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In methane, one part hydrogen combines with three parts carbon by mass. If a sample of a compound containing only carbon and hydrogen contains \(32.0 \mathrm{~g}\) of carbon and \(8.0 \mathrm{~g}\) of hydrogen, could the sample be methane? If the sample is not methane, show that the law of multiple proportions is followed for methane and this other substance.

Short Answer

Expert verified
No, the sample isn't methane; the mass ratios demonstrate the law of multiple proportions.

Step by step solution

01

Determine the Mass Ratio in Methane

Methane has a chemical formula of \(CH_4\). This means it consists of one carbon atom and four hydrogen atoms. The atomic mass of carbon is approximately \(12 \text{ u}\) and for hydrogen it is \(1 \text{ u}\). Therefore, the mass ratio in methane is \(\frac{4 \times 1}{12} = \frac{4}{12} = \frac{1}{3}\). This means for every 1 part of hydrogen by mass, there are 3 parts of carbon by mass.
02

Calculate the Mass Ratio in the Given Sample

The sample contains \(32.0 \text{ g}\) of carbon and \(8.0 \text{ g}\) of hydrogen. So, the mass ratio of hydrogen to carbon is \(\frac{8.0}{32.0} = \frac{1}{4}\). This indicates for every 1 part of hydrogen, there are 4 parts of carbon.
03

Compare Mass Ratios to Determine if the Sample is Methane

In methane, the mass ratio of hydrogen to carbon is \(\frac{1}{3}\). The mass ratio in the sample is \(\frac{1}{4}\). Since \(\frac{1}{3} eq \frac{1}{4}\), the sample cannot be methane, as the proportions of carbon and hydrogen do not match those in methane.
04

Demonstrate the Law of Multiple Proportions

The law of multiple proportions states that if two elements form more than one compound, then the ratios of the masses of the second element that combine with a fixed mass of the first element are ratios of small whole numbers. Compare the two ratios: \(1:3\) (methane) and \(1:4\) (the sample). The ratio of these two ratios \(\frac{3}{4}\) is indeed a small whole number ratio when simplified appropriately.

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Key Concepts

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

Methane
Methane is a simple and significant compound in the field of chemistry. It's often considered the simplest hydrocarbon, consisting of one carbon atom bonded to four hydrogen atoms, represented by the chemical formula \(CH_4\).
Methane is abundantly found in nature and is a primary component of natural gas. It is colorless and odorless but plays a crucial role as a greenhouse gas in Earth's atmosphere.
Understanding methane is fundamental in chemistry because of its structure and composition.
  • Methane showcases a tetrahedral geometry where the carbon atom is at the center, bonded to four hydrogen atoms at the corners of a tetrahedron.
  • This structure is influenced by the electron configuration of carbon, allowing it to form four covalent bonds.
  • In terms of mass, the atomic weight of carbon is approximately \(12\, \text{u}\), and that of hydrogen is \(1\, \text{u}\).
Methane serves as a baseline for understanding other larger and more complex hydrocarbons, making it essential to grasp its chemical properties and reactions.
Chemical Formulas
Chemical formulas are fundamental representations of chemical compounds in which the symbols of the elements are used along with numerical subscripts to denote the number of atoms of each element present in the compound.
They are key to understanding the composition and structure of molecules. For methane, its chemical formula, \(CH_4\), communicates several critical pieces of information.
  • The formula indicates there is one carbon atom and four hydrogen atoms in a molecule of methane.
  • It also helps determine the molecular weight by summing the atomic masses: approx. \(12 \times 1\) for carbon plus \(1 \times 4\) for hydrogen, resulting in \(16 \text{ u}\).
  • Chemical formulas allow chemists to predict properties and reactivity, providing a blueprint for the compound's behavior in reactions.
In essence, chemical formulas are the chemical "shorthand" used to describe a substance's composition succinctly and consistently.
Mass Ratios
In chemistry, mass ratios describe the relative weights of elements in a compound and are crucial for verifying and predicting chemical reactions. Methane demonstrates an interesting mass ratio between its constituent elements.
The mass ratio is calculated by comparing the masses of the atoms involved in the compound. For methane, each molecule contains four hydrogen atoms (mass equals \(4 \times 1 = 4 \)) and one carbon atom (mass equals \(12 \)), giving the mass ratio of hydrogen to carbon as \(\frac{4}{12} = \frac{1}{3}\).
Mass ratios are important for validating the identity of a compound.
  • They allow chemists to determine whether a given sample corresponds to the expected chemical composition of a known compound.
  • For example, a sample with a mass ratio differing from that of methane (\(\frac{1}{4}\) instead of \(\frac{1}{3}\)) indicates it isn't methane but a different compound.
Furthermore, an understanding of mass ratios supports the application of the law of multiple proportions, explaining how compounds can differ in composition even when consisting of the same elements.

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Most popular questions from this chapter

Tetrahydrofuran, an organic substance used as a solvent in many pharmaceutical processes, has the formula \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} .\) In tetrahydrofuran, the four \(\mathrm{C}\) atoms are bonded in a row, each \(\mathrm{C}\) atom is bonded to two \(\mathrm{H}\) atoms, each \(\mathrm{H}\) atom is bonded to one \(\mathrm{C}\) atom, and the \(\mathrm{O}\) atom is bonded to two \(\mathrm{C}\) atoms. Write a structural formula for tetrahydrofuran.

The subscript giving the atomic number of an atom is often left off when writing an isotope symbol. For example, \({ }_{6}^{13} \mathrm{C}\) is often written simply as \({ }^{13} \mathrm{C}\). Why is this allowed?

Classify the following elements as metals, nonmetals, or semimetals: (a) Ar (b) \(\mathrm{Sb}\) (c) Mo (d) \(\mathrm{Cl}\) (e) \(\mathrm{N}\) (f) \(\mathrm{Mg}\)

Ibuprofen was initially synthesized by a process developed by Boots Co. in the \(1960 \mathrm{~s}\). Six reaction steps were utilized, and the percent atom economy was \(40 \%\). For every one mole of ibuprofen produced, 1 mole of Na, 23 moles of \(\mathrm{H}, 1\) mole of \(\mathrm{N}, 7\) moles of \(\mathrm{C}\), 8 moles of \(\mathrm{O}\), and 1 mole of \(\mathrm{Cl}\) were unused and considered as waste. (a) Calculate the mass (g) of each element wasted for every one mole of ibuprofen produced. (b) Calculate the total mass ( \(\mathrm{g}\) ) wasted for every one mole of ibuprofen produced. (c) Yearly production of ibuprofen is approximately 30 million lbs, which is equivalent to \(6.6 \times 10^{7}\) moles. Calculate the total mass (kg) of matter wasted in the annual production ibuprofen.

Correct the error in each of the following statements: (a) The formula of ammonia is \(\mathrm{NH} 3\). (b) Molecules of potassium chloride have the formula \(\mathrm{KCl}\). (c) \(\mathrm{Cl}^{-}\) is a cation. (d) \(\mathrm{CH}_{4}\) is a polyatomic ion.

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