Does \(1 \mathrm{~g}\) of hydrogen molecules contain as many \(\mathrm{H}\) atoms as \(1 \mathrm{~g}\) of hydrogen atoms?

Short Answer

Expert verified
Yes, 1 gram of Hydrogen molecules contains the same number of Hydrogen atoms as 1 gram of Hydrogen atoms, which is 6.022 × 10^23 atoms.

Step by step solution

01

Calculate Number of Moles

First, calculate the number of moles for 1 gram of Hydrogen atoms and Hydrogen molecules. A mole is calculated as mass divided by molecular weight. So, the number of moles of Hydrogen atoms would be 1/1 = 1 mole, and the number of moles of Hydrogen molecules (H2) would be 1/2 = 0.5 moles.
02

Calculate Number of H atoms using Avogadro's Number

Now, calculate the number of Hydrogen atoms in 1 gram of Hydrogen and Hydrogen molecules using Avogadro's number, which is 6.022 × 10^23 entities per mole. Therefore, in 1 gram of Hydrogen atoms (H), there would be 1 mole x 6.022 × 10^23 = 6.022 × 10^23 H atoms.
03

Calculate Number of H atoms in Hydrogen Molecules

In case of Hydrogen molecules (H2), keep in mind that each molecule consists of 2 Hydrogen atoms. So, for 0.5 moles of Hydrogen molecules, there are 0.5 x 6.022 × 10^23 = 3.011 × 10^23 molecules, but since each molecule has 2 Hydrogen atoms, the total number would be 2 x 3.011 × 10^23 = 6.022 × 10^23 H atoms.
04

Compare the number of H atoms

Finally, compare the number of H atoms in both situations. As calculated above, both have 6.022 × 10^23 H atoms. So, 1 gram of Hydrogen molecules contains the same number of H atoms as 1 gram of Hydrogen atoms.

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Key Concepts

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

Mole Concept
The mole concept is a fundamental building block of chemistry, bridging the gap between the microscopic world of atoms and the macroscopic world of grams and liters. A 'mole' is a unit of measurement used in chemistry to express amounts of a chemical substance. It allows chemists to count particles such as atoms, molecules, or ions by weighing them.

According to the definition, one mole is equivalent to Avogadro's Number, which is approximately 6.022 x 1023 entities. It's similar to a dozen, except instead of 12, it's 6.022 x 1023. Think of it as a chemist's dozen. This concept is crucial when we discuss the number of atoms in a given mass of an element or a compound, as seen in the exercise where we compare hydrogen in atomic and molecular forms.
Molecular Weight
Molecular weight, also known as molecular mass, is the total mass of all the atoms in a molecule expressed in atomic mass units (amu). Each atom has a standard atomic weight—for example, a hydrogen atom has a weight close to 1 amu. To find the molecular weight of a molecule, we add up the weights of each constituent atom.

Determining Molecular Weight

In our example with hydrogen, the molecular weight of an individual hydrogen atom (H) is approximately 1 amu. However, for a hydrogen molecule (H2), which consists of two hydrogen atoms, the molecular weight would be roughly 2 amu. This concept is vital in the calculation of moles since the molecular weight determines how much a 'mole' of a substance weighs.
Stoichiometry
Stoichiometry is the field of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is based on the law of conservation of mass where the total mass of the reactants equals the total mass of the products. Stoichiometry involves using the mole concept, molecular weights, and balanced chemical equations to predict the amounts of substances consumed and produced in a reaction.

Using stoichiometry, we can calculate how much product will result from a given amount of reactant or how much reactant is needed to produce a desired amount of product. This process is key in conducting chemical calculations efficiently, especially when it comes to understanding the quantitative aspects of chemical formulas and reactions.
Chemical Calculations
Chemical calculations encompass a variety of mathematical approaches to solving problems in chemistry. These calculations typically involve determining the quantity of reactants or products, concentrations, volumes, and other chemical properties. Tools used in chemical calculations include the mole concept, molecular weights, and stoichiometry, as well as specific principles like gas laws, and solution chemistry.

