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Moles, Equivalents, Formulas, and Stoichiometry
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Moles, Equivalents, Formulas, and Stoichiometry
Moles, Equivalents, Formulas, and Stoichiometry
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1
Question
How does molar mass connect grams and moles?
Answer
Molar mass is the mass of one mole of a substance, expressed in grams per mole. To convert grams to moles, divide the sample mass by its molar mass: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).
2
Question
Why are ionic compounds described with formula units rather than molecules?
Answer
Ionic solids form extended three-dimensional lattices of oppositely charged ions rather than discrete molecular units. Therefore, the empirical formula is represented as a formula unit, and formula weight is used instead of molecular weight.
3
Question
How is the molecular weight of a compound calculated?
Answer
Add the atomic weights of every atom represented in the molecular formula. For \(\text{SOCl}_2\), the calculation is \(32.1 + 16.0 + 2(35.5) = 119.1\) amu per molecule.
4
Question
Why does one mole of a compound have a mass numerically matching its molecular weight?
Answer
A mole contains Avogadro’s number of particles, \(6.022 \times 10^{23}\). The molecular weight gives the mass of one particle in amu, while the corresponding molar mass gives the mass of one mole in grams with the same numerical value.
5
Question
How does Avogadro’s number make chemical amounts manageable?
Answer
Avogadro’s number defines a mole as \(6.022 \times 10^{23}\) particles. This lets chemists count atoms or molecules indirectly by measuring macroscopic amounts such as grams rather than counting individual particles.
6
Question
How would you calculate the moles in a measured mass of magnesium chloride?
Answer
First calculate the molar mass of \(\text{MgCl}_2\) from the atomic weights. Then divide the measured mass, such as \(9.53\) g, by that molar mass using \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).
7
Question
Why can equal gram samples contain different numbers of molecules?
Answer
The number of particles depends on moles, not directly on grams. A gram sample of a compound with lower molar mass contains more moles and therefore more molecules than the same mass of a compound with higher molar mass.
8
Question
How do equivalents extend the meaning of a mole?
Answer
An equivalent counts moles of the particular species relevant to a reaction, such as protons, hydroxide ions, electrons, or other ions. One mole of a compound may produce one, two, or more equivalents depending on how many relevant particles each molecule supplies.
9
Question
Why do equal moles of different acids produce different numbers of equivalents?
Answer
Different acids donate different numbers of protons per molecule. One mole of HCl provides one mole of hydrogen ions, H₂SO₄ provides two, and H₃PO₄ provides three; therefore, their equivalent amounts differ.
10
Question
How does sulfuric acid illustrate the difference between moles and equivalents?
Answer
One mole of H₂SO₄ can donate two moles of hydrogen ions, so it represents two equivalents in an acid–base reaction. Consequently, one-half mole of H₂SO₄ supplies one equivalent of hydrogen ions.
11
Question
How is gram equivalent weight calculated for a reactive compound?
Answer
Gram equivalent weight is calculated by dividing molar mass by \(n\), where \(n\) is the number of relevant particles produced or consumed per molecule: \(\text{GEW} = \frac{\text{molar mass}}{n}\).
12
Question
Why is the gram equivalent weight of an acid smaller than its molar mass when it donates multiple protons?
Answer
If one molecule supplies multiple protons, one mole of the acid supplies multiple equivalents. Therefore, less than one mole—and thus less than the full molar mass in grams—is needed to provide one equivalent: \(\text{GEW} = \frac{\text{molar mass}}{n}\).
13
Question
How are equivalents calculated when the compound mass is known?
Answer
First determine the compound’s gram equivalent weight. Then divide the compound mass by that value: \(\text{equivalents} = \frac{\text{mass of compound}}{\text{gram equivalent weight}}\).
14
Question
How does normality differ from molarity in acid–base chemistry?
Answer
Molarity measures moles of compound per liter, whereas normality measures equivalents of the reaction-relevant species per liter. Because compounds can produce different numbers of equivalents per mole, their normality and molarity may differ.
15
Question
Why does a 1 N carbonic acid solution have lower molarity than a 1 N hydrochloric acid solution?
Answer
HCl is monoprotic, so one mole of HCl supplies one equivalent and a 1 N solution is 1 M HCl. H₂CO₃ is diprotic, so each mole supplies two equivalents and a 1 N solution contains 0.5 M H₂CO₃.
