1. Chemistry Math Crunch Do you have what it takes?

2. Do I Really Have To?!?!?! Believe it or not, there are many good reasons to develop your ability to rearrange equations that are important to Chemistry. It can save time, help you with units and save some brain space! Here are some reasons to develop your equation manipulation skills (in no particular order):  Equations are easier to handle before inserting numbers! And, if you can isolate a variable on one side of the equation, it is applicable to every similar problem that asks you to solve for that variable!  If you know how to manipulate equations, you only have to remember one equation that has all the variables of question in it - you can manipulate it to solve for any other variable! This means less memorization!  Manipulating equations can help you keep track of (or figure out) units on a number. Because units are defined by the equations, if you manipulate, plug in numbers and cancel units, you'll end up with exactly the right units (for a given variable)!

3. Where Do We Use This? To be honest, equation manipulation occurs in almost every aspect of Chemistry. Any time you see a P or T or mL or L (or even =), there is an equation that you could manipulate. Because equations can be used to describe lots of important natural phenomena, being able to manipulate them gives you a powerful tool for understanding the world around you!

4. Important Rules for Rearranging Equations! You probably learned a number of rules for manipulating equations in a previous algebra course. It never hurts to remind ourselves of the rules. So let's review: RULE #1: you can add, subtract, multiply and divide by anything, as long as you do the same thing to both sides of the equals sign. In an equation, the equals sign acts like the fulcrum of a balance: if you add 5 of something to one side of the balance, you have to add the same amount to the other side to keep the balance steady. The same thing goes for an equation - doing the same operation to both sides keeps the meaning of the equation true. RULE #2: to move or cancel a quantity or variable on one side of the equation, perform the "opposite" operation with it on both sides of the equation. For example if you had g-1=w and wanted to isolate g, add 1 to both sides (g-1+1 = w+1). Simplify (because (-1+1)=0) and end up with g = w+1. RULE #2: to move or cancel a quantity or variable on one side of the equation, perform the "opposite" operation with it on both sides of the equation. For example if you had g-1=w and wanted to isolate g, add 1 to both sides (g-1+1 = w+1). Simplify (because (-1+1)=0) and end up with g = w+1.

5. Step By Step Here are some simple steps for manipulating equations. Assess what you have (which of the variables do you have values for?, what units are present?, etc.). DO NOT plug in any numbers yet! Determine which of the variables you want as your answer. (What is the question asking you to calculate? What is the unknown variable?) Rearrange the equation so that the unknown variable is by itself on one side of the equals sign (=) and all the other variables are on the other side. RULE #1: you can add, subtract, multiply and divide by anything, as long as you do the same thing to both sides of the equals sign. NOW plug in the numbers! Replace known variables with their values and don't forget to keep track of units! Determine the value of the unknown variable by performing the mathematical functions. That is, add, subtract, multiply and divide according to the equation you wrote for step 2. Ask yourself whether the answer is reasonable in the context of what you know about the geosciences and how much things should weigh.

6. Practice: Problems P 1 V 1 =P 2 V 2 solve for V 2 PV=nRT solve for R V 1 /T 1 =V 2 /T 2 solve for T 2 Q=mc  T solve for c M 1 V 1 =M 2 V 2 solve for V 1 Answers (P 1 V 1 )\ P 2 =V 2 (PV)\nT=R (V 2 T 1 )/V 1 =T 2 Q\m  T =c V 1 =M 2 V 2 \M 1

7. Rules for Working with Significant Figures: Leading zeros are never significant. Imbedded zeros are always significant. Trailing zeros are significant only if the decimal point is specified. Hint: Change the number to scientific notation. It is easier to see. Addition or Subtraction: The last digit retained is set by the first doubtful digit. Multiplication or Division: The answer contains no more significant figures than the least accurately known number.

8. Sigfig Zeros Leading zeros are never significant.Leading zeros are never significant. Imbedded zeros are always significant.Imbedded zeros are always significant. Trailing zeros are significant only if the decimal point is specified. Hint: Change the number to scientific notation. It is easier to see.Trailing zeros are significant only if the decimal point is specified. Hint: Change the number to scientific notation. It is easier to see.0.006821.072300300.300.0

9. Practice: Sigfigs in Measurements Worksheet

10. Length Matters! (+ & -)

11. Less is More! ( × & ÷ )

12. Notes on Rounding When rounding off numbers to a certain number of significant figures, do so to the nearest value. example: Round to 3 significant figures: 2.3467 x 10 4 (Answer: 2.35 x 10 4 ) example: Round to 2 significant figures: 1.612 x 10 3 (Answer: 1.6 x 10 3 ) What happens if there is a 5? There is an arbitrary rule: If the number before the 5 is odd, round up. If the number before the 5 is even, let it be. The justification for this is that in the course of a series of many calculations, any rounding errors will be averaged out. example: Round to 2 significant figures: 2.35 x 10 2 (Answer: 2.4 x 10 2 ) example: Round to 2 significant figures: 2.45 x 10 2 (Answer: 2.4 x 10 2 ) Of course, if we round to 2 significant figures: 2.451 x 10 2, the answer is definitely 2.5 x 10 2 since 2.451 x 10 2 is closer to 2.5 x 10 2 than 2.4 x 10 2.

13. Self Quiz Give the correct number of significant figures for 4500, 4500., 0.0032, 0.04050 Give the answer to the correct number of significant figures: 4503 + 34.90 + 550 = ? Give the answer to the correct number of significant figures: 1.367 - 1.34 = ? Give the answer to the correct number of significant figures: (1.3 x 10 3 )(5.724 x 10 4 ) = ? Give the answer to the correct number of significant figures: (6305)/(0.010) = ? 2, 4, 2, 4 50880.03 74000000 or 7.4 X 10 7 630000 or 6.3 X 10 5

14. Practice: Sigfig Calculations Worksheet

15. Dimensional Analysis Dimensional Analysis (also called Factor-Label Method or the Unit Factor Method) is a problem-solving method that uses the fact that any number or expression can be multiplied by one without changing its value. It is a useful technique. The only danger is that you may end up thinking that chemistry is simply a math problem - which it definitely is not. Unit factors may be made from any two terms that describe the same or equivalent "amounts" of what we are interested in. For example, we know that 1 inch = 2.54 centimeters Note: Unlike most English-Metric conversions, this one is exact. There are exactly 2.540000000... centimeters in 1 inch. We can make two unit factors from this information:

16. Try It! Now, we can solve some problems. Set up each problem by writing down what you need to find with a question mark. Then set it equal to the information that you are given. The problem is solved by multiplying the given data and its units by the appropriate unit factors so that only the desired units are present at the end. (1) How many centimeters are in 6.00 inches? (2) Express 24.0 cm in inches.

17. Solutions

18. You can also string many unit factors together. (3) How many seconds are in 2.0 years?

19. We also can use dimensional analysis for solving problems. (4) How many atoms of hydrogen can be found in 45 g of ammonia, NH 3 ? We will need three unit factors to do this calculation, derived from the following information: 1 mole of NH 3 has a mass of 17 grams. 1 mole of NH 3 contains 6.02 x 10 23 molecules of NH 3. 1 molecule of NH 3 has 3 atoms of hydrogen in it.

20. Practice: Real Life Chemistry Worksheet