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Chapter 5: Problem 24

Avogadro's number, molar mass, and the chemical formula of a compound are three useful conversion factors. What unit conversions can be accomplished using these conversion factors?

Short Answer

Expert verified
Using Avogadro's number, molar mass, and the chemical formula of a compound as conversion factors, we can accomplish the following unit conversions: 1. Convert the number of particles (atoms, molecules, ions, or electrons) to moles and vice versa using Avogadro's number. 2. Convert mass to moles and vice versa using molar mass. 3. Convert moles of one substance to moles of another substance, and moles of a substance to mass and vice versa using chemical formula and molar mass.

Step by step solution

01

Avogadro's number is a constant and represents the number of particles (atoms, molecules, ions, or electrons) in one mole of a substance. It is equal to approximately 6.022 x 10^23 particles per mole. Using Avogadro's number, we can convert between the number of particles and the amount of substance in moles. #Step 2: Unit conversions using Avogadro's Number#

With Avogadro's number, we can accomplish the following unit conversions: 1. Convert the number of particles (atoms, molecules, ions, or electrons) to moles: Divide the given number of particles by Avogadro's number. 2. Convert moles to the number of particles: Multiply the given number of moles by Avogadro's number. #Step 3: Understanding Molar Mass#
02

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It can be calculated by adding up the atomic masses of all the atoms in a molecule or formula unit of a substance. Molar mass helps us convert between the mass of a substance and the amount of substance in moles. #Step 4: Unit conversions using Molar Mass#

With molar mass, we can accomplish the following unit conversions: 1. Convert mass to moles: Divide the given mass of a substance by its molar mass. 2. Convert moles to mass: Multiply the given number of moles by the molar mass of the substance. #Step 5: Understanding Chemical Formula#
03

The chemical formula of a compound tells us the type and number of each atom present in the compound. It can be used as a conversion factor to relate the moles of different atoms or compounds in a chemical reaction. #Step 6: Unit conversions using Chemical Formula#

With the chemical formula, we can accomplish the following unit conversions: 1. Convert moles of one substance to moles of another substance: Use the stoichiometric coefficients (numbers) in the balanced chemical equation. 2. Convert moles of a substance to mass and vice versa: Use the molar mass of the substance. In summary, the unit conversions that can be accomplished using Avogadro's number, molar mass, and the chemical formula are: - Number of particles to moles and vice versa - Mass to moles and vice versa - Moles of one substance to moles of another substance - Moles of a substance to mass and vice versa.

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Most popular questions from this chapter

Phosphorus can be prepared from calcium phosphate by the following reaction: $$\begin{aligned}2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+& 10 \mathrm{C}(s) \longrightarrow \\\& 6 \mathrm{CaSiO}_{3}(s)+\mathrm{P}_{4}(s)+10 \mathrm{CO}(g) \end{aligned}$$ Phosphorite is a mineral that contains \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) plus other non-phosphorus- containing compounds. What is the maximum amount of \(\mathrm{P}_{4}\) that can be produced from \(1.0 \mathrm{kg}\) of phosphorite if the phorphorite sample is \(75 \% \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) by mass? Assume an excess of the other reactants.

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{g} \mathrm{N}_{2}\) and \(5.00 \times 10^{2} \mathrm{g} \mathrm{H}_{2} ?\) b. What mass of which starting material would remain unreacted?

Commercial brass, an alloy of Zn and 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 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?

Balance the following equations: a. \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) b. \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(A g N O_{3}(a q)+H_{2} S O_{4}(a q) \rightarrow A g_{2} S O_{4}(s)+H N O_{3}(a q)\)

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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