Lecture 12sf

Lecture 12sf - Writing Chemical Equations Magnesium burns...

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Writing Chemical Equations Magnesium burns in air. Burning is a reaction with oxygen, forming an oxide. Magnesium + oxygen → Magnesium oxide Mg + O 2 → MgO (unbalanced) 2Mg + O 2 → 2MgO (balanced) Balancing uses coefficients to change the amounts of different substances reacting. The purpose of balancing is to conserve atoms. In a chemical reaction, atoms are rearranged, but not created or destroyed. Butane, C 4 H 10 , undergoes combustion. When hydrocarbons burn, they react with oxygen, forming carbon dioxide and water. C 4 H 10 + O 2 → CO 2 + H 2 O (unbalanced) 2 C 4 H 10 + 13O 2 → 8CO 2 + 10H 2 O (balanced) Ammonium dichromate decomposes to yield chromium (III) oxide, nitrogen and water (NH 4 ) 2 Cr 2 O 7 → Cr 2 O 3 + N 2 + H 2 O (unbalanced) (NH 4 ) 2 Cr 2 O 7 → Cr 2 O 3 + N 2 + 4H 2 O (balanced)

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Combustion Analysis A hydrocarbon is burned and produces 1.500 g CO 2 and 0.2457 g H 2 O. The molar mass of the hydrocarbon is 128 . What is the molecular formula of this hydrocarbon? 1.500 g CO 2 x x 2 CO mol 1 C mol 1 = 0.03408 mol C 0.2457 g H 2 O x x O H mol 1 H mol 2 2 = 0.02727 mol H Ratio: = = 1.250 Formula cannot be C 1.25 H Look for a multiplier to get to whole numbers Multiplier is 4 (4 x 1.25) = 5 Formula: C 1.25 H → C 5 H 4 The empirical formula is C 5 H 4 . The molar mass is 128 = = 2 Molecular formula is 2(C 5 H 4 ) or C 10 H 8 A 0.1000 g sample of a compound containing carbon, hydrogen, and oxygen burns, producing 0.1953 g CO 2 and 0.1000 g H 2 O.
What is the empirical formula of this compound? If the molar mass of the compound is 90.0 g/mol, what is the molecular formula of the compound? 0.1953 g CO 2 x x 2 CO mol 1 C mol 1 = 0.004438 mol C 0.1000 g H 2 O x x O H mol 1 H mol 2 2 = 0.01110 mol H To get the mol of oxygen, we convert mol carbon and mol H into grams, then subtract the mass of (carbon + hydrogen) from the original mass of the compound. 0.004438 mol C x 12.01 = 0.05330 g C 0.01110 mol H x 1.008 = 0.01119 g H 0.1000 g compound - (0.05330 + 0.01119) = 0.0355 g O 0.0355 g Oxygen x g 00 . 16 mol 1 = 0.00222 mol O We now have: 0.004438 mol C 0.01119 mol H 0.00222 mol O Dividing through by the smallest number to find the simplest whole number ratios: mol C mol H mol O 00222 . 0 004438

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This note was uploaded on 02/20/2012 for the course 160 161 taught by Professor Kim during the Fall '08 term at Rutgers.

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Lecture 12sf - Writing Chemical Equations Magnesium burns...

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