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What is the mass, in kilograms, of an Avogadro's number of people, if the average mass of a person is 160 \(\mathrm{lb}\) ? How does this compare with the mass of Earth, \(5.98 \times 10^{24} \mathrm{kg}\) ?

Short Answer

Expert verified
The total mass of Avogadro's number of people, with each person weighing \(160 \times 0.453592\) kg, is \(160 \times 0.453592 \times 6.022 \times 10^{23}\) kg. Comparing this to the mass of Earth (\(5.98 \times 10^{24}\) kg), the ratio is \(\frac{160 \times 0.453592 \times 6.022 \times 10^{23}}{5.98 \times 10^{24}}\). This ratio is much smaller than 1, indicating that the total mass of Avogadro's number of people is significantly less than the mass of Earth.

Step by step solution

01

Convert the mass of a person from lb to kg

To convert the mass of a person from lb to kg, we can use the conversion factor: 1 lb = 0.453592 kg So, for a person with a mass of 160 lb, we have: \(mass_{kg} = mass_{lb} \times conversion factor\) \(mass_{kg} = 160 \times 0.453592\)
02

Calculate the total mass of Avogadro's number of people

Now that we have the mass of one person in kg, we can find the total mass of Avogadro's number of people by multiplying the mass of one person by Avogadro's number: \(total~mass = mass_{kg} \times Avogadro's~Number\) \(total~mass = 160 \times 0.453592 \times 6.022 \times 10^{23}\)
03

Compare the total mass with the mass of Earth

We are given the mass of Earth as \(5.98 \times 10^{24} \mathrm{kg}\). Now we can compare the total mass of Avogadro's number of people with the mass of Earth: \(ratio = \frac{total~mass}{mass~of~Earth}\) \(ratio = \frac{160 \times 0.453592 \times 6.022 \times 10^{23}}{5.98 \times 10^{24}}\) After calculating the ratio, we can comment on how the total mass compares with the mass of Earth.

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Key Concepts

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

Molar Mass Calculation
Understanding molar mass calculation is essential for students embarking on chemistry-related subjects. Molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). Avogadro's number, which is approximately 6.022 x 1023, indicates the number of entities (atoms, molecules, ions, or electrons) in one mole of any substance.

Let's consider a real-world scenario to illustrate molar mass calculation: the average mass of a human is not commonly a part of standard chemical calculations. However, if we treat the average mass as a 'molar mass' of a sort, we essentially apply chemistry principles to a physics problem. Each 'entity' (in this case, a person) has a mass, and when multiplied by Avogadro's number, would give us the hypothetical 'mass of a mole' of people. This unique approach helps to deepen the students' understanding of molar mass and its implications in a fun and engaging manner.
Unit Conversion
Unit conversion is a basic skill required in many scientific calculations and everyday life. It deals with the ability to convert quantities from one system of measurement to another. In this exercise, we used the conversion factor between pounds and kilograms. Specifically, 1 pound is equivalent to 0.453592 kilograms. This conversion is imperative for students to grasp, not only for this exercise but also as a foundational skill in various scientific and practical applications.

It is crucial to always keep track of units when performing conversions and ensure that the final answer is in the correct units for the context, in this case, kilograms. Pro tip: always double-check units after each step to avoid mistakes that can lead to incorrect answers. Remember, proper unit conversion can make the difference between an accurate result and a mistaken one.
Mass of Earth Comparison
Whereas the first two concepts are more process-oriented, this final concept is about contextualizing our findings. Comparing the total mass calculated from Avogadro's number of people to the actual mass of Earth puts into perspective the enormity of both numbers.

Such comparisons help students to grasp the vastness of the quantities being discussed and can be a valuable tool in understanding the scales involved in scientific calculations. Remember, when we talk about astronomical scales like the mass of Earth, it's easy to lose sight of what those numbers actually represent. Relating them to more tangible ideas, like the number of people it would take to equal that mass, helps to conceptualize otherwise abstract concepts.

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Most popular questions from this chapter

A piece of aluminum foil 1.00 \(\mathrm{cm}^{2}\) and 0.550 -mm thick is allowed to react with bromine to form aluminum bromide. (a) How many moles of aluminum were used? (The density of aluminum is 2.699 \(\mathrm{g} / \mathrm{cm}^{3} .\) ) (b) How many grams of aluminum bromide form, assuming the aluminum reacts completely?

A mixture of \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2}(g)\) reacts in a closed container to form ammonia, \(\mathrm{NH}_{3}(g) .\) The reaction ceases before either reactant has been totally consumed. At this stage \(3.0 \mathrm{mol} \mathrm{N}_{2}, 3.0 \mathrm{mol} \mathrm{H}_{2},\) and 3.0 \(\mathrm{mol} \mathrm{NH}_{3}\) are present. How many moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) were present originally?

(a) Define the terms limiting reactant and excess reactant. (b) Why are the amounts of products formed in a reaction determined only by the amount of the limiting reactant? (c) Why should you base your choice of which compound is the limiting reactant on its number of initial moles, not on its initial mass in grams?

Define the terms theoretical yield, actual yield, and percent yield. (b) Why is the actual yield in a reaction almost always less than the theoretical yield?(c) Can a reaction ever have 110\(\%\) actual yield?

(a) Combustion analysis of toluene, a common organic solvent, gives 5.86 \(\mathrm{mg}\) of \(\mathrm{CO}_{2}\) and 1.37 \(\mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\) . If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O} . \mathrm{A}\) 0.1005 -gsample of mentholis combusted, producing 0.2829 \(\mathrm{g}\) of \(\mathrm{CO}_{2}\) and 0.1159 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for menthol? If menthol has a molar mass of 156 \(\mathrm{g} / \mathrm{mol}\) what is its molecular formula?

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