Consider a gaseous binary compound with a molar mass of \(62.09 \mathrm{~g} / \mathrm{mol}\). When \(1.39 \mathrm{~g}\) of this compound is completely burned in excess oxygen, \(1.21 \mathrm{~g}\) of water is formed. Determine the formula of the compound. Assume water is the only product that contains hydrogen.

Short Answer

Expert verified
The compound has the formula XH_{13}, where X is the other element. This is determined by finding the moles of hydrogen and the other element in the compound, and then finding the simplest whole number ratio between them, which is approximately 6.5, leading to a formula of XH_{13}.

Step by step solution

01

Calculate moles of water

Since we know the mass of water formed and the molar mass of water, we can calculate the moles of water produced in the reaction: Moles of water = \(\frac{mass~of~water}{molar~mass~of~water}\) Moles of water = \(\frac{1.21~g}{18.015~g/mol}\) Moles of water = \(0.0672~mol\)
02

Calculate moles of hydrogen found in the compound

Each water molecule contains two hydrogen atoms, so the amount of hydrogen moles that were found in the compound can be calculated as follows: Moles of hydrogen = Moles of water × 2 Moles of hydrogen = \(0.0672~mol\) × 2 Moles of hydrogen = \(0.1344~mol\)
03

Calculate mass of hydrogen

Now that we know the moles of hydrogen, we can calculate the mass of hydrogen: Mass of hydrogen = Moles of hydrogen × molar mass of hydrogen Mass of hydrogen = \(0.1344~mol\) × \(1.008~g/mol\) Mass of hydrogen = \(0.1354~g\)
04

Calculate mass of the other element

Since the mass of the substance is conserved, we can find the mass of the other element by subtracting the mass of hydrogen from the total mass of compound: Mass of the other element = initial mass of compound - mass of hydrogen Mass of the other element = \(1.39~g\) - \(0.1354~g\) Mass of the other element = \(1.2546~g\)
05

Calculate moles of the other element

To determine the formula of the compound, we need to find the ratio between the elements in the compound. First, let's calculate the moles of the other element: Moles of other element = \(\frac{mass~of~other~element}{molar~mass~of~binary~compound}\) Now we know from the molar mass of the binary compound that \(62.09 \mathrm{~g/mol}\) can be expressed as the sum of the molar mass of hydrogen and the other element. Therefore, the molar mass of the other element is: Molar mass of other element = molar mass of binary compound - molar mass of hydrogen Molar mass of other element = \(62.09~g/mol\) - \(1.008~g/mol\) Molar mass of other element = \(61.082~g/mol\) Now, we can calculate the moles of the other element: Moles of other element = \(\frac{1.2546~g}{61.082~g/mol}\) Moles of other element = \(0.0205~mol\)
06

Determine the formula of the compound

Now that we know the moles of both hydrogen and the other element, we can determine the simplest whole number ratio between them: Ratio = \(\frac{moles ~of ~hydrogen}{moles ~of ~other ~element}\) Ratio = \(\frac{0.1344~mol}{0.0205~mol}\) Ratio ≈ \(6.56\) Since this ratio is close to 6.5, we can assume that the binary compound is composed of 13 moles of hydrogen for every 2 moles of the other element. Therefore, the formula of the compound is: XH_{13} where X is the symbol of the other element (which we do not need to identify specifically for this exercise).

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Key Concepts

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

Chemical Compound Analysis
Chemical compound analysis is a method used to determine the composition and formula of a substance by separating it into its individual elements. In our exercise, the analysis was performed through a combustion reaction where a gaseous binary compound was burned in excess oxygen, yielding water. From the mass of water obtained, one can infer the amount of hydrogen in the original compound.

The key steps in chemical compound analysis include identifying the products formed during a reaction, measuring the mass of these products accurately, and using these masses to calculate the moles of various elements involved. Fundamentally, it's a puzzle that requires accounting for all atoms present before and after a reaction to deduce the formula of the unknown compound. Understanding this concept allows one to unravel the composition of complex substances and determine their molecular formulas.
Molar Mass Calculation
Molar mass calculation is a fundamental concept in chemistry that involves finding the mass of one mole of a substance. The molar mass of an element is numerical equal to its atomic weight from the periodic table and is expressed in grams per mole (g/mol). In the provided exercise, we compute the molar mass of a binary compound and use it as a bridge between the mass of the compound in grams and the amount of substance in moles.

