New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

CH 111, Week 4 Notes

by: Jordyn Meekma

CH 111, Week 4 Notes CH111

Jordyn Meekma

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

This covers combustion reactions, empirical formulas, and an introduction to solutions.
General Chemistry 1
Dr. Tom Getman
Class Notes
25 ?




Popular in General Chemistry 1

Popular in Chemistry

This 2 page Class Notes was uploaded by Jordyn Meekma on Friday September 30, 2016. The Class Notes belongs to CH111 at Northern Michigan University taught by Dr. Tom Getman in Fall 2016. Since its upload, it has received 2 views. For similar materials see General Chemistry 1 in Chemistry at Northern Michigan University.


Reviews for CH 111, Week 4 Notes


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/30/16
CH 111 Notes: Week 4  Combustion Reactions: an organic substance combines with oxygen and is burned, giving off heat energy and creating carbon dioxide and water o Propane (C H 3 8 C 3 +85O → 32CO + 4H O2 2  Empirical Formula: the formula of a molecule with the smallest whole number ratio of atoms o C 5 4 vs. C10 8 → related by a whole number multiple of 2  In a combustion reaction, 135.0 mg of a hydrocarbon reacts with excess oxygen to produce 440.0 mg of carbon dioxide and 135.0 mg water. The hydrocarbon MM = 270 g/mol. Determine the empirical and molecular formulas. o C x +yO → C2 + H O 2 2 -3 0.4400 g CO (1 2ol CO /44.012g)(1 mol C/1 mol CO ) = 9.9977x12 mol C -2 0.1350 g H O 2 1 mol H O/18.22 g)(2 mol H/1 mol H O) = 1.4983210 mol H X = 9.9977x10 mol C/9.9977x10 mol C = 1 (x2) = 2 -2 -3 *Divide C and H by the lower number of Y = 1.4983x10 mol H /9.9977x10 mol C = 1.499 (x2) = 3 Empirical Formula: C H 2 3 moles and multiply to get to the nearest whole number MM C H 2 37.05 g/mol (270 g/mol)/(27.05 g/mol) = 10 C 2 3x10) = … Molecular Formula: C H 20 30  Lithium metal reacts with nitrogen gas to form a compound. If 0.437 g Li are completely reacted to form 0.725 g of the compound, what is the empirical formula? o Li + N →2Li N x y 0.725 g compound – 0.437 g Li = 0.288 g N X = 0.437 g Li (1 mol Li/6.941 g) = 6.296x10 mol Li/2.056x10 mol N -2 = 3.06 Y = 0.288 g N (1 mol N/14.01 g) = 2.056x10 mol N/2.056x10 mol N -2 = 1 Empirical Formula: Li N 3  A compound containing C, H, and O is burned. If 125.0 mg of the compound produces 177.5 mg of Co2 and 108.8 mg of H2O, what are the empirical and molecular formulas? MM of the compound = 62.08 g/mol. C ? O?+?O → C2 + H O 2 2 -3 o C: 0.1775 g CO (1 m2l CO /44.01 2)(1 mol C/1 mol CO ) = 4.033x10 2 mol C -3 -2 4.033x10 mol C (12.01 g/1 mol C) = 4.844x10 g C  4.033x10 mol C/4.03x10 mol O = 1.002 -2 o H: 0.1087 g H O (2 mol H O/18.02 g)(2 mol H/1 mol H O) = 1.206x12 mol H 1.206x10 mol H (1.01 g/1 mol H) = 1.216x10 g H -2 -2 -3  1.206x10 mol H/4.03x10 mol O = 3.00 o O: 0.1250 g compound - 4.844x10 g C - 1.216x10 g H = 6.44x10 g -2 O 6.44x10 g O (1 mol O/16.00 g) = 4.03x10 mol O -3 -3 -3  4.03x10 mol O/4.03x10 mol O = 1 Empirical Formula: CH O M3 = 31.04 g/mol Molecular Formula: (62.08 g/mol)/(31.04 g/mol) = 2.000 CH 3 (x2) = C 2 O6 2 Chapter 4: Solutions Solvent: the component in a solution with the greatest quantity (of moles) Solute: the component in a solution that is dissolved in the solvent Molarity: (M) moles of a component divided by the volume in liters (mol/L) o How many grams of sucrose (C H O ) are 12qu22e11to make 250.0 mL of a 1.25 M sucrose solution? MM C H O 12 22 11 = 342.34 g/mol 1.25 mol C H12 22 L110.2500 L)(342.34 g/1 mol C H O ) = 112 22 11 C 12O22 11 o Using this 1.25 M solution, how could we prepare 1.00 L of a 0.20 M sucrose solution? *M 1 =1M V *2 2 dilution equation (1.25 M)(V )1= (0.20 M)(1.00 L) V1= 0.260 L Spectrophotometry: using light (absorbance) to determine the concentration of a solute in solution o Beer’s Law *You will be using this concept in lab* Absorbance (no unit) Concentration of solute


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Amaris Trozzo George Washington University

"I made $350 in just two days after posting my first study guide."

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.