Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 3 - Sections 3.1-3.12 - Exercises - Cumulative Problems - Page 134: 121a

Answer

$K_{2}CrO_{4}$ K = 40.26% Cr = 26.78% O = 32.96%

Work Step by Step

We can tell from the name that the compound is made of potassium and a chromate ion. Since chromate ($CrO_{4}^{-2}$) has a $-2$ charge and potassium ($K^{+1}$) has a $+1$ charge then we need to potassium to cancel the charge of the chromate. Therefore the formula is $K_{2}CrO_{4}$. Adding up the molar masses of the constituent elements we get $\frac{194.185g}{mol}$. So to find the percent composition we just multiply the molar mass of each element (found on the periodic table) by how many times it occurs, divide by the molar mass of the whole molecule and multiply by 100%. Here is $K$ as an example: $$\frac{2(39.098\frac{g}{mol})}{194.185\frac{g}{mol}}\times100\%= 40.26\%$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.