Unit 2 Measurements

Matter - anything that occupies space and has mass

mass – measure of the quantity of matter

SI unit of mass is the kilogram (kg)

1 kg = 1000 g = 1 x 103 g

weight – force that gravity exerts on an object

A 1 kg bar will weigh1 kg on earth

0.1 kg on moon

weight = mass g (F = ma)

on earth, g = 1.0

on moon, g ~ 0.1

La Grande K 1 Kg Pt/Ir alloy

World’s Roundest Objecthttps://www.youtube.com/watch?v=ZMByI4s-D-Y

International System of Units (SI) Base Units

Used in this class and should have memorized

All other units are derived from these units and are known as Derived Units

Velocity: m/sForce: 1 Newton = 1 kg•m/s2

Volume: m3

Used most often in this class, be sure to memorize.

Prefixes can be used to simplify for extremely large or small quantities of base units

“mu”

Prefix examples

Driving 321,000 meters to LR = 321 kilometers

Radio station 90.9 MHz = 90,900,000 Hz

A mosquito weighs 2.5 milligrams (mg) = 0.0025 grams (g)

A dust mite’s length 0.0002 meters = 200 micrometers (mm)

10-6 ML = 1 L10-3 kL = 1 L

Liter (base)102 cL = 1 L

103 mL = 1 L106 mL = 1 L109 nL = 1 L

Conversion factors can be written/used 2 ways

1 Mm = 106 L1km = 103 LLiter (base)

1 cL = 10-2 L1 mL = 10-3 L1 mL = 10-6 L1 nL = 10-9 L

Or

I favor the forms using (+) exponents

Volume – SI derived unit for volume is cubic meter (m3)

1 cm3 = 1 mL

(cm3 is commonly used)

Density – SI derived unit for density is kg/m3

1 g/cm3 = 1 g/mL = 1000 kg/m3

density = mass

volume

d = mV

*more commonly used

2 L of Os = 100 lbs

Example 1.1

Gold is a precious metal that is chemically unreactive.It is used mainly in jewelry, dentistry, and electronic devices.

A piece of gold ingot with a mass of 301 g has a volume of 15.6 cm3. Calculate the density of gold.

gold ingots

Example 1.2

The density of mercury, the only metal that is a liquid at room temperature, is 13.6 g/mL. Calculate the mass of 5.50 mL of the liquid.

d = mV

Chemistry In Action

On 9/23/99, $125M Mars Climate Orbiter entered Mars’ atmosphere 100 km (62 miles) lower than planned and was destroyed by heat.

1 lb = 1 N

1 lb = 4.45 N

“This is going to be the cautionary tale that will be embedded into introduction to the metric system in elementary school, high school, and college science courses till the end of time.”

Failed to convert English to

metric units

Scientific NotationThe number of atoms in 12 g of carbon:

602,200,000,000,000,000,000,000

6.022 x 1023

The mass of a single carbon atom in grams:

0.0000000000000000000000199

1.99 x 10-23

N x 10nN is the base number between 1 and 10

Exponent (n) is a positive or negative integer

We can factor out powers of 10 to simplify very large or small numbers

Scientific Notation

• Base x 10exponent

• Base number ≥ 1 and < 10

= 3.2 x 105 Decimal moved left so (+)

= 7.4 x 10-5 Decimal moved right so (–)

.

.

320,000

0.000074

Scientific Notation Practice

Write these in long notation

• 2.0 x 103

• 3.58 x 10-4

• 4.651 x 107

• 9.87 x 10-2

Write these in scientific notation

• 0.00578

• 579

• 96,000

• 0.0140

Mathematics in Scientific NotationAddition or Subtraction: Must have same exponent

