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Chapter 3: Problem 21

Which of the following has more atoms: \(1.10 \mathrm{~g}\) of hydrogen atoms or \(14.7 \mathrm{~g}\) of chromium atoms?

Short Answer

Expert verified
1.10 g of hydrogen (with \(6.62 \times 10^{23}\) atoms) has more atoms than 14.7 g of chromium (with \(1.70 \times 10^{23}\) atoms).

Step by step solution

01

Calculate the number of moles of hydrogen

To calculate the number of moles of hydrogen, use the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]For hydrogen, this comes out to: \[ \text{moles}_H = \frac{1.10 \, \mathrm{g}}{1.00 \, \mathrm{g/mol}} = 1.10 \, \mathrm{mol} \]
02

Calculate the number of atoms in the given mass of hydrogen

Multiply the number of moles by Avogadro's number to find the total number of atoms: \[ \text{atoms}_H = \text{moles}_H \times \text{Avogadro's number} = 1.10 \, \mathrm{mol} \times 6.022 \times 10^{23} \, \mathrm{atoms/mol} = 6.62 \times 10^{23} \, \mathrm{atoms} \]
03

Calculate the number of moles of chromium

Use the molar mass of chromium to find: \[ \text{moles}_C = \frac{14.7 \, \mathrm{g}}{52 \, \mathrm{g/mol}} = 0.283 \, \mathrm{mol} \]
04

Calculate the number of atoms in the given mass of chromium

The total number of chromium atoms is found by: \[ \text{atoms}_C = \text{moles}_C \times \text{Avogadro's number} = 0.283 \, \mathrm{mol} \times 6.022 \times 10^{23} \, \mathrm{atoms/mol} = 1.70 \times 10^{23} \, \mathrm{atoms} \]
05

Compare the number of atoms

Compare the number of atoms found in steps 2 and 4. The higher number indicates which substance has more atoms.

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Most popular questions from this chapter

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 \(\mathrm{NH}_{3} ?\)

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The empirical formula of a compound is \(\mathrm{CH}\). If the molar mass of this compound is about \(78 \mathrm{~g}\), what is its molecular formula?
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