In the formation of carbon monoxide, CO, it is found that \(2.445 \mathrm{~g}\) of carbon combine with \(3.257 \mathrm{~g}\) of oxygen. What is the atomic mass of oxygen if the atomic mass of carbon is 12.01 amu?

Short Answer

Expert verified
To find the atomic mass of oxygen, you need to divide the mass of oxygen by the determined number of moles from carbon, following the 1:1 ratio stoichiometry of the reaction between carbon and oxygen in the formation of carbon monoxide.

Step by step solution

01

Identify the given quantities

The mass of carbon involved in the reaction is \(2.445 \mathrm{~g}\) and the atomic mass of carbon is \(12.01 \mathrm{~amu}\). The total mass of oxygen is \(3.257 \mathrm{~g}\). We need to find the atomic mass of oxygen.
02

Calculate the number of moles of Carbon

The number of moles of carbon can be calculated using the formula: \[\text{Number of moles} = \frac {\text{given mass}} {\text{molar mass}}\]. Plug in the known values, \[\text{Number of moles of Carbon} = \frac {2.445 \mathrm{~g}} {12.01 \mathrm{~g/mol}}\]
03

Apply stoichiometry of the reaction

In the reaction, one atom of Carbon reacts with one atom of oxygen to form one molecule of carbon monoxide. This implies that number of moles of Carbon will be equal to the number of moles of Oxygen since the reaction is in a 1:1 ratio.
04

Calculate the atomic mass of oxygen

Now, we know that the number of moles of oxygen equals the number of moles of carbon. Therefore, we can find the atomic mass of oxygen using the formula: \[ \text{Molar mass} = \frac {\text{given mass}} {\text{Number of moles}} \]. Substitute the known values and solve for oxygen's molar mass or atomic mass.

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Key Concepts

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

Stoichiometry
Stoichiometry is the section of chemistry that involves the calculation of reactants and products in chemical reactions. It's a kind of 'recipe' that tells you how much of each substance is needed or produced. The stoichiometry of a reaction is derived from the balanced chemical equation, showcasing the relationship between the amounts of reactants and products.

For instance, in the given exercise, the formation of carbon monoxide (CO) shows a 1:1 stoichiometry between carbon (C) and oxygen (O); meaning for every atom of carbon, you need one atom of oxygen to form a molecule of CO. When solving stoichiometric problems, the first step is often to balance the chemical equation and then use conversion factors, such as the mole ratio, to solve for unknown quantities.
Mole Concept
The mole concept is a fundamental aspect of chemistry that enables chemists to count atoms, molecules, and ions by weighing. A mole is defined as the amount of substance that contains as many entities (atoms, molecules, or ions) as there are atoms in 12 grams of pure carbon-12. This number is known as Avogadro's number and is approximately equal to \(6.022 \times 10^{23}\).

In the context of the given exercise, this concept allows us to convert between the mass of a substance and the number of moles present. By knowing the molar mass of carbon (12.01 amu or g/mol in this case), we can calculate the number of moles of carbon involved in the reaction, which in turn will tell us how many moles of oxygen are required considering the 1:1 ratio.
Atomic Mass Unit (amu)
The atomic mass unit (amu) is a standard unit of mass that quantifies mass on an atomic or molecular scale. It is defined as one-twelfth of the mass of a carbon-12 atom, making it a relative scale to compare the masses of different atoms and molecules. One amu is equivalent to \(1.660539 \times 10^{-24}\) grams.

To answer the exercise, we use the known atomic mass of carbon (12.01 amu) as a reference to find the atomic mass of oxygen. In real-world applications, the atomic masses of elements are used to calculate the molar masses of compounds, and these values are critical when performing stoichiometric calculations.
Chemical Reaction
A chemical reaction involves the transformation of one or more substances into one or more new substances. Reactions are governed by the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction. Thus, the total mass of the reactants equals the total mass of the products.

In the exercise, we explore the reaction in which carbon (C) reacts with oxygen (O) to form carbon monoxide (CO). We use the stoichiometry of the reaction and the concept of moles to deduce the atomic mass of oxygen given the mass of each reactant. Understanding chemical reactions is not only crucial for solving stoichiometric problems but also for predicting the outcomes of reactions and for the synthesis of new materials.

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Most popular questions from this chapter

Balance the following equations using the method outlined in Section 3.7 . (a) \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}\) (b) \(\mathrm{KNO}_{3} \longrightarrow \mathrm{KNO}_{2}+\mathrm{O}_{2}\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\) (e) \(\mathrm{NaHCO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) (f) \(\mathrm{P}_{4} \mathrm{O}_{10}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}\) (g) \(\mathrm{HCl}+\mathrm{CaCO}_{3} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) (h) \(\mathrm{Al}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2}\) (i) \(\mathrm{CO}_{2}+\mathrm{KOH} \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}\) (j) \(\mathrm{CH}_{4}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (k) \(\mathrm{Be}_{2} \mathrm{C}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Be}(\mathrm{OH})_{2}+\mathrm{CH}_{4}\) (l) \(\mathrm{Cu}+\mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\) \((\mathrm{m}) \mathrm{S}+\mathrm{HNO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (n) \(\mathrm{NH}_{3}+\mathrm{CuO} \longrightarrow \mathrm{Cu}+\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Why is the actual yield of a reaction almost always smaller than the theoretical yield?

A sample containing \(\mathrm{NaCl}, \mathrm{Na}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaNO}_{3}\) gives the following elemental analysis: Na: 32.08 percent; O: 36.01 percent; Cl: 19.51 percent. Calculate the mass percent of each compound in the sample.

A certain sample of coal contains 1.6 percent sulfur by mass. When the coal is burned, the sulfur is converted to sulfur dioxide. To prevent air pollution, this sulfur dioxide is treated with calcium oxide \((\mathrm{CaO})\) to form calcium sulfite \(\left(\mathrm{CaSO}_{3}\right)\). Calculate the daily mass (in kilograms) of CaO needed by a power plant that uses \(6.60 \times 10^{6} \mathrm{~kg}\) of coal per day.

Write balanced equations for the following reactions described in words. (a) Pentane burns in oxygen to form carbon dioxide and water. (b) Sodium bicarbonate reacts with hydrochloric acid to form carbon dioxide, sodium chloride, and water. (c) When heated in an atmosphere of nitrogen, lithium forms lithium nitride. (d) Phosphorus trichloride reacts with water to form phosphorus acid and hydrogen chloride. (e) Copper(II) oxide heated with ammonia will form copper, nitrogen gas, and water.

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