The ability to calculate molar mass is crucial as it assists in converting between the mass of a substance and the number of moles, allowing for quantitative analysis of chemical reactions. For example, in the solution steps, the molar masses of water and hydrogen were used to calculate the moles from the given masses. This forms the crux of stoichiometric calculations, as it allows chemists to scale reactions correctly and convert easily between mass and moles.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is based on the law of conservation of mass and the concept of moles. Stoichiometry enables us to predict the amounts of substances consumed and produced in a reaction.

In the case of our exercise, once we knew the moles of water produced, we could determine the moles of hydrogen in the original compound. This information, paired with the given molar mass of the compound and the law of conservation of mass, allowed us to find the moles and then the mass of the other element in the compound. Finally, we related the moles of hydrogen to the moles of the other element to deduce the simplest whole number ratio, leading us to the empirical formula. Stoichiometry is not only fundamental in chemistry but also in fields where chemical processes are involved, such as engineering, environmental science, and pharmacology.

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Most popular questions from this chapter

Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of \(47.6 \mathrm{mg}\) cumene produces some \(\mathrm{CO}_{2}\) and \(42.8 \mathrm{mg}\) water. The molar mass of cumene is between 115 and \(125 \mathrm{~g} / \mathrm{mol}\). Determine the empirical and molecular formulas.

Consider the following unbalanced reaction: $$ \mathrm{P}_{4}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{3}(g) $$ What mass of \(\mathrm{F}_{2}\) is needed to produce \(120 . \mathrm{g}\) of \(\mathrm{PF}_{3}\) if the reaction has a \(78.1 \%\) yield?

Gallium arsenide, GaAs, has gained widespread use in semiconductor devices that convert light and electrical signals in fiber-optic communications systems. Gallium consists of \(60 . \%^{69} \mathrm{Ga}\) and \(40 . \%^{71} \mathrm{Ga}\). Arsenic has only one naturally occurring isotope, \({ }^{75}\) As. Gallium arsenide is a polymeric material, but its mass spectrum shows fragments with the formulas GaAs and \(\mathrm{Ga}_{2} \mathrm{As}_{2}\). What would the distribution of peaks look like for these two fragments?

The space shuttle environmental control system handles excess \(\mathrm{CO}_{2}\) (which the astronauts breathe out; it is \(4.0 \%\) by mass of exhaled air) by reacting it with lithium hydroxide, LiOH. pellets to form lithium carbonate, \(\mathrm{Li}_{2} \mathrm{CO}_{3}\), and water. If there are seven astronauts on board the shuttle, and each exhales 20\. L of air per minute, how long could clean air be generated if there were \(25,000 \mathrm{~g}\) of \(\mathrm{LiOH}\) pellets available for each shuttle mission? Assume the density of air is \(0.0010 \mathrm{~g} / \mathrm{mL}\).

The aspirin substitute, acetaminophen \(\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{~N}\right)\), is produced by the following three-step synthesis: I. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{~N}(s)+3 \mathrm{H}_{2}(g)+\mathrm{HCl}(a q) \longrightarrow\) $$ \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{ONCl}(s)+2 \mathrm{H}_{2} \mathrm{O}(t) $$ II. \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+\mathrm{NaOH}(a q) \longrightarrow\) $$ \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{H}_{2} \mathrm{O}(t)+\mathrm{NaCl}(a q) $$ III. \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}(l) \longrightarrow\) $$ \mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{~N}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(l) $$ The first two reactions have percent yields of \(87 \%\) and \(98 \%\) by mass, respectively. The overall reaction yields 3 moles of acetaminophen product for every 4 moles of \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{O}_{3} \mathrm{~N}\) reacted. a. What is the percent yield by mass for the overall process? b. What is the percent yield by mass of Step III?

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