WebLecture

- The Mathematics We Need For Chemistry
- Practice with Concepts
- Discussion Questions
- Optional Website Reading

We can clarify which right side zeroes are significant by using scientific notation. 716.0 * 10 indicates unambiguously that the 0 is significant, since it is mentioned. The usual practice is to reduce a number to 1 decimal place and the appropriate power of ten. The error is always then plus or minus 1 tenth of the represented answer.

There are different ways of representing scientific notation. One common way is the one shown above: 593.1 * 10^{3} would work out to 593100. Another way to represent this number, and one which is common to calculators, is 953.1E3. You may use either.

If I need to calculate real error ranges, then I should take the extreme amounts indicated by my error estimates, and work out the areas.

Just to review, here are the measurements I got before:

Actual measurements (cm) |
High possible values |
Low possible values |

15.21 |
15.22 |
15.20 |

* 10.12 |
* 10.13 |
* 10.11 |

152.9252 |
153.6720 |
154.1786 |

The difference is 152.9252 cm^{2} plus .2534cm^{2} or minus .2532cm^{2}; in other words, we have a possible error of .253 cm^{2} or so in our final answer. An easy way to determine whether this is good or bad is to look at the difference as a percent of the "actual" measurements. In this case, we would do the operation .253/152.9252 = 1.6544035E-3 (on my calculator). This works out to .0016 or about .2%, or two parts per thousand. Whether this error is significant depends on the circumstances. A miscalculation of .2% in planetary orbit calculations may mean that your very expensive exploration satellite misses the planet entirely. In the case of my postcard, however, I probably won't notice the difference in area.

Doing the actual calculation, I find that

$$152.1\mathrm{mm}\text{}\ast \text{}101.2\text{}\mathrm{mm}\text{}=\text{}15392.52\text{}{\mathrm{mm}}^{2}$$But I didn't actually measure out the 2.52 part, and I can't claim to be accurate to 7 places. As with any measurement and calculation, I need to determine how many digits of my answer are significant, that is, how many carry real and reliable information, and how many are artifacts of the way I manipulated them to get the area. This question is especially important if I use a calculator, since it may (in division, for example) carry out an operation to a numerical accuracy far beyond any physical accuracy in the input information.

There are two rules for determining significant figures after an arithmetic operation, depending on the operation involved.

*When numbers are manipulated by addition or subtraction*, the result can have only as many**decimal places**as the number with the fewest decimal digits. For example, 31.15 + 22.953 + 16.4 can have only one decimal digit, because 16.4 has only one. The appropriate answer is to round the straight mathematical answer of 70.503 to one decimal place: 70.5.

31.15 |

22.953 |

16.4 |

-------- |

70.5 |

*When numbers are manipulated by multiplication or division*, the result can have only as many**digits**as the number with the fewest digits. So our postcard area can have only 4 significant figures, because each of the numbers which went into it had only 4. The answer with the appropriate amount of information is 15390 mm^{2}.

Now, I happen to prefer looking at units for such areas in centimeters rather than millimeters. In order to convert the units, I need to realize that 1 cm^{2} = 100 mm^{2}, so I multiply 15390 mm^{2} * 1 cm ^{2}/100 mm^{2} = 153.9 cm^{2}. Notice that my units cancel--all mention of mm^{2} goes away. Notice also that zeros on the right of this number are not significant: we eliminated the 2.52 as excess.

There are special rules for zeroes.

*Zeroes to the right of the decimal place are significant*: 7.160 has four significant figures and indicates that we have an value accurate to .160 plus or minus .001.*Zeroes to the left of a number are not significant*: 0.135 has 3 significant figures.*Non-decimal zeroes to the right of a number may or may not be significant*: 7160 may be accurate to plus or minus 1 (zero is not significant), or to plus or minus .1 (zero is significant).

In measuring the postcard, we moved from an actual measurement to a derived quantity. Some physical quantities are always derived. The seven base quantities recognized in the Systeme Internationale (the internationally used metric system) are

Type of quantity | Unit |
---|---|

Amount of substance | mole |

Electric current | ampere |

Length | meter |

Luminous intensity | candela |

Mass | kilogram |

Temperature | kelvin (or centigrade) |

Time | second |

All other quantities can be expressed as some derivation of two or more of these units. Velocity is length/time, and density is area (a derived quantity from length)/mass.

Each of these units can be expanded by a prefix which tells how many base units are needed for the new unit. A kilogram is 1000 grams. A millimeter is 1/1000 of a meter. You should become familiar with the prefixes and the base units in various systems which are listed inside the back cover of your text.

- Why do we need agreement about the units we use for measurement or counting?

Need more drill work on significant figures? Scientific notation? Concepts of mass, density, and volume? Percents? ...You name it, you can probably work on it at the Chemistry drill site, courtesy Scott Van Bramer at Widener University.

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