1. Write each quantity with the same exponent n

2. Combine N1 and N2

3. The exponent, n, remains the same

4.31 x 104 + 3.9 x 103 =

4.31 x 104 + 0.39 x 104 =

4.70 x 104

1.36 x 10-1 – 4 x 10-3 =

1.36 x 10-1 – 0.04 x 10-1 =

= 1.32 x 10-1

Tip: change smaller number to match larger exponent

More practice:1.45 x 10601 + 2.4 x 10600

Multiplication: Add exponents

1. Multiply N1 and N2

2. Add exponents n1 and n2

(4.0 x 10-5) x (7.0 x 103) =(4.0 x 7.0) x (10-5+3) =

28 x 10-2 =2.8 x 10-1

Division: Subtract exponents

1. Divide N1 and N2

2. Subtract exponents n1 and n2

8.5 x 104 ÷ 5.0 x 109 =(8.5 ÷ 5.0) x 104-9 =

1.7 x 10-5

Mathematics in Scientific Notation

Practice: (3 x 10250) (7.2 x 10200 )

Practice: 9.5 x 1045) ÷ (3.7 x 1050)

Bell Ringer

• 8,705,000 m

• 0.00237 sec

• 0.0000045 L

• 9,300 g

b) Rewrite above numbers using the nearest SI prefix

c) Perform the below mathematics in Sci. Notation

a) Write in scientific notation

• (9.01 x 103 g) + (3.8 x 102 g)

• (3.98 x 10-2 m) – (8.2 x 10-3 m)

• (2.61 x 107 m) x (9.87 x 10-2 m)

• (8.4 x 109 g) ÷ (2.0 x 104 mL)

Precision indicates to what degree we know our measurement. (Arithmetic precision)

A measurement of 18.0 grams could be made on an average countertop food scale (balance). (~$20)

A high-precision milligram scale could weigh the same sample with a much higher precision (18.0235 grams) (~$1,500)

Every measurement is limited by the equipment’s level of precision. (Never exact)

e.g. mass of hydrogen atom: 0.0000000000000000000000017 grams

Significant Figures: Used to prevent uncertainty from rounding of various measured quantities with various levels of precision.

1) Any digit that is not zero is significant1.234 kg 4 significant figures34,000 mm 2 significant figures

2) Zeros between nonzero digits are significant606 cm 3 significant figures50,050 s 4 significant figures

3) Zeros to the left of the first nonzero digit are not significant0.08 mL 1 significant figure

0.00054 ML 2 significant figures

4) If a number is greater than 1, then all zeros to the right of the decimal point are significant

2.0 mg 2 significant figures20.000 g 5 significant figures

5) If a number is less than 1, then only the zeros at the end are significant0.00420 g 3 significant figures0.1000 g 4 significant figures

Significant Figures

Every significant figure is shown when using Scientific notation.

0.001400 m____ 4 significant figures

1.400 x 10-3 Not 1.4 x 10-3

500 mL__ 2 significant figures

5.0 x 102 Not 5 x 102

Example

1.4 Unit Conversions

(a) 478 cm

(b) 600,001 g

(c) 0.85 m

(d) 0.0430 kg

(e) 1.310 × 1022 atoms

(f) 7000 mL

Determine the number of significant figures in the following measurements:

Example 1.4 Solution

(a) 478 cm -- Three, because each digit is a nonzero digit.

(b) 600,001- Six, because zeros between nonzero digits are significant.

(c) 0.825 m -- Three, because zeros to the left of the first nonzero digit do not count as significant figures.

(d) 0.0430 kg -- Three. The zero after the nonzero is significant because the number is less than 1.

(e) 1.310 × 1022 atoms -- Four, because the number is greater than one so all the zeros written to the right of the decimal point count as significant figures.

Example 1.4 solution

(f)7000 mL -- This is an ambiguous case. The number of significant figures may be four (7.000 × 103), three (7.00 × 103), two (7.0 × 103), or one (7 × 103).

This example illustrates why scientific notation must be used to show the proper number of significant figures.

If no decimal is present it is usually assumed only non-zeros are significant. If a decimal is present, than all zero’s are significant.

7,000 mL ≠ 7,000. mL

They display differing degrees of precision.

Significant Figures

Addition or Subtraction

The answer cannot have more digits to the right of the decimal point than any of the original numbers. Use the least precise number.

