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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%.

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Molecular Formula

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  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

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