New version page

UM CHEM 1110 - Dimensional Analysis

This preview shows page 1-2-3 out of 9 pages.

View Full Document
View Full Document

End of preview. Want to read all 9 pages?

Upload your study docs or become a GradeBuddy member to access this document.

View Full Document
Unformatted text preview:

Dimensional Analysis 1Preview The lectures in this unit cover an introduction to chemistry and matter, working with numbers and units, and an introduction to atoms and the periodic table. This lecture covers mathematical problem solving, including formulas and using dimensional analysis. Dimensional Analysis Our last lecture covered a basic understanding of the numbers and the units we work with. Here we will look at the method we use to solve most of our math problems in this class: Dimensional Analysis. There are two basic methods of doing calculations and conversions in the sciences: using formulas and following dimensions. The latter method is called Dimensional Analysis, or the Factor-Label method. Though you are likely familiar with the first method, the second may be new to you – and therefore difficult at first. But understanding Dimensional Analysis is critical to your success in this class and as a medical professional. Lots of practice will help you feel more comfortable with this method. I. Formulaic Method To solve a math problem using the formulaic method you must first find the formula which suits the problem at hand. Then you perform algebraic conversions to solve for the unknown item you are looking for. Finally you ‘plug in’ the numbers you are given to get the correct answer. Once you memorize the formula, the conversion is fairly trivial. For example: The conversion between degrees Fahrenheit and degrees Celsius is as follows: °F = (9/5)°C + 32 The formula which relates density, mass, and volume we saw from the last lecture: Density = Mass/Volume Each of these requires you to remember a formula and plug in the values as necessary. If I asked you the density of an object with a mass of 452.1 g and a volume of 292 mL you would remember that density = mass/volume, then plug 452.1 g into the mass spot and 292 mL into the volume slot. So density = (452.1 g)/(292 mL) = 1.55 g/mL. Pretty easy! Lots of students like the formulaic method because it eliminates a lot of thinking. As long as you can remember the formula, solve for the variable, and plug in your numbers correctly, you will get the correct answer.Dimensional Analysis 2II. Dimensional Analysis Dimensional analysis is a little bit harder because we do not use memorized formulas. But as a result the method is much more robust, allowing us to do conversions we could not do otherwise. There are many different types of units in common use. It would take hundreds of pre-made formulas to solve every possible unit conversion with a memorized formula. Sure, your book has conversion factors for feet to miles and feet to meters. You could memorize those. But what about feet to micrometers? What about miles/hour to cm/sec? It is not practical to memorize a formula for each potential conversion. Instead we follow the units – that’s another word for dimensions – and create conversion factors which make logical sense for each problem we do. In the previous lecture we said units are an integral part of a measurement. No measurement is complete without the units. This is especially important in dimensional analysis where we are essentially doing math with the units – and the numbers get to tag along. Here is how it works: conversion factors are based on equivalencies. Any mathematical statement which is written as “something equals something else” is an equivalency. You may already be familiar with the following: 1 in = 2.54 cm 1 lb = 453.59 g 1 L = 1.0567 quarts Basic algebra teaches us that any number divided by itself is 1. So, 22 = 1. And 100100 = 1. And if we really believe 1 inch = 2.54 centimeters: or that those things are equivalent, that means 1 in2.54 cm = 1. It also means that 2.54 cm 1 in = 1. Please note that we are NOT saying that 12.54 = 1. This math only works if you leave the units in place. Each of these statements (1 in2.54 cm and 2.54 cm 1 in ) is a conversion factor. Whenever we multiply a number by a conversion factor like 1 in2.54 cm or 1 lb453.59 g or 1 L1.0567 qt we can be confident that we have not actually changed that number. This is because each of these conversion factors is equal to 1! We have only changed the units. Please note that any equivalency you come across can be used as a conversion factor. So let’s do math like this. What if I want to know how many centimeters there are in a foot? Well, we do not have a nice handy formulaic conversion for centimeters to feet. But we do have several conversions which are useful. We know that:Dimensional Analysis 31 foot = 12 inches and 1 inch = 2.54 cm. That gives us four conversion factors: 12 in1 foot = 1 and 1 in2.54 cm = 1 1 foot 12 in = 1 and 2.54 cm 1 in = 1 Our job now is to figure out how to put together these conversion factors to go from 1 foot to centimeters. We start with the number of feet we are given, put in the conversion factors to cancel out the undesired dimensions, and we get our answer: 1 foot * 12 in1 foot * 2.54 cm 1 in = 30.48 cm Every * means multiplication, but sometimes this expression is shown as follows where every vertical line indicates multiplication. Both ways are appropriate but you should be sure to understand each! 1 foot 12 inches 2.54 cm = 30.48 cm 1 foot 1 inch Please confirm that I used two conversion factors which are both equal to 1, so I have not changed my original value. The original 1 foot is the same as 30.48 cm, or in other words: 1 foot = 30.48 cm. I am not saying 1 = 30.48 in some kind of bogus math scheme. You should all recognize that 1 foot is the same length as 30.48 cm, and so long as we all understand what units are being used, they both refer to the same length value. Something else I want to bring to your attention: I started out with five length values on the left hand side: 1 foot, 12 inches, 1 foot, 2.54 cm, and 1 inch. I ended up with only one length value on the right: 30.48 cm. Where did they all go? I did math with the numerical values (1 *12*2.54) = 30.48, that is clear. But I also did math with the unit values. This was easier: we started out with (feet * inches*cm)/(foot*inch). All together foot/foot = 1: that cancels out. Inches/inches = 1, that cancels out. So I am left with cm. I keep track of that by canceling out the dimensions as I go so I am sure I end up with the right units. Here I


View Full Document
Loading Unlocking...
Login

Join to view Dimensional Analysis and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Dimensional Analysis and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?