89.392 L1.1+

90.492 round off to 90.5

one significant figure after decimal point

3.70-2.91330.7867

two significant figures after decimal point

round off to 0.79

± 50 mL

± 1.0 mL

XX

XX

Significant Figures

Multiplication or Division

The number of significant figures in the result is set by the original number that has the smallest number of significant figures.

4.51 x 3.0006 = 13.532706 = 13.5

3 sig figsround to 3 sig figs

6.8 ÷ 112.04 = 0.0606926

2 sig figs round to 2 sig figs

= 0.061

Example 1.5Carry out the following arithmetic operations to the correct number of significant figures:

(a) 11,254.1 g + 0.1983 g

(b) 66.59 L − 3.113 L

(c) 8.16 m × 5.1355 kg

(d) 0.0154 kg ÷ 88.3 mL

(e) (2.64 × 103 cm) + (3.27 × 102 cm)

Example 1.5 Solution

Solution In addition and subtraction, the number of decimal places in the answer is determined by the number having the lowest number of decimal places.

(a)

(b)

Example 1.5 Solution

(c)

(d)

(e) First we change 3.27 × 102 cm to 0.327 × 103 cm and then carry out the addition (2.64 cm + 0.327 cm) × 103. Following the procedure in (a), we find the answer is 2.97 × 103 cm.

In multiplication and division, the significant number of the answer is determined by the number having the smallest number of significant figures.

Bell Ringer a) Perform the below mathematics in Sci. Notation. using

Significant Figures in your answer.

1. (9.8 x 105 g) + (6.75 x 104 g)

2. (5.98 x 10-6 m) – (7 x 10-8 m)

3. (2.612 x 1010 m) x (9.87 x 10-3 m)

4. (7 x 102 mg) ÷ (1.875 x 104 mL)

b) Rewrite the first 2 solutions using the nearest SI prefix

Significant Figures

Exact NumbersNumbers from definitions or numbers of objects are considered to have an infinite number of significant figures.

• The average of three measured lengths: 6.64, 6.68 and 6.70?

6.64 + 6.68 + 6.703

= 6.67333 = 7

Because 3 is an exact number, not a measured number; It is not used for sigfigs.

= 6.673

• How many feet are in 6.82 yards?

6.82 yards x 3 ft/yard

1 yard = exactly 3 ft by definition

= 20.5 ft = 20 ft

Accuracy – how close a measurement is to the true value

Precision – how close a set of measurements are to each other

accurate&

precise

precisebut

not accurate

not accurate&

not precise

Percent Error

A way to determine how accurate your measurements are to a known value.

|Obtained value – Actual value| x 100% Actual Value

Ranges between 0 and 100%

Ex. I weigh a 3 kg block on three different scales:3.2 kg, 3.0 kg, 3.1 kg = 3.1 kg average

3.1 – 3.0

3.0x 100% = 3.3% error

Dimensional Analysis of Solving Problems (Train-Tracks)

1. Determine which unit conversion factors are needed

2. Carry units through calculation

3. If all units cancel except for the desired unit(s), then the problem was solved correctly.

given quantity x conversion factor = desired quantity

given unit x = desired unit desired unit

given unit

Train Track ExampleHow many inches are in 3.0 miles?

Identify beginning information

Write measurement as a fraction

3 miles

Draw a train track

Train Track Example

How many inches are in 3.0 miles?

• We are going from a larger measurement to a smaller one.• Find a conversion factor you know that changes miles into

something smaller.

Conversion Factor: 1 mile = 5,280 feet

3 miles

5280 feet 1 mile

• Write your conversion factor on the track so that miles cancels out and you are left with the unit feet.

Always need same units on opposite sides to cancel out

3.0 miles

5280 feet 1 mile

12 inches 1 foot

Train Track Example

How many inches are in 3.0 miles?

We now need another conversion factor between Feet and Inches: 1 foot = 12 inches

Again, place conversion factor so that the previous unit cancels out.

Train Track Unit Conversions

How many inches are in 3.0 miles?

