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Dimensional analysisand Units of Measurements

Dimensional analysis

• Dimensional analysis uses conversion factors to convert from one unit to another.

• Also called Factor Label (and railroad tracks)

• You do this in your head all the time– How many quarters are in 4 dollars?

Dimensional analysis practice

3 Big Mac = 7 salads9 salads = 2 slices of pepperoni pizza

22 slices of pepperoni pizza = 27 Sonic cokes Ex. 1) What number of Big Macs equal 365.4 salads? Ex. 2) How many sonic cokes do you have to drink to

equal 11 salads?

Units of MeasurementMeter m

Liter L

Celsius C

Mass is the amount of matter, weight is a measure of the gravitational pull on matter

SI UnitsPrefix Symbol Scientific

notationFactor Example

Mega M 1 x 106 1,000,000 megagram (Mg)

Kilo k 1 x 103 1,000 kilometer (km)

Hecto h 1 x 102 100 hectoliter (hL)

Deka da or (D) 1 x 101 10 dekagram (Dg)

BASE UNIT 1 x 100 1 meter

Deci d 1 x 10-1 .1 deciliter (dL)

Centi c 1 x 10-2 .01 centimeter (cm)

Milli m 1 x 10-3 .001 milligram (mg)

Micro u 1 x 10-6 .000001 microgram (ug)

Nano n 1 x 10-9 .000000001 nanometer (nm)

Pico p 1 x 10-12 .000000000001 picogram (pg)

PracticeIn each pair below, circle the larger

Millimeter Centimeter

picometer Micrometer

kilogram Hectogram

deciliter millileter

Basic SI UnitsQuantity Base unitLength meter (m)Mass gram (g)Time second (s)Volume Liter (L)Temperature Kelvin (K) or Celsius (C)Amount of substance mole (mol)Heat & Energy joule (J)

Metric Conversions Practice

Ex. 3) 2.435 g __________________cg

Ex. 4) 23.8 mL = ________________kL

Ex. 5) 23.5 cs = ________________ns

Some Useful Conversions

Length:1 in = 2.54 cm1 mi = 5280 ft

Volume:1 cm3 = 1 mL1 L = 1.06 qt

Weight:1 kg = 2.2 lb16 oz = 1 lb1 ton = 2000 lb

Temperature

• 20°C = K

Use both the Kelvin and Celsius scale, to convert

Celsius + 273 = Kelvin

Kelvin -273 = Celsius

Temperature

• 20°C = 293 K




Temperature

• 20°C = 293 K• 373 K = °C




Temperature

• 20°C = 293 K• 373 K = 100 °C




Volume: measured in cubic centimeters (cm3) or liters

• 1 liter (L) = 1 cubic decimeter (dm3) = 1000 millileters (mL)• 1 mL= 1 cm3

• Volume can be measure by Length x x or the Water Displacement method

• Volume can be measure by Length x width x or the Water Displacement method

• Volume can be measure by Length x width x height or the Water Displacement method


Know the relationship between the following volume units…

L = mL = cm3 (or cc in medical lingo)



1 L = mL = cm3 (or cc in medical lingo)



1 L = 1000 mL = cm3 (or cc in medical lingo)



1 L = 1000 mL = 1000 cm3 (or cc in medical lingo)

Density

• Is the ratio of mass per unit of volume. How much matter is packed into a given amount of space

• Density = mass/volume

• D= m/v

• The Density of a substance stays regardless of the size of the sample. For example: if you cut a block of copper in half, you have decreased both the mass and volume, the ratio of the 2 stays the same. This is called an Intensive Physical Property.

• The Density of a substance stays constant regardless of the size of the sample. For example: if you cut a block of copper in half, you have decreased both the mass and volume, the ratio of the 2 stays the same. This is called an Intensive Physical Property.

• The appropriate units of density are:

• for solids• for liquids


• g/cm3 for solids• for liquids


• g/cm3 for solids• g/mL for liquids

Example problems:

• A sample of aluminum metal has a mass of 8.4g. The volume of the sample is 3.1 cm3. Calculate the Density of aluminum.

Example problems:

• A sample of aluminum metal has a mass of 8.4g. The volume of the sample is 3.1 cm3. Calculate the Density of aluminum.

• 8.4 g/3.1 cm3 =

Example problems:

• A sample of aluminum metal has a mass of 8.4 g. The volume of the sample is 3.1 cm3. Calculate the Density of aluminum.

• 8.4 g/3.1 cm3 = 2.7 g/cm3

Example problems:

• Diamond has a density of 3.26 g/cm3. What is the mass of a diamond that has a volume of 0.350 cm3?

Example problems:


• 3.26 g/cm3 x 0.350 cm3 =

Example problems:


• 3.26 g/cm3 x 0.350 cm3 = 1.14 g

Example problems:

• What is the volume of a sample of liquid mercury that has a mass of 76.2 g, given that the density of mercury is 13.6 g/mL?

Example problems:


76.2 g = 13.6 g/mL

Example problems:


76.2 g = 5.60 mL13.6 g/mL

Reliable Measurements

• refers to the closeness of the measure value is to the , or real, value.

• refers to how a series of measurements are to one another.


• Accuracy refers to the closeness of the measure value is to the , or real, value.



• Accuracy refers to the closeness of the measure value is to the accepted, or real, value.




• Precision refers to how a series of measurements are to one another.



• Precision refers to how close a series of measurements are to one another.

• is calculated by subtracting the value from the value.

• Error is calculated by subtracting the experimental value from the accepted value.

• The is the ratio of an error to an accepted value.

• The percent error is the ratio of an error to an accepted value.

% error = error x 100 = accepted value – experimental value x 100 accepted value accepted value

Example

• An experiment finds the density of lead to be 10.95 g/cm3. The literature value for the density of lead is 13.34 g/cm3.

The error: accepted value – experimental value=• 13.34 – 10.95 =

An experiment finds the density of lead to be 10.95 g/cm3. The literature value for the density of lead is 13.34 g/cm3.

The error: accepted value – experimental value=

• 13.34 – 10.95 = 2.39

An experiment finds the density of lead to be 10.95 g/cm3. The literature value for the density of lead is 13.34 g/cm3.

The error: accepted value – experimental value=• 13.34 – 10.95 = 2.39

The % error: error x 100 = accepted value

2.39 x 100 =13.34

The error: accepted value – experimental value=• 13.34 – 10.95 = 2.39

The % error: error x 100 = accepted value

2.39 x 100 = 17.9%13.34

Practice

• Sara’s lab shows the atomic mass of aluminum to be 28.9. What is her percent error if the accepted value is 27.0?

Practice


• 28.9 – 27.0 =

Practice


• 28.9 – 27.0 = 1.90

Practice


• 28.9 – 27.0 = 1.90• 1.90/27.0 x 100% =

Practice


• 28.9 – 27.0 = 1.90• 1.90/27.0 x 100% = 7.04%

Practice

• What is the percent error in a measurement of the boiling point of bromine if the textbook value is 60.8 °C and the lab value is 40.6 °C?

Practice


• 60.8 °C – 40.6 °C =

Practice


• 60.8 °C – 40.6 °C = 20.2 °C

Practice


• 60.8 °C – 40.6 °C = 20.2 °C• 20.2 °C / 60.8 °C x 100% =

Practice


• 60.8 °C – 40.6 °C = 20.2 °C• 20.2 °C / 60.8 °C x 100% = 33.2%

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