Significant Figure Rules

Chemistry 20 Math Review Booklet Significant Figure Rules

There are three rules on determining how many significant figures are in a number:

1. Non-zero digits are always significant. 2. Any zeros between two significant digits are significant. 3. All trailing zero’s are significant

Focus on these rules and learn them well. They will be used extensively throughout the remainder of this course.

Please remember that, in science, all numbers are based upon measurements (except for a very few that are defined). Since all measurements are uncertain, we must only use those numbers that are meaningful. A common ruler cannot measure something to be 22.4072643 cm long. Not all of the digits have meaning (significance) and, therefore, should not be written down. In science, only the numbers that have significance (derived from measurement) are written.

Rule 1: Non-zero digits are always significant.

Hopefully, this rule seems rather obvious. If you measure something and the device you use (ruler, thermometer, triple-beam balance, etc.) returns a number to you, then you have made a measurement decision and that ACT of measuring gives significance to that particular numeral (or digit) in the overall value you obtain.

Hence a number like 26.38 would have four significant figures and 7.94 would have three. The problem comes with numbers like 0.00980 or 28.09.

Rule 2: Any zeros between two significant digits are significant.

Suppose you had a number like 406. By the first rule, the 4 and the 6 are significant. However, to make a measurement decision on the 4 (in the hundred's place) and the 6 (in the one’s place), you HAD to have made a decision on the ten's place.

Rule 3: All final zeros or trailing zeros are significant. Here are two examples of this rule with the zeros this rule affects in boldface:

0.00500 0.03040

Here are two more examples where the significant zeros are in boldface:

2.30 x 10¯5 4.500 x 1012

What Zeros are Not Discussed Above

Zero Type #1: Space holding zeros on numbers less than one.

Here are the first two numbers from just above with the digits that are NOT significant in boldface:

0.00500 0.03040

These zeros serve only as space holders. They are there to put the decimal point in its correct location. They DO NOT involve measurement decisions. Upon writing the numbers in scientific notation (5.00 x 10¯3 and 3.040 x 10¯2), the non-significant zeros disappear.

Zero Type #2: the zero to the left of the decimal point on numbers less than one.

When a number like 0.00500 is written, the very first zero (to the left of the decimal point) is put there by convention. Its sole function is to communicate unambiguously that the decimal point is a decimal point. If the number were written like this, .00500, there is a possibility that the decimal point might be mistaken for a period. Many students omit that zero. They should not.

Zero Type #3: leading zeros in a whole number.

00250 has two significant figures. 005.00 x 10¯4 has three.

Exact Numbers Exact numbers, such as the number of people in a room, have an infinite number of significant figures. Exact numbers are counting up how many of something are present, they are not measurements made with instruments. Another example of this are defined numbers, such as 1 foot = 12 inches. There are exactly 12 inches in one foot. Therefore, if a number is exact, it DOES NOT affect the accuracy of a calculation nor the precision of the expression. Some more examples:

There are 100 years in a century.

2 molecules of hydrogen react with 1 molecule of oxygen to form 2 molecules of water.

There are 500 sheets of paper in one ream.

Interestingly, the speed of light is now a defined quantity. By definition, the value is 299,792,458 meters per second.

Practice Problems

Identify the number of significant figures:

1) 3.0800

2) 0.00418

3) 7.09 x 10¯5

4) 91,600

5) 0.003005

6) 3.200 x 109

7) 250

8) 780,000,000

9) 0.0101

10) 0.00800 Chemistry 20: Scientific Notation Practice Problems Part A: Express in correct scientific notation:

1. 61500

2. 0.000568

3. 321

4. 64960000

5. 0.07085

Part B: Express in standard form:

6. 1.09 x 103

7. 4.22715 x 108

8. 3.078 x 10-4

9. 9.004 x 10-2

10. 5.1874 x 102 Part C: Multiply or divide as indicated and express in correct scientific notation:

11. (2.85 x 107) (3.16 x 10-3)

12. (8.09 x 10-5) / (3.46 x 106)

13. (1.16 x 103) (5.09 x 10-7) / (2.45 x 102)

14. (4.06 x 10-5) (7.19 x 103) / (6.57 x 10-4)

15. (9.25 x 10-11) / (4.98 x 1026) (7.58 x 10-15)

Chemistry 20: Rearrangement of formulas Note: When rearranging formulas it is important to remember what you do to one side you must do to the other. Also, sometimes it simplifies the problem if you flip the equations over.