16
Question
How are molarity and normality converted when the equivalence factor is known?
Answer
Normality equals molarity multiplied by the number of relevant particles produced or consumed per mole: \(N = M n\). Rearranging gives \(M = \frac{N}{n}\).
17
Question
Why can normality compare acids and bases more directly than molarity?
Answer
Normality expresses both solutions in equivalents of the species that actually reacts, such as hydrogen or hydroxide ions. One equivalent of acid neutralizes one equivalent of base even when their compound molarities differ.
18
Question
How does the diprotic nature of carbonic acid affect neutralization?
Answer
Each H₂CO₃ molecule can provide two protons under the stated completion assumption. Therefore, two equivalents of base are required to neutralize one mole of carbonic acid completely.
19
Question
How do empirical and molecular formulas differ in the information they provide?
Answer
An empirical formula gives the simplest whole-number ratio of elements. A molecular formula gives the exact number of atoms in a molecule and is either identical to or a whole-number multiple of the empirical formula.
20
Question
Why can an ionic compound such as sodium chloride have only an empirical formula?
Answer
Ionic compounds do not consist of discrete molecules in the solid state. Their formulas represent the simplest ratio of ions in the lattice, so NaCl is treated as an empirical formula or formula unit.
21
Question
How does the law of constant composition constrain samples of a compound?
Answer
Every pure sample of a given compound contains the same elements in the same mass ratio. For water, the stated ratio is one gram of hydrogen to eight grams of oxygen, regardless of where the sample is obtained.
22
Question
How can percent composition by mass be calculated from a formula?
Answer
Determine the total mass contributed by the element in the formula, divide it by the compound’s molar mass, and multiply by 100 percent: \(\%\text{ composition} = \frac{\text{mass of element in formula}}{\text{molar mass}} \times 100\%\).
23
Question
Why can either an empirical or molecular formula be used for percent composition?
Answer
Both formulas preserve the same relative ratio of the elements. Multiplying every subscript by the same factor multiplies both the element mass and total formula mass proportionally, leaving the percentage unchanged.
24
Question
How is an empirical formula determined from mass percentages?
Answer
Assume a 100-gram sample so each percentage becomes grams. Convert each element’s grams to moles, divide all mole values by the smallest value, and multiply by an integer when necessary to obtain whole-number subscripts.
25
Question
How is a molecular formula obtained after finding the empirical formula?
Answer
Calculate the empirical formula weight, divide the compound’s molar mass by that weight, and use the resulting whole-number multiplier on every empirical-formula subscript. For the stated carbohydrate example, \(\text{C}_3\text{H}_4\text{O}_3\times 3\) gives \(\text{C}_9\text{H}_{12}\text{O}_9\).
26
Question
What breaks when empirical-formula subscripts are rounded independently without checking the ratio?
Answer
Independent rounding can produce a molecular formula inconsistent with the simplest mole ratio. In the carbohydrate example, an apparent \(\text{C}_9\text{H}_{13}\text{O}_9\) is rejected in favor of \(\text{C}_9\text{H}_{12}\text{O}_9\), whose ratio reduces consistently to \(\text{C}_3\text{H}_4\text{O}_3\).
27
Question
How does a combination reaction differ from a decomposition reaction?
Answer
A combination reaction joins two or more reactants to form one product, represented as \(A+B\rightarrow C\). A decomposition reaction starts with one reactant and produces two or more products, represented as \(A\rightarrow B+C\).
28
Question
Why is combustion classified as a special reaction type?
Answer
Combustion involves a fuel, usually a hydrocarbon, reacting with an oxidant, normally oxygen. Hydrocarbon combustion commonly produces carbon dioxide and water, although other fuels can produce different products.
29
Question
How does a single-displacement reaction change the participating compounds?
Answer
An atom or ion in one compound is replaced by an atom or ion from another element. In the stated copper–silver example, copper displaces silver ions, producing copper nitrate and elemental silver.
30
Question
Why is the copper–silver reaction also classified as oxidation–reduction?
Answer
Silver ions gain an electron when they become elemental silver, so silver is reduced. Copper loses an electron when it joins the nitrate ion, so copper is oxidized.