\) b. \(135 \mathrm{mg}\) of \(\mathrm{Li}_{2} \mathrm{O}\) c. \(1.21 \mathrm{~kg}\) of \(\mathrm{Cr}\) d. 62.5 g of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) e. \(42.7 \mathrm{~g}\) of \(\mathrm{C}_{6} \mathrm{H}_{6}\) f. 135 g of \(\mathrm{H}_{2} \mathrm{O}_{2}\)

Short Answer

Expert verified
The calculated number of moles for each substance is as follows: a. Approximately \(0.44 \mathrm{~mol}\) of \(\mathrm{MgCl}_{2}\) b. Approximately \(0.00452 \mathrm{~mol}\) of \(\mathrm{Li}_{2}\mathrm{O}\) c. Approximately \(23.27 \mathrm{~mol}\) of \(\mathrm{Cr}\) d. Approximately \(0.637 \mathrm{~mol}\) of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) e. Approximately \(0.546 \mathrm{~mol}\) of \(\mathrm{C}_{6}\mathrm{H}_{6}\) f. Approximately \(3.97 \mathrm{~mol}\) of \(\mathrm{H}_{2}\mathrm{O}_{2}\)

Step by step solution

01

Calculate the Molar Mass of Each Substance

For each substance, we need to find the molar mass by adding up the molar masses of its constituent elements. You can find the molar masses of elements in the periodic table. a. \(\mathrm{MgCl}_{2}\): Molar mass = 1(\(\mathrm{Mg}\)) + 2(\(\mathrm{Cl}\)) = 24.31 + 2(35.45) = 95.21 g/mol b. \(\mathrm{Li}_{2} \mathrm{O}\): Molar mass = 2(\(\mathrm{Li}\)) + 1(\(\mathrm{O}\)) = 2(6.94) + 16.00 = 29.88 g/mol c. \(\mathrm{Cr}\): Molar mass = 51.996 g/mol d. \(\mathrm{H}_{2} \mathrm{SO}_{4}\): Molar mass = 2(\(\mathrm{H}\)) + 1(\(\mathrm{S}\)) + 4(\(\mathrm{O}\)) = 2(1.008) + 32.07 + 4(16.00) = 98.08 g/mol e. \(\mathrm{C}_{6} \mathrm{H}_{6}\): Molar mass = 6(\(\mathrm{C}\)) + 6(\(\mathrm{H}\)) = 6(12.01) + 6(1.008) = 78.12 g/mol f. \(\mathrm{H}_{2} \mathrm{O}_{2}\): Molar mass = 2(\(\mathrm{H}\)) + 2(\(\mathrm{O}\)) = 2(1.008) + 2(16.00) = 34.02 g/mol
02

Calculate the Number of Moles

Now we can use the calculated molar masses and the given masses of each substance to calculate the number of moles. a. $$\frac{41.5 \mathrm{~g}}{95.21 \mathrm{~g/mol}} \approx 0.44 \mathrm{~mol} $$ of \(\mathrm{MgCl}_{2}\) b. $$\frac{135 \mathrm{~mg}}{29.88 \mathrm{~g/mol}} \times \frac{1 \mathrm{~g}}{1000 \mathrm{~mg}} \approx 0.00452 \mathrm{~mol}$$ of \(\mathrm{Li}_{2}\mathrm{O}\) c. $$\frac{1.21 \mathrm{~kg}}{51.996 \mathrm{~g/mol}} \times \frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \approx 23.27 \mathrm{~mol}$$ of \(\mathrm{Cr}\) d. $$\frac{62.5 \mathrm{~g}}{98.08 \mathrm{~g/mol}} \approx 0.637 \mathrm{~mol}$$ of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) e. $$\frac{42.7 \mathrm{~g}}{78.12 \mathrm{~g/mol}} \approx 0.546 \mathrm{~mol}$$ of \(\mathrm{C}_{6}\mathrm{H}_{6}\) f. $$\frac{135 \mathrm{~g}}{34.02 \mathrm{~g/mol}} \approx 3.97 \mathrm{~mol}$$ of \(\mathrm{H}_{2}\mathrm{O}_{2}\)

