Chemical Reactions

Beyond the notes below, this link offer some useful info: http://www.files.chem.vt.edu/RVGS/ACT/notes/Types_of_Equations.html

Formula weight

Formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a formula by the number of atoms of that element present in the formula and then adding all of these products together. For example, the formula weight of water (H2O) is two times the atomic weight of hydrogen plus one times the atomic weight of oxygen. Numerically, this is (2×1.00797)+(1×15.9994) = 2.01594+15.9994 = 18.01534. If the formula used in computing the formula weight is the molecular formula, the formula weight computed is the molecular weight.
The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100. For example, the weight percentage of hydrogen in water is determined by taking two times the atomic weight of hydrogen, dividing it by the formula weight of water, and multiplying by 100. Numerically, this is 100×(2×1.00797)/18.01534 = 11.19% hydrogen in water by weight.
Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. For example, it is known that two molecules of hydrogen gas, H2, react with one molecule of oxygen gas, O2, to form two molecules of water, H2O. This reaction may be represented by the chemical equation 2H2+O2→2H2O. The formula weight of hydrogen gas is 2.01594, that of oxygen gas 31.9998, and that of water 18.01534. Our chemical equation is numerically equivalent to 2×2.01594+31.9998 = 2×18.01534 or 4.03188+31.9998 = 36.03068 if the formula weight of each reactant is substituted for the formula of that reactant. From this equation we know, for example, that 4.03188 grams of hydrogen gas will react with 31.9998 grams of oxygen gas to yield 36.03068 grams of water. The relative proportions by weight of these reactants is the same in any reaction of hydrogen and oxygen to form water. These relative weights computed from the chemical equation are sometimes called equation weights.

Examples of chemical changes

A silver spoon tarnishes. The silver reacts with sulfur in the air to make silver sulfide, the black material we call tarnish.
2 Ag + S Ag2S
An iron bar rusts. The iron reacts with oxygen in the air to make rust.
4 Fe + 3 O2 2 Fe2O3
Methane burns. Methane combines with oxygen in the air to make carbon dioxide and water vapor.
CH4 + 2 O2 CO2 + 2 H2O
An antacid (calcium hydroxide) neutralizes stomach acid (hydrochloric acid).
Ca(OH)2 + 2 HCl CaCl2 + 2 H2O
Glucose (simple sugar) ferments to ethyl alcohol and carbon dioxide. The sugar in grapes or from grain ferments with yeast to make the alcohol and carbon dioxide. The carbon dioxide is the gas that bubbles out of beer or champaign.
C6H12O6 (glucose) 2 C2H5OH (ethyl alcohol) + 2 CO2
Alcohol plus oxygen becomes vinegar and a molecule of water. As in the fermentation of glucose, this is a more complex reaction than it appears here because it is a biochemical reaction.
C2H5OH + O2 HC2H3O2 + H2O

Examples of physical changes

Here are some examples of changes that are NOT chemical reactions. In each case, the original material or materials may be reclaimed by physical processes.
- Water boils out of a kettle or condenses on a cold glass.
- An aluminum pot is put on a burner and gets hot.
- Dry ice goes from a solid to a gaseous form of carbon dioxide (sublimation).
- Gold melts or solidifies.
- Sand is mixed in with salt.
- A piece of chalk is ground to dust.
- Glass breaks.
- An iron rod gets magnetized.
- A lump of sugar dissolves in water.

Balancing equations

Sample problem: If 1.0 mole ethane is mixed with 5.0 mole O₂ and the 1.0 mole ethane is completely consumed during the reaction, how many moles of O₂ will be left at the end of the reaction?
Solution: We first write the balanced equation for this reaction:
C₂H₆ + 3.5 O₂ → 2 CO₂ + 3 H₂O
To balance the equation, we first balance the carbon in the carbon dioxide, then the hydrogen in the water, and finally the oxygen on the left. The equation tells us that 1 mole C₂H₆ requires 3.5 moles O₂ for complete combustion. Since we started with 5.0 mole O₂, 5.0 - 3.5 = 1.5 mole O₂ will be left at the end of the reaction.

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