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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free