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The following math concepts are basic tools that you will need to solve chemistry problems.

**Definition and Examples**

Exponents are a short-hand notation for writing very large or very small numbers. The exponent gives the number of times that a number should be multiplied to itself. For example, 10^{3}, where the exponent is the superscript 3, indicates that 10 should be multiplied together three times:

10^{3} = 10 * 10 * 10 = 1,000.

Using exponents is often referred to as "raising a number to the power of..." In the above example, 10 is raised to the power of 3.

When working with the number 10, changing the exponent by one is the same as moving the decimal point one place:

1.000x10^{3} = 10.00x10^{2} = 100.0x10^{1} = 1,000

A negative exponent indicates that the result is the inverse of the number raised to the exponent. For example:

8^{-2} = 1/8^{2} = 1/64 = 0.0156.

As with positive exponents, when working with the number 10, changing the exponent by one is the same as moving the decimal point one place:

3.3x10^{-2} M = 0.33x10^{-1} M = 0.033 M

**Multiplication and Division**

When multiplying or dividing numbers with exponents, the exponents are added or subtracted. For example:

10^{8} * 10^{4} = 10^{(8+4)} = 10^{12}

(5.0x10^{-4} moles/L) * (1.0x10^{-3} L) = (5.0*1.0)x10^{-4-3} moles = (5.0x10^{-7} moles

10^{9}--- = 10^{(9-3)}= 10^{6}10^{3}

Raising a number with an exponent to another power is the same as working with any other number.

(10^{3})^{3} = 10^{3} * 10^{3} * 10^{3} = 10^{9}

Note that the result is the same as multiplying the two exponents together:

(10^{3})^{3} = 10^{(3*3)} = 10^{9}

Square roots are written as x^{1/2}, cube roots as x^{1/3}, fourth roots as x^{1/4}, and so on. These exponents are treated like any other exponent. For example, the square root of 10^{6} is:

(10^{6})^{1/2} = 10^{(6*1/2)} = 10^{6/2} = 10^{3}

**Addition and Subtraction**

Be careful when adding and subtracting numbers with exponents. It is easiest to writh the numbers with the same exponent. For example:

4.00x10^{4}g - 5.00x10^{3}g = 40.0x10^{3}g - 5.00x10^{3}g = 35.0x10^{3}g = 3.50x10^{4}g = 40,000 g - 5,000 g = 35,000 g

The logarithm of a number is the exponent of a base number that equals the original number raised to that exponent. There are two common base numbers that are useful in science, base 10 and base e (e=2.71828).

When working with base 10, the logarithm is called the common logarithm and it is abbreviated as log_{10} or log. When working with base e, the logarithm is called the natural logarithm and it is abbreviated as log_{e} or ln. Be sure that you know with which logarithm you are working before beginning a calculation. Examples:

log(1000) = log(10^{3}) = 3

log(100) = log(10^{2}) = 2

log(10) = log(10^{1}) = 1

log(1) = log(10^{0}) = 0

log(0.1) = log(10^{-1}) = -1

log(0.01) = log(10^{-2}) = -2

Note that the log of 1000 and 100 differ by one. A logarithmic scale increases by an order of magnitude for each unit increase. A common log scale is pH (p is an abbreviation for -log, so pH = -log[H^{+}]). The [H^{+}], hydrogen ion concentration, usually ranges from 1.0 to 1.0x10^{-14} in aqueous solution (water).

very very acidic neutral basic +----------------+----------------+ pH 0 7 14 [H^{+}] 1.0 M 1.0x10^{-7}M 1.0x10^{-14}M

Note that a change in pH by one is a change in [H^{+}] by a factor of ten.

Solving chemistry problems often requires using algebra to simplify mathematical expressions to solve for an unknown. The key rule of algebra is: **Anything you do to one side of an expression, you must also do to the other side of the expression.**

Using this rule the two sides of the expression remain equal. You can add, subtract, multiply, divide, raise to a power, or take the logarithm.

Example 1, simplify the following expression: | 17x^{2} + 9x = 80 -3x + x^{2} |

Subtract 1x^{2} from each side: | 16x^{2} + 9x = 80 -3x |

Add 3x to each side: | 16x^{2} + 12x = 80 |

Subtract 80 from each side: | 16x^{2} + 12x - 80 = 0 |

Divide each side by the largest common demoninator: | 4x^{2} + 3x - 20 = 0 |

Example 2, find the pH of a solution that has a [H^{+}] of 1.32x10^{-4} M (pH = -log[H^{+}]).

You are given: | [H^{+}] = 1.32x10^{-4} M |

Take the log of each side: | log[H^{+}] = log(1.32x10^{-4} M) |

Simplify: | log[H^{+}] = -3.879 |

Multiply each side by -1: | -log[H^{+}] = 3.879 |

Simplify: | pH = 3.879 |

Results are generally reported with one uncertain significant figure (usually ±1). Example: reporting a mass of 12.554 g indicates that the experimenter is certain that the true mass lies in the range 12.553 - 12.555 g. If the uncertainty in the last significant figure is greater than ± 1 it should be written explicitly: 12.554 g ± 0.003 g.

**Addition and Subtraction**

The answer must be rounded at the same least significant decimal place as the number in the problem that has the least significant figure from the right.

Example:

8.9444 g +18.52 g -------- 27.46 g

**Multiplication and Division**

The answer has the same number of significant figures as the factor with the smallest number of significant figures.

Example:

8.9 g / 12.01 g/mol = 0.74 mol

It is common to carry one additional significant figure through extended calculations and to round off the final answer at the end.

A first step in problem solving is to identify what is given and what is requested. A valuable tool in this aspect of the process is to identify the units of given and requested information, and keeping track of units as you try to get from the given to the requested information. Keeping track of units as you work a problem can help you avoid simple mistakes, or proceeding down a dead-end approach.

Example: What weight of KI (mol wt.= 166.0 g/mol) do you need if you want to prepare 0.500 L of a 0.100 M solution? (M is the abbreviation for molarity, a unit of concentration in moles/L).

Dissecting this problem, you must first find the number of moles of KI needed to make the requested solution. Looking at the units, you should multiply the desired volume (0.500 L) by the desired concentration (0.100 moles KI/L H_{2}O): (0.500 L H_{2}O)*(0.100 moles KI/L H_{2}O) = 0.0500 moles KI. Check units: L * moles/L = moles.

To finish the calculation you determine the weight of KI that equals 0.0500 moles of KI. Looking at the units, you must take the number of moles you need and multiply by the mol. wt.:

weight of KI = (0.0500 moles KI) * (166.0 g KI/mol KI) = 8.30 g KI.

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