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Chapter 3: Problem 32

What is the difference between an empirical formula and a molecular formula? Can they ever be the same?

Short Answer

Expert verified
An empirical formula gives the simplest whole-number ratio of elements, while a molecular formula gives the exact number of atoms in a molecule. They can be the same if the simplest ratio and actual count are identical.

Step by step solution

01

Understanding Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of elements in the compound. It does not give information on the exact number of atoms, just the proportion. For example, the empirical formula for glucose is CH₂O, indicating that for every carbon atom, there are two hydrogen atoms and one oxygen atom.
02

Understanding Molecular Formula

The molecular formula of a compound indicates the exact number of each type of atom in a molecule. It provides the actual count and composition. For example, the molecular formula for glucose is C₆H₁₂O₆, showing that one molecule of glucose contains six carbon atoms, twelve hydrogen atoms, and six oxygen atoms.
03

Comparing Empirical and Molecular Formulas

The molecular formula is always a multiple of the empirical formula. Sometimes, the empirical formula and molecular formula can be the same if the compound's simplest ratio and actual atom count per molecule are identical, such as in the case of benzene (C₆H₆) where both the empirical and molecular formulas are the same.

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Key Concepts

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

Empirical Formula Definition
The empirical formula is a way to denote the basic elemental composition of a compound. It represents the simplest whole-number ratio of the different atoms present. The focus here is on the proportion, not the exact count of atoms.

For instance, consider glucose. Its empirical formula is CH₂O. This shows that for every carbon atom, there are two hydrogen atoms and one oxygen atom.

It's important to note:
  • Empirical formulas do not show the exact number of atoms in a molecule.
  • They are more about the relative proportions of the elements.
This simplification is particularly useful for understanding the basic makeup of a compound without needing the full molecular details.
Molecular Formula Definition
Unlike the empirical formula, the molecular formula gives the detailed number of each type of atom present in a single molecule of a compound. It gives a complete picture of the molecule's actual composition.

For example, glucose's molecular formula is C₆H₁₂O₆. This tells us precisely that one molecule of glucose consists of six carbon atoms, twelve hydrogen atoms, and six oxygen atoms.

Key points about molecular formulas include:
  • They provide the exact count of atoms in a molecule.
  • They are crucial for understanding the molecule's full structure and makeup.
  • Molecular formulas can be multiples of empirical formulas.
Difference Between Empirical and Molecular Formulas
The main difference between empirical and molecular formulas lies in the amount of detail they provide.

Empirical formulas are about the ratio, while molecular formulas provide the exact count of each atom in a compound.

Some essential points include:
  • The molecular formula is always a whole-number multiple of the empirical formula.
  • In some cases, they can be the same. This occurs when the ratio and the actual count match perfectly, such as in benzene (C₆H₆).
Understanding these differences helps in fields such as chemistry and biochemistry, where the detailed knowledge of molecular structures is crucial for further study and applications.

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Most popular questions from this chapter

When powdered zinc is heated with sulfur, a violent reaction occurs, and zinc sulfide forms: $$ \mathrm{Zn}(s)+\mathrm{S}_{8}(s) \longrightarrow \mathrm{ZnS}(s)[\text { unbalanced }] $$ Some of the reactants also combine with oxygen in air to form zinc oxide and sulfur dioxide. When \(83.2 \mathrm{~g}\) of Zn reacts with \(52.4 \mathrm{~g}\) of \(\mathrm{S}_{8}\), \(104.4 \mathrm{~g}\) of \(\mathrm{ZnS}\) forms. (a) What is the percent yield of \(\mathrm{ZnS}\) ? (b) If all the remaining reactants combine with oxygen, how many grams of each of the two oxides form?

Calculate the molar mass of each of the following: (a) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}\) (b) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{BrF}_{3}\)

The multistep smelting of ferric oxide to form elemental iron occurs at high temperatures in a blast furnace. In the first step, ferric oxide reacts with carbon monoxide to form \(\mathrm{Fe}_{3} \mathrm{O}_{4}\). This substance reacts with more carbon monoxide to form iron(II) oxide, which reacts with still more carbon monoxide to form molten iron. Carbon dioxide is also produced in each step. (a) Write an overall balanced equation for the iron- smelting process. (b) How many grams of carbon monoxide are required to form 45.0 metric tons of iron from ferric oxide?

High-temperature superconducting oxides hold great promise in the utility, transportation, and computer industries. (a) One superconductor is \(\mathrm{La}_{2-x} \mathrm{Sr}_{x} \mathrm{CuO}_{4} .\) Calculate the molar masses of this oxide when \(x=0, x=1,\) and \(x=0.163\). (b) Another common superconducting oxide is made by heating a mixture of barium carbonate, copper(II) oxide, and yttrium(III) oxide, followed by further heating in \(\mathrm{O}_{2}\) : \(4 \mathrm{BaCO}_{3}(s)+6 \mathrm{CuO}(s)+\mathrm{Y}_{2} \mathrm{O}_{3}(s) \longrightarrow\) $$ \begin{array}{c} 2 \mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{6.5}(s)+4 \mathrm{CO}_{2}(g) \\\ 2 \mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{6.5}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{7}(s) \end{array} $$ When equal masses of the three reactants are heated, which reactant is limiting? (c) After the product in part (b) is removed, what is the mass \(\%\) of each reactant in the remaining solid mixture?

Metal hydrides react with water to form hydrogen gas and the metal hydroxide. For example, $$ \operatorname{SrH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \operatorname{Sr}(\mathrm{OH})_{2}(s)+2 \mathrm{H}_{2}(g) $$ You wish to calculate the mass (g) of hydrogen gas that can be prepared from \(5.70 \mathrm{~g}\) of \(\mathrm{SrH}_{2}\) and \(4.75 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) (a) What amount (mol) of \(\mathrm{H}_{2}\) can be produced from the given mass of \(\mathrm{SrH}_{2} ?\) (b) What amount (mol) of \(\mathrm{H}_{2}\) can be produced from the given mass of \(\mathrm{H}_{2} \mathrm{O} ?\) (c) Which is the limiting reactant? (d) How many grams of \(\mathrm{H}_{2}\) can be produced?
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