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Write the empirical formula corresponding to each of the following molecular formulas: \((\mathbf{a}) \mathrm{Al}_{2} \mathrm{Br}_{6},(\mathbf{b}) \mathrm{C}_{8} \mathrm{H}_{10},\) (c) \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\) (d) \(\mathrm{P}_{4} \mathrm{O}_{10}\) (e) \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2},\) (f) \(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6}\).

Short Answer

Expert verified
AlBr₃, C₄H₅, C₂H₄O, P₂O₅, C₃H₂Cl, BNH₂.

Step by step solution

01

Define Empirical Formula

The empirical formula represents the simplest whole-number ratio of the elements in a compound. It may be different from the molecular formula which provides the exact number of atoms of each element in the compound.
02

Simplify Al₂Br₆ to Empirical Formula

For \(\mathrm{Al}_2\mathrm{Br}_6\), divide the subscripts by the greatest common divisor (GCD), which is 2. This gives \(\mathrm{AlBr}_3\).
03

Simplify C₈H₁₀ to Empirical Formula

For \(\mathrm{C}_8\mathrm{H}_{10}\), divide both subscripts by the GCD, which is 2. This results in \(\mathrm{C}_4\mathrm{H}_5\).
04

Simplify C₄H₈O₂ to Empirical Formula

For \(\mathrm{C}_4\mathrm{H}_8\mathrm{O}_2\), the GCD of the subscripts is 2. Divide each by 2 to get \(\mathrm{C}_2\mathrm{H}_4\mathrm{O}\).
05

Simplify P₄O₁₀ to Empirical Formula

In \(\mathrm{P}_4\mathrm{O}_{10}\), the GCD is 2. Divide the subscripts by 2 to get \(\mathrm{P}_2\mathrm{O}_5\).
06

Simplify C₆H₄Cl₂ to Empirical Formula

For \(\mathrm{C}_6\mathrm{H}_4\mathrm{Cl}_2\), the GCD is 2, giving \(\mathrm{C}_3\mathrm{H}_2\mathrm{Cl}\) as the empirical formula.
07

Simplify B₃N₃H₆ to Empirical Formula

The molecular formula \(\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6\) can be divided by the GCD of 3. This results in the empirical formula \(\mathrm{BNH}_2\).

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Key Concepts

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

Molecular Formula
A molecular formula is an essential representation of a chemical compound in chemistry. It specifies the exact number of atoms of each element that are present in a molecule of the compound. Unlike the empirical formula, which provides only the simplest whole-number ratio of these atoms, the molecular formula provides a more comprehensive picture.
Imagine working with water—the molecular formula is \(\mathrm{H}_{2}\mathrm{O}\), indicating each molecule of water contains exactly 2 hydrogen atoms and 1 oxygen atom. This specificity helps chemists understand and predict the behaviors and reactions of the compound at a molecular level.
  • The molecular formula tells you how many atoms are in a single molecule.
  • It is especially important when considering the molecular weight and when calculating moles during reactions.
  • In some cases, the molecular formula and the empirical formula might be identical, but often they are not.
Understanding the difference between molecular and empirical formulas is crucial for accurately solving chemistry problems.
Chemical Compound
A chemical compound is a substance formed when two or more elements chemically bond together. These compounds possess unique properties that differ from their constituent elements.
For example, water is a chemical compound made from hydrogen and oxygen, yet it exhibits characteristics distinct from either of these gases in their individual states.
  • Compounds can be molecular (covalent) or ionic. Molecular compounds are formed via sharing electrons, while ionic compounds arise from electrical attractions between ions.
  • The properties of a compound can vastly differ from the properties of the elements that make up the compound.
  • Compounds are represented by chemical formulas which denote their elemental composition.
In chemistry, understanding chemical compounds is vital for grasping how different substances interact and transform under various conditions.
Stoichiometry
Stoichiometry is a branch of chemistry that focuses on measuring and quantifying substances involved in chemical reactions. It is utilized to calculate amounts of reactants and products and is rooted in the mole concept.
The basic idea is to start with the balanced chemical equation and use ratios to determine the amounts for any substance involved.
  • Stoichiometry uses coefficients from balanced equations as conversion factors to move between moles of reactants and products.
  • It allows chemists to calculate yields of reactions as well as to determine limiting reactants, which are dead ends in chemical reactions.
  • Balancing equations is the foundation for stoichiometry, ensuring reactions respect the law of conservation of mass.
By mastering stoichiometry, students can predict and quantify the outcomes of chemical reactions accurately.
Greatest Common Divisor
The greatest common divisor (GCD) is a key mathematical concept used in chemistry when simplifying molecular formulas to empirical formulas. The GCD of a set of numbers is the largest positive integer that evenly divides each of the numbers.
In chemical contexts, it helps reduce the subscripts in a molecular formula to their simplest form without altering the compound's identity.
  • To find the empirical formula, divide all subscripts in the molecular formula by their GCD.
  • This process does not affect the chemical nature of the compound but represents it in its simplest terms.
  • Applying the GCD ensures that the empirical formula adheres to the basic principle of depicting the smallest ratio of elements.
Grasping how to apply the GCD effectively aids chemists and students in simplifying chemical structures consistent with stoichiometric principles.

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Most popular questions from this chapter

How many hydrogen atoms are in each of the following: (a) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (b) \(\mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}\right)_{2}\) (c) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} ?\)

Write the molecular and structural formulas for the compounds represented by the following models: $$ \begin{array}{l|c|c|c|c} \hline \text { Symbol } & { }^{58} \mathrm{Fe}^{2+} & & & \\ \hline \text { Protons } & & 50 & & 40 \\ \text { Neutrons } & & 68 & 78 & 50 \\ \text { Electrons } & & & 54 & 38 \\ \text { Net charge } & & 4+ & 2- & \\ \hline \end{array} $$

The explosion of an atomic bomb releases many radioactive isotopes, including strontium-90. Considering the location of strontium in the periodic table, suggest a reason for the fact that this isotope is particularly dangerous for human health.

There are two different isotopes of bromine atoms. Under normal conditions, elemental bromine consists of \(\mathrm{Br}_{2}\) molecules, and the mass of a \(\mathrm{Br}_{2}\) molecule is the sum of the masses of the two atoms in the molecule. The mass spectrum of \(\mathrm{Br}_{2}\) consists of three peaks: $$ \begin{array}{lc} \hline \text { Mass (u) } & \text { Relative Size } \\ \hline 157.836 & 0.2569 \\ 159.834 & 0.4999 \\ 161.832 & 0.2431 \\ \hline \end{array} $$ (a) What is the origin of each peak (of what isotopes does each consist)? (b) What is the mass of each isotope? (c) Determine the average molecular mass of a \(\mathrm{Br}_{2}\) molecule. (d) Determine the average atomic mass of a bromine atom. (e) Calculate the abundances of the two isotopes.

A cube of gold that is \(1.00 \mathrm{~cm}\) on a side has a mass of 19.3 g. A single gold atom has a mass of \(197.0 \mathrm{u} .(\mathbf{a})\) How many gold atoms are in the cube? (b) From the information given, estimate the diameter in \(\AA\) of a single gold atom. (c) What assumptions did you make in arriving at your answer for part (b)?

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