Download presentation

Presentation is loading. Please wait.

Published byEgbert Hubbard Modified over 5 years ago

1
Warm-Up: To be turned in 3.6 mol NaNO 3 = ______ g 228.50 g MgCl 2 = ______ mol

2
Empirical & Molecular Formulas

3
Empirical formula Formula that shows the smallest whole- number ratio of elements in a compound Does not always indicate the actual # of atoms in the compound

4
Determining Empirical Formulas If given % composition: – Assume you have 100.0 g total – Percentages become the # of g for each element – Calculate # of mol for each atom – Divide all mol values by smallest mol value Round to nearest whole number If given grams: – Skip steps 1 and 2

5
Example A compound contains 78.1% B and 21.9% H. 78.1g x ___1__ = 7.22 mol B 10.81g B 21.9g x ___1__ = 21.7 mol H 1.01g H 7.22 mol B : 21.7 mol H = 1 mol B: 3.01 mol H 7.22 mol 7.22 mol BH 3

6
Practice Determine the empirical formula for a compound containing the following: 40.0% C, 6.7% H, 53.3% O

7
Molecular Formula Actual formula of a compound Contains the same ratio of atoms as the empirical formula Must know the molar mass of the compound

8
Determining Molecular Formulas Determine the empirical formula Calculate molar mass of empirical formula Divide actual molar mass by empirical molar mass – Determines the multiplier Multiply empirical formula by multiplier

9
Example 85.64% C and 14.36% H, mol mass= 42.08 g/mol 85.64g x _1mol_ = 7.13 mol C 7.13mol: 14.21mol 12.01g7.13 7.13 14.36g x 1mol = 14.21 mol H CH 2 1.01g________________________________ 12.01 + (2x 1.01)= 14.03g/mol CH 2 42.08/ 14.03= 2.99 3 CH 2 x 3= C 3 H 6 Empirical formula

10
Practice 2 Determine the empirical and molecular formula for a compound that contains the following: 5.9% H, 94.1% O, molar mass= 34.02g/mol

Similar presentations

© 2021 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google