1 / 9

Molecular Formula

Molecular Formula. number and type of atoms. covalent compounds. non-metals. glucose. C 6 H 12 O 6. mix. C(s). H 2 (g). O 2 (g). 1 atom of C.  2 x 10 -23 g.  12 amu. 1 atom of 12 C. (6 protons + 6 neutrons). C. =12.011 amu. different isotopes. 12 C 98.9% 13 C 1.10%.

aelwen
Download Presentation

Molecular Formula

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Molecular Formula number and type of atoms covalent compounds non-metals glucose C6H12O6 mix C(s) H2(g) O2(g) 1 atom of C  2 x 10-23 g 12 amu 1 atom of 12C (6 protons + 6 neutrons) C =12.011 amu different isotopes 12C 98.9% 13C 1.10% ave. = (0.989 x 12) + (0.001 x 13.00335) = 12.011

  2. relative masses C =12.011 amu exact masses (g) O =15.9994 amu H 12 g12C =1.0079 amu = 6.022 x 1023 atoms12C Avogadro’s number (NA) = mole 12 g 12C 6.022 x 1023 atoms = 12 g 12C 6.022 x 1023 atoms mole mol C =12.011 g/mol O =15.9994 g/mol H =1.0079 g/mol

  3. C =12.011 g/mol O =15.9994 g/mol H =1.0079 g/mol Start with 15.43 g carbon How much H2(g) and O2(g) ? 1 mol 15.43 g C = 1.284655732 mol C = 1.285 mol C 12.011 g 2.569 mol H 1.0079 g H = 2.589609 g H = 2.590 g H 1 mol H 1.285 mol O 15.9994 g O = 20.55922 g O = 20.56 g H 1 mol O C (s) H2 (g) O2 (g) C6H12O6 2.59 g 20.56 g 15.43 g 1 atom 2 atoms 1 atom 1 mol C 2 mol H 1 mol O 1.285 mol C 2.569 mol H 1.285 mol O

  4. How many grams of glucose will be produced? 15.43 + 2.59 + 20.56 = 38.58 g glucose How many moles of glucose is this? 6 mol C + 12 mol H + 6 mol O = 180.16 g 1 mol glucose = 6 x 12.011 12 x 1.0079 6 x 15.9994 38.58 g glucose 1 mol = 0.2141 mol glucose 180.16 g C (s) + H2 (g) + O2 (g) C6H12O6 2.59 g 20.56 g 15.43 g

  5. Percent composition C6H12O6 % (by mass) of each element molar mass = 180.16 g/mol 6 mol C = (6 x 12.01 g/mol) = 0.40001 = 40.00% C mol C6H12O6 (1 x 180.16 g/mol) = 6.72% H 12 mol H = (12 x 1.008 g/mol) = 0.06717 mol C6H12O6 (1 x 180.16 g/mol) 6 mol O = (6 x 15.9994 g/mol) = 0.53283 = 53.28% O mol C6H12O6 (1 x 180.16 g/mol)

  6. Used to determine formula of unknown 40.92% carbon 4.58% hydrogen 54.50% oxygen 1. Assume you have 100 grams 40.92 g C 4.58 g H 54.50 g O 2. Calculate moles of elements 3. Determine empirical formula 3.41 mol C 4.54 mol H C H O 3.41 4.54 3.41 3.41 mol O C H O 1 1.33 1 C H O 3 4 3

  7. C H O C H O C H O 3 4 3 3 6 4 8 3 6 C H O 9 12 9 88.06 g/mol 176.13 g/mol 264.19 g/mol need molecular weight Combustion reaction CnHnOn CO2 + O2 + H2O 11.2 g of an unknown compound is burned, producing and 13.5 g H2O. 22.0 g CO2 1. Change grams to moles

  8. 11.2 g of an unknown compound is burned, producing and 13.5 g H2O. 22.0 g CO2 1. Change grams to moles 0.500 moles CO2 0.500 mol C mol H 0.75 moles H2O 1.50 2. How much O came from unknown? mass unknown = 11.2 g mass unknown = mass C + mass H + mass O mass unknown = 11.2 g = 6.00 g + 1.51 g + mass O 4.0 g mass O = 4.0 g O = mol O 0.25

  9. 0.500 mol C mol H 1.50 mol O 0.25 C H O .5 1.5 .25 C H O 2 6 Empirical formula

More Related