determination of chemical formula

# determination of chemical formula - Date Performed Date...

This preview shows pages 1–3. Sign up to view the full content.

Date Performed: 2/25/2010 Name: Joe Staba Date Submitted: 3/4/2010 Name: Alyssa S. Instructor: Zhibing Xu Determination of a Chemical Formula Objective To determine the percentage composition and simplest formula for an oxide of copper by its reduction with methane gas (CH4) at approximately 500°C. This experiment is a study of stoichiometry. Experimental Data Mass of Test Tube = 20.002 ± .01g Mass of Test Tube + Copper Oxide = 22.258 ± .01g Mass of Copper Oxide = 2.256 ± .01g Sample Calculations Mass of Copper Oxide 22.258 ± .01g – 20.002 ± .01g = 2.256 ± .01g Uncertainty = √((.01)^2 + (.01)^2) = .01g Mass of Copper 21.805 ± .01g – 20.002 ± .01g = 1.803 ± .01g Uncertainty = √((.01)^2 + (.01)^2) = .01g Mass % Copper 1.803g Copper / 2.256g Copper Oxide (100) = 79.9% Copper Uncertainty = (Sr/.799)^2 = (.01/1.803)^2 + (.01/2.256)^2 = .006 Mass of Oxygen 2.256 ± .01g – 1.803 ± .01g = .453 ± .01g Uncertainty = √((.01)^2 + (.01)^2) = .01 Mass % Oxygen .453g Oxygen / 2.256g Copper Oxide (100) = 20.1% Uncertainty = (Sr/.201)^2 = (.01/.453)^2 + (.01/2.256)^2 = .005

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Moles of Copper 1.803g Copper / 63.55 g/mol Copper (molar mass) = .0284 mol Copper Uncertainty = (Sr/.0284)^2 = (.01/1.803)^2 + (.01/63.55)^2 = .0002 Moles of Oxygen .453g Oxygen / 16.0 g/mol Oxygen (molar mass) = .0283 mol Oxygen Uncertainty = (Sr/.0283)^2 = (.01/.453)^2 + (.01/16.0)^2 = .0006
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern