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Name ________ DIMENSIONAL ANALYSIS (FACTOR LAB EL METHOD) Using this method, it is possible to solve many problems by using the relationship of one unit to another . For example, 12 inches =one foot. Since these two numbers represent the same value , the fractions 12 in/1 ft and 1 ft/12 in are both equal to one. When you multiply another number by the number one, you do not change its value . However, you may change its unit. Example 1: Co nvert 2 miles to inches. 2 miles x 5.280 ft x 12 inches = 126,720 in 1 mile 1 ft (Using significant figures, 2 mi = 100,000 in.) Example 2: How many seconds are in 4 days? 4 days x 24 hrs x 60 min x 60 sec = 345,600 sec 1 day 1 hr 1 min (Using significant figures, 4 days = 300,000 sec.) Solve the following problems. Write the answers in significant figures. 1. 3 hrs = sec 2. 0.035 mg = cg 3. 5.5 kg = Ibs 4. 2.5 yds = in 5. 1.3 yrs = hr (1 yr =365 days) 6. 3 moles = molecules (1 mole = 6. 02 x 10 23 molecules) 7. 2.5 x 10 24 molecules = ___ moles 8. 5 moles = ___ liters (1 mole = 22.4 liters) 9. 100. liters = ___ moles 10. 50. liters = ___ molecules 11. 5.0 x 10 24 molecules = liters 12. 7.5x103 mL = liters Chemistry f F8766 6 ©Instructional Fair, Inc.

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Page 1: DIMENSIONAL ANALYSIS - Physicsmsbuenconsejo.weebly.com/uploads/2/2/6/9/22695210/chem_i_summer... · DIMENSIONAL ANALYSIS Name _____ (FACTOR LABEL METHOD) Using this method, it is

Name ________DIMENSIONAL ANALYSIS (FACTOR LABEL METHOD) Using this method it is possible to solve many problems by using the relationship of one unit to another For example 12 inches =one foot Since these two numbers represent the same value the fractions 12 in1 ft and 1 ft12 in are both equal to one When you multiply another number by the number one you do not change its value However you may change its unit

Example 1 Convert 2 miles to inches 2 miles x 5280 ft x 12 inches = 126720 in

1mile 1 ft (Using significant figures 2 mi = 100000 in)

Example 2 How many seconds are in 4 days 4 days x 24 hrs x 60 min x 60 sec = 345600 sec

1 day 1hr 1 min (Using significant figures 4 days =300000 sec)

Solve the following problems Write the answers in significant figures

1 3 hrs = sec

2 0035 mg = cg

3 55 kg = Ibs

4 25 yds = in

5 13 yrs = hr (1 yr =365 days)

6 3 moles = molecules (1 mole =602 x 1023 molecules)

7 25 x 1024 molecules = ___ moles

8 5 moles = ___ liters (1 mole =224 liters)

9 100 liters = ___ moles

10 50 liters = ___ molecules

11 50 x 1024 molecules = liters

12 75x103mL = liters

Chemistry fF8766 6 copyInstructional Fair Inc

- -

Name _---______METRICS AND MEASUREMENT In the chemistry classroom and lab the metric system of measurement is used so it is important to be able to convert from one unit to another

mega kilo hecto deca

(M) (k) (da)

1000000 1 000 100 10

106 103 102 10 _ _--

Factor Label Method 1 Write the given number and unilt

Basic Unit gram (g) liter (L)

meter (m)

deci centi milli micro

(d) (c) (m) (p)

1 01 001 000001 102 103 10610

2 Set up a conversion factor (fraction used to convert one unit to another) a Place the given unit as denominator of conversion factor b Place desired unit as numerator c Place a 1 in front of the larger unit d Determine the number of smaller units needed to make 1 of the larger unit

3 Cancel units Solve the problem

Example 1 55 mm =__ m Example 2 88 km =__ m

55 rnm 11m =0055 m I 1000 m = 88000 m 1000Jrull 1jmi

Example 3 7000 em =__ hm Example 4 8 dol = __ dL

7000)Rr1 1JgtI I 1 hm = 07 hm 8 de1t I lOx 10 dL =800 dL 100cm lOOpgt( 1 dert 1)

The factor label method can be used to solve virtually any problem including changes in units It is espeCially useful in making complex conversions dealing with concentrations and derived units Convert the following

1 35 mL = dL 6 4500 mg = g

2 950 g = kg 7 25cm = mm

3 275 mm = cm 8 0005 kg = dag

4 1000 L = kL 9 0075 m = cm

5 1000 mL = L 10 15 g = mg

Chemistry IF8766 7 copyInstructlonal Fair Inc

Name _________SCI NTIFIC NOTATION Scientists very often deal with very small and very large numbers which can lead to a lot of confusion when counting zeros We have learned to express these numbers as powers of 10 Scientific notation takes the form of M x 10n where 1 M lt 10 and nil represents the number of decimal places to be moved Positive n indicates the standard form is a large number Negative n indicates a number between zero and one

Example 1 Convert 1500000 to scientific notation We move the decimal point so that there is only one digit to its left a total of 6 places

1500000 = 15 x 106

Example 2 Convert 0000025 to scientific notation For this we move the decimal point 5 places to the right

0000025 = 25 x 1Q-5 (Note that when a number starts out less than one the exponent is always negative)

Convert the following to scientific notation

1 0005 = 6 025 = 2 5050 = 7 0025 = 3 00008 = 8 00025 =

4 1000 = 9 500 =

5 1000000 = 10 5000 =

Convert the following to standard notation

1 15 X 103 = 6 335 X 10-1 =

2 15 X 10-3 = 7 12 x 10-4 =

3 375 X 10-2 = 8 lxl()4=

4 375 x 102 = 9 1 X 10-1 =

5 22x lOS = 10 4xl00 =

Chemistry IF8766 8 copyInstructional Fair Inc

Name _-----_-----___---SIGNIFICANT -FIGURES A measurement can only be as accurate and precise as the instrument that produced it A scientist must be able to express the accuracy of a number not just its numerical value We can determine the accuracy of a number by the number of significant figures it contains

1) All digits 1-9 inclusive are significant Example 129 has 3 significant figures

2) Zeros between significant digits are always significant Example 5007 has 4 significant figures

3) Trailing zeros in a number are significant QDjy if the number contains a decimal point Example 1000 has 4 significant figures

100 has 1 significant figure 4) Zeros in the beginning of a number whose only function

is to place the decimal point are not significant Example 00025 has 2 significant figures

5) Zeros following a decimal significant figure are significant Example 0000470 has 3 significant figures

047000 has 5 significant figures

Determine the number of significant figures in the following numbers 1 002 6 5000

2 0020 7 605100

3 501 B 00005

4 5010 9 01020

5 5000 10 10001

Determine the location of the last significant place value by placing a bar over the digit (Example 1700)

1 B040 6 90100

2 00300 7 47 x 10-8

3 6995 B 10BOO000

4 2000 x 102 9 301 X 1021

5 090100 10 0000410

Chemistry IF8766 9 copyInstructional Fair Inc

Name _________CALCULATIONS USING SIGNIFICANT FIGURES When multiplying and dividing limit and round to least number of significant figures in any of the factors

Example 1 230 cm x 432 cm x 19 cm = 188784 cm3

The answer is expressed as 190000 cm3 since 19 cm has only two significant figures

When adding and subtracting limit and round your answer to the least number of decimal places in any of the numbers that make up your answer

Example 2 12325 mL + 460 mL + 86257 mL = 255507 mL II The answer is expressed as 2555 mL since 460 ml has only lone decimal place

Perform the following operations expressing the answer in the correct number of significant figures

1 135 m x 2467 m = ____

2 1035 m2 + 42 m =____

3 1201 mL + 352 mL + 6 mL = ____

4 5546 g - 289 g =

5 021 cm x 32 cm x 1001 cm =

6 015cm + 115 cm + 2051 cm =

7 150L3 + 4L =

8 505 kg - 45025 kg =

9 1252 mm x 0115 mm x 0012 mm =

10 1278 x 103 m2 + 1 4267 x 102 m =

Chemistry IF8766 10 copyInstructional Fair Inc

  • 6 Demensional Analysis (Factor Label Method)
  • 7 Metrics and Measurement
  • 8 Scientific Notation
  • 9 Significant Figures
  • 10 Calculations Using Significant Figures
Page 2: DIMENSIONAL ANALYSIS - Physicsmsbuenconsejo.weebly.com/uploads/2/2/6/9/22695210/chem_i_summer... · DIMENSIONAL ANALYSIS Name _____ (FACTOR LABEL METHOD) Using this method, it is

- -

Name _---______METRICS AND MEASUREMENT In the chemistry classroom and lab the metric system of measurement is used so it is important to be able to convert from one unit to another

mega kilo hecto deca

(M) (k) (da)

1000000 1 000 100 10

106 103 102 10 _ _--

Factor Label Method 1 Write the given number and unilt

Basic Unit gram (g) liter (L)

meter (m)

deci centi milli micro

(d) (c) (m) (p)

1 01 001 000001 102 103 10610

2 Set up a conversion factor (fraction used to convert one unit to another) a Place the given unit as denominator of conversion factor b Place desired unit as numerator c Place a 1 in front of the larger unit d Determine the number of smaller units needed to make 1 of the larger unit

3 Cancel units Solve the problem

Example 1 55 mm =__ m Example 2 88 km =__ m

55 rnm 11m =0055 m I 1000 m = 88000 m 1000Jrull 1jmi

Example 3 7000 em =__ hm Example 4 8 dol = __ dL

7000)Rr1 1JgtI I 1 hm = 07 hm 8 de1t I lOx 10 dL =800 dL 100cm lOOpgt( 1 dert 1)

The factor label method can be used to solve virtually any problem including changes in units It is espeCially useful in making complex conversions dealing with concentrations and derived units Convert the following

1 35 mL = dL 6 4500 mg = g

2 950 g = kg 7 25cm = mm

3 275 mm = cm 8 0005 kg = dag

4 1000 L = kL 9 0075 m = cm

5 1000 mL = L 10 15 g = mg

Chemistry IF8766 7 copyInstructlonal Fair Inc

Name _________SCI NTIFIC NOTATION Scientists very often deal with very small and very large numbers which can lead to a lot of confusion when counting zeros We have learned to express these numbers as powers of 10 Scientific notation takes the form of M x 10n where 1 M lt 10 and nil represents the number of decimal places to be moved Positive n indicates the standard form is a large number Negative n indicates a number between zero and one

Example 1 Convert 1500000 to scientific notation We move the decimal point so that there is only one digit to its left a total of 6 places

1500000 = 15 x 106

Example 2 Convert 0000025 to scientific notation For this we move the decimal point 5 places to the right

0000025 = 25 x 1Q-5 (Note that when a number starts out less than one the exponent is always negative)

Convert the following to scientific notation

1 0005 = 6 025 = 2 5050 = 7 0025 = 3 00008 = 8 00025 =

4 1000 = 9 500 =

5 1000000 = 10 5000 =

Convert the following to standard notation

1 15 X 103 = 6 335 X 10-1 =

2 15 X 10-3 = 7 12 x 10-4 =

3 375 X 10-2 = 8 lxl()4=

4 375 x 102 = 9 1 X 10-1 =

5 22x lOS = 10 4xl00 =

Chemistry IF8766 8 copyInstructional Fair Inc

Name _-----_-----___---SIGNIFICANT -FIGURES A measurement can only be as accurate and precise as the instrument that produced it A scientist must be able to express the accuracy of a number not just its numerical value We can determine the accuracy of a number by the number of significant figures it contains

1) All digits 1-9 inclusive are significant Example 129 has 3 significant figures

2) Zeros between significant digits are always significant Example 5007 has 4 significant figures

3) Trailing zeros in a number are significant QDjy if the number contains a decimal point Example 1000 has 4 significant figures

100 has 1 significant figure 4) Zeros in the beginning of a number whose only function

is to place the decimal point are not significant Example 00025 has 2 significant figures

5) Zeros following a decimal significant figure are significant Example 0000470 has 3 significant figures

047000 has 5 significant figures

Determine the number of significant figures in the following numbers 1 002 6 5000

2 0020 7 605100

3 501 B 00005

4 5010 9 01020

5 5000 10 10001

Determine the location of the last significant place value by placing a bar over the digit (Example 1700)

1 B040 6 90100

2 00300 7 47 x 10-8

3 6995 B 10BOO000

4 2000 x 102 9 301 X 1021

5 090100 10 0000410

Chemistry IF8766 9 copyInstructional Fair Inc

Name _________CALCULATIONS USING SIGNIFICANT FIGURES When multiplying and dividing limit and round to least number of significant figures in any of the factors

Example 1 230 cm x 432 cm x 19 cm = 188784 cm3

The answer is expressed as 190000 cm3 since 19 cm has only two significant figures

When adding and subtracting limit and round your answer to the least number of decimal places in any of the numbers that make up your answer

Example 2 12325 mL + 460 mL + 86257 mL = 255507 mL II The answer is expressed as 2555 mL since 460 ml has only lone decimal place

Perform the following operations expressing the answer in the correct number of significant figures

1 135 m x 2467 m = ____

2 1035 m2 + 42 m =____

3 1201 mL + 352 mL + 6 mL = ____

4 5546 g - 289 g =

5 021 cm x 32 cm x 1001 cm =

6 015cm + 115 cm + 2051 cm =

7 150L3 + 4L =

8 505 kg - 45025 kg =

9 1252 mm x 0115 mm x 0012 mm =

10 1278 x 103 m2 + 1 4267 x 102 m =

Chemistry IF8766 10 copyInstructional Fair Inc

  • 6 Demensional Analysis (Factor Label Method)
  • 7 Metrics and Measurement
  • 8 Scientific Notation
  • 9 Significant Figures
  • 10 Calculations Using Significant Figures
Page 3: DIMENSIONAL ANALYSIS - Physicsmsbuenconsejo.weebly.com/uploads/2/2/6/9/22695210/chem_i_summer... · DIMENSIONAL ANALYSIS Name _____ (FACTOR LABEL METHOD) Using this method, it is

Name _________SCI NTIFIC NOTATION Scientists very often deal with very small and very large numbers which can lead to a lot of confusion when counting zeros We have learned to express these numbers as powers of 10 Scientific notation takes the form of M x 10n where 1 M lt 10 and nil represents the number of decimal places to be moved Positive n indicates the standard form is a large number Negative n indicates a number between zero and one

Example 1 Convert 1500000 to scientific notation We move the decimal point so that there is only one digit to its left a total of 6 places

1500000 = 15 x 106

Example 2 Convert 0000025 to scientific notation For this we move the decimal point 5 places to the right

0000025 = 25 x 1Q-5 (Note that when a number starts out less than one the exponent is always negative)

Convert the following to scientific notation

1 0005 = 6 025 = 2 5050 = 7 0025 = 3 00008 = 8 00025 =

4 1000 = 9 500 =

5 1000000 = 10 5000 =

Convert the following to standard notation

1 15 X 103 = 6 335 X 10-1 =

2 15 X 10-3 = 7 12 x 10-4 =

3 375 X 10-2 = 8 lxl()4=

4 375 x 102 = 9 1 X 10-1 =

5 22x lOS = 10 4xl00 =

Chemistry IF8766 8 copyInstructional Fair Inc

Name _-----_-----___---SIGNIFICANT -FIGURES A measurement can only be as accurate and precise as the instrument that produced it A scientist must be able to express the accuracy of a number not just its numerical value We can determine the accuracy of a number by the number of significant figures it contains

1) All digits 1-9 inclusive are significant Example 129 has 3 significant figures

2) Zeros between significant digits are always significant Example 5007 has 4 significant figures

3) Trailing zeros in a number are significant QDjy if the number contains a decimal point Example 1000 has 4 significant figures

100 has 1 significant figure 4) Zeros in the beginning of a number whose only function

is to place the decimal point are not significant Example 00025 has 2 significant figures

5) Zeros following a decimal significant figure are significant Example 0000470 has 3 significant figures

047000 has 5 significant figures

Determine the number of significant figures in the following numbers 1 002 6 5000

2 0020 7 605100

3 501 B 00005

4 5010 9 01020

5 5000 10 10001

Determine the location of the last significant place value by placing a bar over the digit (Example 1700)

1 B040 6 90100

2 00300 7 47 x 10-8

3 6995 B 10BOO000

4 2000 x 102 9 301 X 1021

5 090100 10 0000410

Chemistry IF8766 9 copyInstructional Fair Inc

Name _________CALCULATIONS USING SIGNIFICANT FIGURES When multiplying and dividing limit and round to least number of significant figures in any of the factors

Example 1 230 cm x 432 cm x 19 cm = 188784 cm3

The answer is expressed as 190000 cm3 since 19 cm has only two significant figures

When adding and subtracting limit and round your answer to the least number of decimal places in any of the numbers that make up your answer

Example 2 12325 mL + 460 mL + 86257 mL = 255507 mL II The answer is expressed as 2555 mL since 460 ml has only lone decimal place

Perform the following operations expressing the answer in the correct number of significant figures

1 135 m x 2467 m = ____

2 1035 m2 + 42 m =____

3 1201 mL + 352 mL + 6 mL = ____

4 5546 g - 289 g =

5 021 cm x 32 cm x 1001 cm =

6 015cm + 115 cm + 2051 cm =

7 150L3 + 4L =

8 505 kg - 45025 kg =

9 1252 mm x 0115 mm x 0012 mm =

10 1278 x 103 m2 + 1 4267 x 102 m =

Chemistry IF8766 10 copyInstructional Fair Inc

  • 6 Demensional Analysis (Factor Label Method)
  • 7 Metrics and Measurement
  • 8 Scientific Notation
  • 9 Significant Figures
  • 10 Calculations Using Significant Figures
Page 4: DIMENSIONAL ANALYSIS - Physicsmsbuenconsejo.weebly.com/uploads/2/2/6/9/22695210/chem_i_summer... · DIMENSIONAL ANALYSIS Name _____ (FACTOR LABEL METHOD) Using this method, it is

Name _-----_-----___---SIGNIFICANT -FIGURES A measurement can only be as accurate and precise as the instrument that produced it A scientist must be able to express the accuracy of a number not just its numerical value We can determine the accuracy of a number by the number of significant figures it contains

1) All digits 1-9 inclusive are significant Example 129 has 3 significant figures

2) Zeros between significant digits are always significant Example 5007 has 4 significant figures

3) Trailing zeros in a number are significant QDjy if the number contains a decimal point Example 1000 has 4 significant figures

100 has 1 significant figure 4) Zeros in the beginning of a number whose only function

is to place the decimal point are not significant Example 00025 has 2 significant figures

5) Zeros following a decimal significant figure are significant Example 0000470 has 3 significant figures

047000 has 5 significant figures

Determine the number of significant figures in the following numbers 1 002 6 5000

2 0020 7 605100

3 501 B 00005

4 5010 9 01020

5 5000 10 10001

Determine the location of the last significant place value by placing a bar over the digit (Example 1700)

1 B040 6 90100

2 00300 7 47 x 10-8

3 6995 B 10BOO000

4 2000 x 102 9 301 X 1021

5 090100 10 0000410

Chemistry IF8766 9 copyInstructional Fair Inc

Name _________CALCULATIONS USING SIGNIFICANT FIGURES When multiplying and dividing limit and round to least number of significant figures in any of the factors

Example 1 230 cm x 432 cm x 19 cm = 188784 cm3

The answer is expressed as 190000 cm3 since 19 cm has only two significant figures

When adding and subtracting limit and round your answer to the least number of decimal places in any of the numbers that make up your answer

Example 2 12325 mL + 460 mL + 86257 mL = 255507 mL II The answer is expressed as 2555 mL since 460 ml has only lone decimal place

Perform the following operations expressing the answer in the correct number of significant figures

1 135 m x 2467 m = ____

2 1035 m2 + 42 m =____

3 1201 mL + 352 mL + 6 mL = ____

4 5546 g - 289 g =

5 021 cm x 32 cm x 1001 cm =

6 015cm + 115 cm + 2051 cm =

7 150L3 + 4L =

8 505 kg - 45025 kg =

9 1252 mm x 0115 mm x 0012 mm =

10 1278 x 103 m2 + 1 4267 x 102 m =

Chemistry IF8766 10 copyInstructional Fair Inc

  • 6 Demensional Analysis (Factor Label Method)
  • 7 Metrics and Measurement
  • 8 Scientific Notation
  • 9 Significant Figures
  • 10 Calculations Using Significant Figures
Page 5: DIMENSIONAL ANALYSIS - Physicsmsbuenconsejo.weebly.com/uploads/2/2/6/9/22695210/chem_i_summer... · DIMENSIONAL ANALYSIS Name _____ (FACTOR LABEL METHOD) Using this method, it is

Name _________CALCULATIONS USING SIGNIFICANT FIGURES When multiplying and dividing limit and round to least number of significant figures in any of the factors

Example 1 230 cm x 432 cm x 19 cm = 188784 cm3

The answer is expressed as 190000 cm3 since 19 cm has only two significant figures

When adding and subtracting limit and round your answer to the least number of decimal places in any of the numbers that make up your answer

Example 2 12325 mL + 460 mL + 86257 mL = 255507 mL II The answer is expressed as 2555 mL since 460 ml has only lone decimal place

Perform the following operations expressing the answer in the correct number of significant figures

1 135 m x 2467 m = ____

2 1035 m2 + 42 m =____

3 1201 mL + 352 mL + 6 mL = ____

4 5546 g - 289 g =

5 021 cm x 32 cm x 1001 cm =

6 015cm + 115 cm + 2051 cm =

7 150L3 + 4L =

8 505 kg - 45025 kg =

9 1252 mm x 0115 mm x 0012 mm =

10 1278 x 103 m2 + 1 4267 x 102 m =

Chemistry IF8766 10 copyInstructional Fair Inc

  • 6 Demensional Analysis (Factor Label Method)
  • 7 Metrics and Measurement
  • 8 Scientific Notation
  • 9 Significant Figures
  • 10 Calculations Using Significant Figures