Example 1: Solve for displacement (d)

Example 2: Solve for time (t)

Rearrange the following equations: 1. Solve for force (F)

2. Solve for work (W)

3. Solve for mass (m) 4. Solve for velocity (v)

5. Solve for specific heat capacity (c)

6. Solve for molar mass (M)

7. Solve for volume (V) 8. Solve for the second volume (V2)

9. Solve for radius (r)

10. Solve for the initial velocity (Vi) Chemistry 20: Unit Conversions

The base unit for any measurement (distance, mass, etc…) has no prefix and is equal to 100 which is equal to 1.

When moving up the chart move the decimal to the left the number of spaces equal to the difference between the exponents for the two units involved. Remember as unit gets bigger the number gets smaller and as the unit gets smaller the number gets bigger.

ex: How many kg’s are in 500g?

g = 100 and kg = 103, therefore the difference between exponents is 3 so we move the decimal 3 places to the left.

500g = 0.500kg

When moving down the chart move the decimal to the right the number of spaces equal to the difference between the exponents for the two units involved. ex: How many millimeters in 2.00km?

mm = 10-3 and km = 103, therefore the difference between the exponents is 6 so we move the decimal 6 places to the right.

2.00km = 2000000mm or 2.00 x 106m Practice Problems

a) 100nm = ______m b) 420ng = ______g

c) 5.5 x 10-9m = ______nm d) 3.4 x 10-10g = ______µg

e) 6.6 x 10-12g = ______ng f) 120nm = ______cm

g) 6.1 x 10-7cm = ______nm h) 3.2 x 10-6cm = ______nm Converting non-SI Units

To convert non-SI units you must set up a conversion factor using known values that are equal to each other. For example, 1m = 100cm, therefore, the conversion factor is 1m/100cm or 100cm/1m depending on what you are trying to convert into.

You must multiply the term given by the conversion factor, making sure that the units desired are on the top of the conversion factor. This method will also work for converting SI units. This is sometimes called “need over got times got”. ex: How many minutes are in 432 seconds? Given: s = 432s Required: min = x Analysis: Conversion factor = 1min/60s 432s * 1min/60s = 7.2min Solution: 7.2min Paraphrase: 432 seconds is equal to 7.2 minutes

Convert the following units:

A. 3.2 min = ______s

B. 5600 min = ______days

C. 37 000 000s = ______years

D. 2.5hrs = ______s Dimensional Analysis

Dimensional Analysis is not for everyone. But it's probably for you.

Who should avoid Dimensional Analysis (DA)? Reasons for not using Dimensional Analysis

1. You're super-intelligent and enjoy solving relatively simple problems in the most complex manner.

2. You're tired of always getting the correct answers.

3. You're an arty type and you can't be confined by the structure of DA. You like messy solutions scribbled all over the page in every which direction. It's not that you want to make a mistake. But you really don't care that much about the answer. You just like the abstract design created by the free-wheeling solution... and the freedom from being confined by structure.

4. You have no interest in going to the prom or making the soccer team, and you don't mind being unpopular, unattractive, ignorant, insecure, uninformed, and unpleasant.

Otherwise,

You Need Dimensional Analysis!

Testimonials

"I was a Wilton High School student who dozed off while Mr. Laptick taught us dimensional analysis in physical science. I never quite got the hang of it. It irritated me... all of those fractions. I never really liked fractions. Although my grades had been pretty high, I got a D in physical science and subsequently dropped out of chemistry in the first quarter of my junior year. It was not long before I started on drugs, and then crime to support my drug habit. I have recently learned dimensional analysis and realize how simply it could have solved all of my problems. Alas, it is too late. I won't get out of prison until 2008 and even then, my self image is permanently damaged. I attribute all of my problems to my unwillingness to learn dimensional analysis." Jane

"I thought I knew everything and that sports was the only thing that mattered in high school. When Mr. Hoogenboom taught our class dimensional analysis, I didn't care about it at all. I was making plans for the weekend with my girlfriend who loved me because I was a running back and not because of physical science. While other kids were home solving dimensional analysis problems, I was practicing making end sweeps. Then one day I was hit hard. Splat. My knee was gone. I was despondent. My girl friend deserted me. My parents, who used to brag about my football stats, started getting on my case about grades. I decided to throw myself into my school work. But I couldn't understand anything. I would get wrong answers all of the time. I now realize that my failure in school came from never having learned dimensional analysis. Alas, I thought everyone else was smarter. After the constant humiliation of failing I finally gave up. I am worthless. I have no friends, no skills, no interests. I have now learned dimensional analysis, but it is too late." Bill

The evidence: Studies in Wilton High School from 1994-99 show that 100% of high school students who do not use and understand dimensional analysis are seriously insecure by their junior year. Damage done from this deprivation in the first two years of high school is probably permanent and cannot be overcome by learning the method later in life. We recommend mastering this skill before your junior year.

