Molecular Formula number and type of atoms covalent compounds

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Molecular Formula number and type of atoms covalent compounds non-metals glucose C6H12O6 mix C(s) H2(g) O2(g) Start with 15.43 g carbon How many grams H2(g) and O2(g) ?

How many grams H2(g) and O2(g) ? C =12.01 g/mol O =16.00 g/mol H =1.008 g/mol Start with 15.43 g carbon How many grams H2(g) and O2(g) ? 15.43 g C 1 mol = 1.284655732 mol C = 1.285 mol C 12.01 g 2.569 mol H 1.008 g H = 2.589609 g H = 2.590 g H 1 mol H 1.285 mol O 16.00 g O = 20.55922 g O = 20.56 g O 1 mol O C (s) H2 (g) O2 (g) C6H12O6 15.43 g 2.590 g 20.56 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

How many grams of glucose will be produced? 15.43 + 2.590 + 20.56 = 38.58 g glucose How many moles of glucose is this? 1 mol glucose = 6 mol C + 12 mol H + 6 mol O = 180.157 g 180.16 6 x 12.01 12 x 1.008 6 x 16.00 38.58 g glucose 1 mol = 0.2141 mol glucose 180.16 g C (s) + H2 (g) + O2 (g) C6H12O6 15.43 g 2.590 g 20.56 g

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) 12 mol H = (12 x 1.008 g/mol) = 0.06717 = 6.717% H mol C6H12O6 (1 x 180.16 g/mol) 6 mol O = (6 x 16.00 g/mol) = 0.53283 = 53.28% O mol C6H12O6 (1 x 180.16 g/mol) Used to determine formula of unknown

Used to determine formula of unknown 40.92% carbon 4.58% hydrogen 1. Assume you have 100 grams 54.50% oxygen 40.92 g C 2. Calculate moles of elements 4.58 g H 54.50 g O 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

C H O 3 4 C H O 3 4 88.06 g/mol C H O 6 8 176.13 g/mol C H O 9 12 264.19 g/mol need molecular weight Where does mass % come from? Combustion reaction

Combustion reaction C?H?O? + O2 n CO2 + m H2O 11.5 g of an unknown compound is burned, producing 22.0 g CO2 and 13.5 g H2O. 1. Change grams to moles 0.500 moles CO2 0.500 mol C 0.75 moles H2O 1.50 mol H

11.5 g of an unknown compound is burned, producing 22.0 g CO2 and 13.5 g H2O. C?H?O? + O2 n CO2 + m H2O mol C mol H 1.50 0.500 mol O 0.25 2. How much O came from unknown? mass unknown = 11.5 g = mass C + mass H + mass O 11.5 g = 6.00 g + 1.51 g + mass O mass O = 4.0 g 4.0 g O = mol O 0.25

11.5 g of an unknown compound is burned, producing 22.0 g CO2 and 13.5 g H2O. mol C mol H 1.50 0.500 C?H?O? + O2 n CO2 + m H2O mol O 0.25 C H O .5 1.5 .25 C H O 2 6 Empirical formula