Applying Chemical Calculations

In our exercise, we used chemical calculations to determine that 1 gram of hydrogen atoms and 1 gram of hydrogen molecules contain the same number of hydrogen atoms. This is done by first figuring out the number of moles, and then using Avogadro's Number to translate those moles into a count of atoms. Through these calculations, students can understand how to quantitatively compare different forms of the same element—in this case, atomic versus molecular hydrogen.

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Most popular questions from this chapter

(a) For molecules having small molecular masses, mass spectrometry can be used to identify their formulas. To illustrate this point, identify the molecule that most likely accounts for the observation of a peak in a mass spectrum at 16 amu, 17 amu, 18 amu, and 64 amu. (b) Note that there are (among others) two likely molecules that would give rise to a peak at 44 amu, namely, \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{CO}_{2} .\) In such cases, a chemist might try to look for other peaks generated when some of the molecules break apart in the spectrometer. For example, if a chemist sees a peak at 44 amu and also one at 15 amu, which molecule is producing the 44 -amu peak? Why? (c) Using the following precise atomic masses \(-\mathrm{H}\) ( 1.00797 amu), \({ }^{12} \mathrm{C}(12.00000 \mathrm{amu}),\) and \({ }^{16} \mathrm{O}(15.99491 \mathrm{amu})-\) how precisely must the masses of \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{CO}_{2}\), be measured to distinguish between them?

Balance the following equations using the method outlined in Section 3.7 : (a) \(\mathrm{C}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}\) (b) \(\mathrm{CO}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}\) (c) \(\mathrm{H}_{2}+\mathrm{Br}_{2} \longrightarrow \mathrm{HBr}\) (d) \(\mathrm{K}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{KOH}+\mathrm{H}_{2}\) (e) \(\mathrm{Mg}+\mathrm{O}_{2} \longrightarrow \mathrm{MgO}\) (f) \(\mathrm{O}_{3} \longrightarrow \mathrm{O}_{2}\) (g) \(\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (h) \(\mathrm{N}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{NH}_{3}\) (i) \(\mathrm{Zn}+\mathrm{AgCl} \longrightarrow \mathrm{ZnCl}_{2}+\mathrm{Ag}\) (j) \(\mathrm{S}_{8}+\mathrm{O}_{2} \longrightarrow \mathrm{SO}_{2}\) (k) \(\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (l) \(\mathrm{Cl}_{2}+\mathrm{NaI} \longrightarrow \mathrm{NaCl}+\mathrm{I}_{2}\) \((\mathrm{m}) \mathrm{KOH}+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (n) \(\mathrm{CH}_{4}+\mathrm{Br}_{2} \longrightarrow \mathrm{CBr}_{4}+\mathrm{HBr}\)

Heating \(2.40 \mathrm{~g}\) of the oxide of metal \(\mathrm{X}\) (molar mass of \(\mathrm{X}=55.9 \mathrm{~g} / \mathrm{mol}\) ) in carbon monoxide (CO) yields the pure metal and carbon dioxide. The mass of the metal product is \(1.68 \mathrm{~g}\). From the data given, show that the simplest formula of the oxide is \(\mathrm{X}_{2} \mathrm{O}_{3}\) and write a balanced equation for the reaction.

Ammonia is a principal nitrogen fertilizer. It is prepared by the reaction between hydrogen and nitrogen.$$3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$In a particular reaction, 6.0 moles of \(\mathrm{NH}_{3}\) were produced. How many moles of \(\mathrm{H}_{2}\) and how many moles of \(\mathrm{N}_{2}\) were reacted to produce this amount of \(\mathrm{NH}_{3} ?\)

Carbohydrates are compounds containing carbon, hydrogen, and oxygen in which the hydrogen to oxygen ratio is \(2: 1 .\) A certain carbohydrate contains 40.0 percent carbon by mass. Calculate the empirical and molecular formulas of the compound if the approximate molar mass is \(178 \mathrm{~g}\)

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