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Is each of the following statements true or false? Correct any that are false. (a) A mole of one substance has the same number of atoms as a mole of any other substance. (b) The theoretical yield for a reaction is based on the balanced chemical equation. (c) A limiting-reactant problem is being stated when the available quantity of one of the reactants is given in moles. (d) The empirical and molecular formulas of a compound are always different.

Short Answer

Expert verified
(a) True, (b) True, (c) False - more info needed, (d) False - can be same

Step by step solution

01

- Statement (a)

A mole of any substance, per Avogadro's number, contains the same number of elementary entities (atoms, molecules, etc.). Therefore, statement (a) is true.
02

- Statement (b)

The theoretical yield for a reaction is indeed based on the stoichiometry revealed by the balanced chemical equation, which tells how much product can theoretically be made from given amounts of reactants. Therefore, statement (b) is true.
03

- Statement (c)

A limiting-reactant problem typically requires the quantities of all reactants to determine which one limits the reaction. Thus, merely stating the quantity of one reactant in moles does not constitute a limiting-reactant problem on its own. Therefore, statement (c) is false.
04

- Statement (d)

The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula provides the exact number of atoms of each element in a molecule. For some compounds, the empirical and molecular formulas can be the same (like in simple compounds such as H2O). Therefore, statement (d) is false.

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Key Concepts

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

Avogadro's Number
Avogadro's Number is a fundamental constant in chemistry. It is the number of elementary entities (such as atoms, molecules, or ions) in one mole of any substance.
This number is approximately \(6.022 \times 10^{23}\).
Because of this, a mole of any substance will have the same number of individual units, making it a universal measure. For example, one mole of carbon atoms has the same number of atoms as one mole of hydrogen molecules.

Understanding this concept helps chemists count and compare very large numbers of tiny particles efficiently. Whether dealing with large molecules or single atoms, Avogadro's Number ensures consistency in measurements.
Theoretical Yield
Theoretical yield is the amount of product that can be formed in a chemical reaction, based on the stoichiometric relationships established by a balanced chemical equation.
To calculate the theoretical yield, you begin with the balanced equation, which provides the mole ratio of reactants to products. Using the amount of limiting reactant (the reactant that determines how much product is made), you can predict the maximum amount of product that could be formed if the reaction proceeds perfectly.
It’s important to note that the theoretical yield is often higher than the actual yield obtained in practice due to losses in the process or incomplete reactions.
Limiting Reactant
The limiting reactant is the reactant in a chemical reaction that is completely consumed first, thus limiting the amount of products formed.
To identify the limiting reactant, you need the quantities of all reactants involved. By comparing the molar proportions required (as given in the balanced equation) with the actual amounts of reactants available, you can determine which reactant will run out first.
This helps in calculating the theoretical yield because once the limiting reactant is used up, the reaction cannot proceed further, even if other reactants are still available.
  • Example: If you have 5 moles of A and 8 moles of B and the reaction requires 1 mole of A to react with 2 moles of B, A is the limiting reactant as it will run out after forming 5 moles of product.
Empirical and Molecular Formulas
The empirical formula of a compound gives the simplest whole-number ratio of atoms of each element in the compound.
For example, the empirical formula of hydrogen peroxide (H\(_2\)O\(_2\)) is HO, as the simplest ratio of atoms is 1:1.
The molecular formula, however, provides the actual number of each type of atom in a molecule of the compound.
In the case of hydrogen peroxide, the molecular formula is H\(_2\)O\(_2\), showing that each molecule consists of two hydrogen atoms and two oxygen atoms.
The relationship between these formulas is crucial because some compounds have the same empirical formula but different molecular formulas, such as glucose (C\(_6\)H\(_12\)O\(_6\)) and acetic acid (C\(_2\)H\(_4\)O\(_2\)), which have different properties and structures despite having the same empirical formula (CH\(_2\)O).
  • Empirical formulas help in identifying the basic composition, while molecular formulas give insights into the actual structure and amount of elements in a molecule.

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Most popular questions from this chapter

Hydrocarbon mixtures are used as fuels. (a) How many grams of \(\mathrm{CO}_{2}(g)\) are produced by the combustion of \(200 . \mathrm{g}\) of a mixture that is \(25.0 \% \mathrm{CH}_{4}\) and \(75.0 \% \mathrm{C}_{3} \mathrm{H}_{8}\) by mass? (b) A \(252-\mathrm{g}\) gaseous mixture of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{3} \mathrm{H}_{8}\) burns in excess \(\mathrm{O}_{2},\) and \(748 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) gas is collected. What is the mass \(\%\) of \(\mathrm{CH}_{4}\) in the mixture?

Calculate each of the following quantities: (a) Amount (mol) of Mn atoms in \(62.0 \mathrm{mg}\) of Mn (b) Amount (mol) for \(1.36 \times 10^{22}\) atoms of \(\mathrm{Cu}\) (c) Mass (g) of \(8.05 \times 10^{24} \mathrm{Li}\) atoms

Nicotine is a poisonous, addictive compound found in tobacco. A sample of nicotine contains \(6.16 \mathrm{mmol}\) of \(\mathrm{C}, 8.56 \mathrm{mmol}\) $$ \text { of } \mathrm{H} \text { , and } 1.23 \mathrm{mmol} \text { of } \left.\mathrm{N}[1 \mathrm{mmol} \text { ( } 1 \text { millimole })=10^{-3} \mathrm{~mol}\right] \text { . } $$ $$ \text { What is the empirical formula of nicotine? } $$

Chromium(III) oxide reacts with hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) gas to form chromium(III) sulfide and water: $$ \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2} \mathrm{~S}(g) \longrightarrow \mathrm{Cr}_{2} \mathrm{~S}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ To produce \(421 \mathrm{~g}\) of \(\mathrm{Cr}_{2} \mathrm{~S}_{3},\) how many (a) moles of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) and (b) grams of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) are required?

Chlorine gas can be made in the laboratory by the reaction of hydrochloric acid and manganese(IV) oxide: $$ 4 \mathrm{HCl}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{MnCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2}(g) $$ When 1.82 mol of \(\mathrm{HCl}\) reacts with excess \(\mathrm{MnO}_{2}\), how many (a) moles of \(\mathrm{Cl}_{2}\) and (b) grams of \(\mathrm{Cl}_{2}\) form?

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