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Chapter 3: Problem 81

Write empirical formulas for the following compounds. (a) \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) (d) \(\mathrm{As}_{2} \mathrm{O}_{6}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (e) \(\mathrm{H}_{2} \mathrm{O}_{2}\) (c) \(\mathrm{NH}_{3}\)

Short Answer

Expert verified
The empirical formulas are (a) \(\mathrm{SCl}\), (d) \(\mathrm{AsO}_{3}\), (b) \(\mathrm{CH}_{2}O\), (e) \(\mathrm{H}_{2}O\) (which represents hydrogen peroxide in this context), and (c) \(\mathrm{NH}_{3}\).

Step by step solution

01

Understanding Empirical Formulas

An empirical formula represents the simplest whole-number ratio of atoms of each element present in a compound. It may or may not be the same as the actual molecular formula of a compound, which shows the exact number of atoms of each element in a molecule.
02

Determining the Empirical Formula for \(\mathrm{S}_{2} \mathrm{Cl}_{2}\)

The molecular formula \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) is already the simplest integer ratio (1:1 ratio of sulfur to chlorine atoms), hence the empirical formula is the same as the molecular formula: \(\mathrm{SCl}\).
03

Determining the Empirical Formula for \(\mathrm{As}_{2} \mathrm{O}_{6}\)

The molecular formula has a 2:6 ratio of arsenic to oxygen atoms, which can be simplified to a 1:3 ratio. Therefore, the empirical formula is \(\mathrm{AsO}_{3}\).
04

Determining the Empirical Formula for \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

The molecular formula shows a 6:12:6 ratio of carbon, hydrogen, and oxygen atoms which simplifies to a 1:2:1 ratio. The empirical formula is therefore \(\mathrm{CH}_{2}O\).
05

Determining the Empirical Formula for \(\mathrm{H}_{2} \mathrm{O}_{2}\)

The molecular formula \(\mathrm{H}_{2} \mathrm{O}_{2}\) is the simplest whole-number ratio (1:1 ratio of hydrogen to oxygen atoms), so the empirical formula is the same as the molecular formula: \(\mathrm{H}_{2}O\). Note that \(\mathrm{H}_{2}O\) is also the empirical and molecular formula for water, but in this context, it is used to indicate hydrogen peroxide.
06

Determining the Empirical Formula for \(\mathrm{NH}_{3}\)

The molecular formula \(\mathrm{NH}_{3}\) is already in the simplest integer ratio (1:3 ratio of nitrogen to hydrogen atoms), meaning the empirical formula is the same as the molecular formula: \(\mathrm{NH}_{3}\).

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Key Concepts

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

Chemical Composition
Chemical composition is a fundamental concept in chemistry that describes the arrangement, type, and ratio of atoms in molecules of chemical substances. It is crucial information as it helps to understand the properties, behaviors, and reactivity of the substance.

The chemical composition is typically represented by a chemical formula, which uses symbols for the elements present and subscripts to indicate the number of atoms of each element in the compound. For example, in the molecular formula \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\), carbon (C), hydrogen (H), and oxygen (O) are present in a 6:12:6 ratio.
Molecular Formula
A molecular formula indicates the exact number of atoms of each element found in a single molecule of a compound. Unlike empirical formulas, which show the simplest whole-number ratio, the molecular formula highlights the actual composition of a molecule.

For instance, \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\) is the molecular formula for glucose, specifying that each molecule contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. To understand the relationship between molecular and empirical formulas, it is vital to note that the empirical formula of glucose is \(\mathrm{CH}_2\mathrm{O}\) since the 6:12:6 ratio can be reduced to 1:2:1.
Stoichiometry
Stoichiometry is the section of chemistry that deals with the calculation of the quantities of reactants and products in chemical reactions. It is based on the conservation of mass and the concept of the mole, a unit that measures the amount of substance.

By using the empirical formula, stoichiometric calculations can determine the amount of each element needed to form a compound. In the exercise, understanding stoichiometry enables us to simplify the number ratios of atoms, like from \(\mathrm{As}_2\mathrm{O}_6\) to its empirical formula \(\mathrm{AsO}_3\) by simplifying the ratio from 2:6 to 1:3.
Chemical Nomenclature
Chemical nomenclature is a systematic way of naming chemical substances. Each compound's name is unique and standard across the scientific community, making communication more precise and avoiding misunderstandings.

For example, \(\mathrm{H}_2\mathrm{O}\) is commonly known as water, while \(\mathrm{H}_2\mathrm{O}_2\) is called hydrogen peroxide. Despite having similar empirical formulas, these compounds have distinct names to reflect their different chemical structures and properties. Using appropriate nomenclature is crucial for identifying the correct substance, especially when the empirical formula alone might not provide enough information.

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Most popular questions from this chapter

How many atoms of carbon are combined with 4.13 moles of hydrogen atoms in a sample of the compound propane, \(\mathrm{C}_{3} \mathrm{H}_{8} ?\) (Propane is used as the fuel in gas barbecues.)

Lead compounds are often highly colored and are toxic to mold, mildew, and bacteria, properties that in the past were useful for paints used before 1960 . Today we know lead is very hazardous and it is not used in paint; however, old paint is still a problem. If a certain lead-based paint contains \(14.5 \% \mathrm{PbCr}_{2} \mathrm{O}_{7}\) and \(73 \%\) of the paint evaporates as it dries, what mass of lead will be in a paint chip that weighs \(0.15 \mathrm{~g}\) ?

The atomic mass of aluminum is 26.98 . What specific conversion factors does this value make available for relating a mass of aluminum (in grams) and a quantity of aluminum given in moles?

Which contains more molecules: \(2.5 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) or \(2.5 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) ? Which contains more atoms? Which weighs more?

The combustion of methyl alcohol in an abundant excess of oxygen follows the equation $$ 2 \mathrm{CH}_{3} \mathrm{OH}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O} $$ When \(6.40 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) was mixed with \(10.2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and ignited, \(6.12 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) was obtained. What was the percentage yield of \(\mathrm{CO}_{2} ?\)
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