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Key Concepts

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

Molar Mass
Molar mass is a fundamental concept in chemistry that relates the mass of a substance to the number of its particles, specifically moles. It is the mass in grams of one mole of a chemical substance.
In practical terms, to calculate the molar mass of a compound, you sum the molar masses of the individual elements that make up the compound, accounting for their respective quantities within a single molecule. The unit for molar mass is grams per mole (g/mol).
This calculation is the cornerstone of stoichiometry as it establishes a quantifiable link between mass and moles, enabling chemists to predict how much of a reactant is needed or how much of a product will be created in a chemical reaction.
Stoichiometry
Stoichiometry is essentially the mathematics behind chemistry. It involves using the balanced chemical equation to calculate the relative amounts of reactants and products involved in a chemical reaction. The ratios provided by the coefficients in a balanced equation tell us the proportions of reactants and products, often measured in moles.
Using stoichiometry, one can calculate how much of a reactant is needed to completely create a product and vice versa. It requires a firm understanding of the mole concept, molar masses, and Avogadro's number to transform between atoms and grams or liters and moles when dealing with gases.
Molecular Weight
Molecular weight is synonymous with molar mass and is also expressed as grams per mole (g/mol). Molecular weight can be calculated by taking the atomic masses of the elements present in the molecule and adding them together based on their ratio in the formula. For instance, the molecular weight of water, H2O, is the sum of twice the atomic mass of hydrogen (approximately 1.008 g/mol each) and the atomic mass of oxygen (approximately 16.00 g/mol), resulting in approximately 18.02 g/mol.
Understanding molecular weight is critical because it helps us relate a quantifiable amount of a substance (in grams) to its number of molecules or formula units, vital for quantitative chemical analysis and reactions.
Avogadro's Number
Avogadro's number, approximately 6.022 x 1023, is the number of atoms, ions, or molecules in one mole of any substance. It provides a bridge between the macroscopic scale that we can measure (grams, liters, etc.) and the molecular scale (atoms, molecules).
This constant is particularly significant when converting between the number of particles and the amount of substance in moles. When it comes to calculations involving gases at standard temperature and pressure (STP), Avogadro's number lets us know that one mole of any gas occupies 22.4 liters.
For example, if a problem asks for the number of atoms in 2 moles of carbon, you would use Avogadro's number to find the answer: 2 moles x 6.022 x 1023 atoms/mole = 1.2044 x 1024 atoms of carbon.

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Most popular questions from this chapter

a. How many atoms of carbon are present in \(1.0 \mathrm{~g}\) of \(\mathrm{CH}_{4} \mathrm{O} ?\) b. How many atoms of carbon are present in \(1.0 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} ?\) c. How many atoms of nitrogen are present in \(25.0 \mathrm{~g}\) of \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2} ?\)

Calculate the mass in grams of each of the following samples. a. 10,000,000,000 nitrogen molecules b. \(2.49 \times 10^{20}\) carbon dioxide molecules c. 7.0983 moles of sodium chloride d. \(9.012 \times 10^{-6}\) moles of 1,2 -dichloroethane, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\)

A compound has the following percentage composition by mass: copper, \(33.88 \%\) nitrogen, \(14.94 \%\); oxygen, \(51.18 \%\). Determine the empirical formula of the compound.

Calculate the mass in grams of each of the following samples. a. 1.25 moles of aluminum chloride b. 3.35 moles of sodium hydrogen carbonate c. 4.25 millimoles of hydrogen bromide ( 1 millimole \(=1 / 1000\) mole \()\) d. \(1.31 \times 10^{-3}\) moles of uranium e. 0.00104 mole of carbon dioxide f. \(1.49 \times 10^{2}\) moles of iron

When a 2.118 -g sample of copper is heated in an atmosphere in which the amount of oxygen present is restricted, the sample gains \(0.2666 \mathrm{~g}\) of oxygen in forming a reddishbrown oxide. However, when \(2.118 \mathrm{~g}\) of copper is heated in a stream of pure oxygen, the sample gains \(0.5332 \mathrm{~g}\) of oxygen. Calculate the empirical formulas of the two oxides of copper.
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