83% of the students who went to the senior proms from 1994-99 admitted that they enjoyed solving problems with Dimensional Analysis in order to impress and confuse their parents. Of the remaining 17%, 11% were home from the senior prom before 11 PM and 7 % went home alone.

What's the method?

Example 1 Example 2

This is a structured way of Convert 6.0 cm to km Convert 4.17 kg/m2 to g/cm2 helping you to convert units. With this method, you can easily and automatically convert very complex units if you have the conversion formulas. The method involves the following steps

1. Write the term to be 6.0 cm 4.17 kg converted, (both number and m2 unit)

2. Write the conversion 100 cm = .00100 km 1.00 m = 100 cm formula(s) 1.00 kg = 1000 g 3. Make a fraction of the .00100 km 1000 g 1.00 m 1.00 m conversion formula, such that 100 cm 1.00 kg 100 cm 100 cm the numerator is the unit you need and the denominator is the unit you have (got) previously

4. Multiply the term in step 1 6.0 cm * 00100 km 4.17kg 1000g 1.00m 1.00m by the fraction in step 3. Since 100 cm 1m2 1.00 kg 100 cm 100 cm the fraction equals 1, you can multiply by it without changing the size of the term.

5. Cancel units 6.0 cm .00100 km 4.17 kg 1000 g 1.00 m 1.00 m 100 cm 1 m2 1.00 kg 100 cm 100 cm

6. Perform the indicated .000060 km or 0.417 g -5 2 calculation rounding the 6.0 E -5 km (6.0 x 10 ) cm answer to the correct number of significant figures.

Dimensional Analysis Problem Set

You must use the formal method of dimensional analysis as taught in this class in order to get credit for these solutions (one point for each correct solution). Later in the course you may use any method of dimensional analysis to solve this type of problem.

1. Every three times I clean my bedroom, my mother makes me an apple pie. I cleaned my bedroom 9 times. How many apple pies does she owe me? (What? Your mother doesn't reward you for cleaning your bedroom? Aren't there child labor laws? To make up for that injustice, you may have this very easy extra credit problem.) 2. A chemistry teacher working at a golf camp during the summer found a liquid, which caused him to slice ball after ball into the water without disturbing him at all. He thought that this was an important liquid to identify so he set out to determine its density. He found that a sample of the liquid had a mass equal to 455 golf balls and occupied a volume of 620 water cups that he obtained at the 7th hole. Each golf ball massed 50 g and the water cups at the 7th hole of the golf course held 45 mL each. What is the density (g/mL) of the unknown liquid?

3. A Wilton High School senior was applying to college and wondered how many applications she needed to send. Her counselor explained that with the excellent grade she received in chemistry she would probably be accepted to one school out of every three to which she applied. [3 applications = 1 acceptance] She immediately realized that for each application she would have to write 3 essays, [1 application = 3 essays] and each essay would require 2 hours work [1 essay = 2 hours]. Of course writing essays is no simple matter. For each hour of serious essay writing, she would need to expend 500 calories [1 hour = 500 calories] which she could derive from her mother's apple pies [1 pie = 1000 calories]. How many times would she have to clean her room in order to gain acceptance to 10 colleges? Hopefully you didn't skip problem No 1. I'll help you get started.... 10 acceptances [ ] [ ] etc. 4. Because you never learned dimensional analysis, you have been working at a fast food restaurant for the past 35 years wrapping hamburgers. Each hour you wrap 184 hamburgers. you work 8 hours per day. you work 5 days a week. you get paid every 2 weeks with a salary of $840.34. How many hamburgers will you have to wrap to make your first one million dollars? [You are in a closed loop again. If you can solve the problem, you will have learned dimensional analysis and you can get a better job. But, since you won't be working there any longer, your solution will be wrong. If you can't solve the problem, you can continue working which means the problem is solvable, but you can't solve it. We have decided to overlook this impasse and allow you to solve the problem as if you had continued to wrap hamburgers.]