3.0 miles

5,280 feet 1 mile

Multiply all numbers on the topDivide all numbers on the bottom

3.0 x 5,280 x 12 1 x 1

12 inches 1 foot

= 190,080 inches

Inches are the only remaining unit ✔

= 1.9 x 105 inches(2 sig figs)

More practice:Convert 1.40 x 10-6 g to mg

Metric to Metric Conversion Problems

Convert 2.79 x 105 mm to km

2.79 x 105 mm

1 m 103 mm

2.79 x 105 103 x 103

1 km 103 m

= 2.79 x 10(5-3-3)

Kilometers are the only remaining units ✔

= 2.79 x 10-1 km(3 sig figs)

Don’t try to convert directly from mm to km. Go to the base unit (m) first

Conversion factors: 1,000 mm = 1 m; 1,000 m = 1 km

More practice:Convert 3.4 x 103 cg to mg

Example

Example

A person’s average daily intake of glucose (a form of sugar) is 0.0833 pound (lb). What is this mass in milligrams (mg)? (1 lb = 453.6 g.)

A metric conversion is needed to convert grams to milligrams (1 mg = 1 × 10−3 g)

(Or we could write: 1,000 mg = 1 g)Either Conversion factor will work

0.0833 lb 453.6 g

1 lb

103 mg

1 g = 3.78 x 104 mg(3 sig figs)

= 37,784.88 mg

2-D Conversion Problems (Unit1/Unit2)

Convert 70.0 miles/hour to m/s

70.0 miles 1 hour

1609 meter 1 mile

70.0 x 1,609 x 1 x 1 1 x 1 x 60 x 60

1 hour 60 min

= 31.286 m/s

Meter/sec are the only remaining units ✔

= 31.3 m/s

1 min 60 sec

We convert one unit at a time, followed by the other

Conversion factors: 1 mile = 1,609 meters; 1 hour = 60 min; 1 min = 60 sec

*Note: to cancel out hours (on bottom) it must appear again on the top

More practice:Convert 3.4 kg/L to g/mL

Example

2-D Conversion Problems (Unit#)

Convert 2.5 x 10-5 m3 to mm3

2.5 x 10-5 m3 1000 mm 1 m

2.5 x 10-5 x 109 = 2.5 x 104 mm3

Conversion factors: 1 m = 1,000 mm

More practice:Convert 34 yd2 to ft2

1 m3 ≠ 1,000 mm3

1 m3 = (1,000)3 mm3 ✔

1. Write the 1-D units first (1m = 103m)2. Add exponent to entire conversion factor

32.5 x 10-5 m3 109 mm3

1 m3=

Example

2-D Conversion Problems (Unit#)

Convert 6.70 x 103 ft2 to inches2

6.70 x 103 ft2 12 in. 1 ft

= 9.65 x 105 in2

Conversion factors: 1 ft = 12 in

More practice:Convert 34 yd3 to ft3

1 ft2 ≠ 12 in2

1 ft2 = 122 in2✔

1. Write the 1-D units first (1ft = 12 in)

Alternate 2nd Method2. Write the same conversion factor again

until they cancel

12 in. 1 ft

Example

Example

An average adult has 5.2 L of blood. What is the volume of blood in m3?

5.2 L 103 mL

1 L

1 cm3

1 mL

1 m3

1003 cm3

1 mL = 1 cm3

= 5.2 x 10-3 m3

Example 2-D Conversion Problem

The circumference of the earth is approximately 2.49 x 104 miles long.

If the speed of sound travels at 760 mph, how many days would it take a sound wave to circulate Earth’s circumference.

2.49 x 104 miles 1 hour

760 miles

1 day

24 hours

Conversion factor: 760 miles = 1 hour

= 1.4 days

Starting with length and velocity; needing time

More Practice Conversion problems

• Convert 3.0 mL to ounces (33.8 oz = 1 L)

• 1.67 Mm to mm

• 2.35 x 1012 inches to cm (1 ft = 0.305 m)

• 3.50 x 104 mL to cL

ReviewUnit Conversion & Significant

Figures: Crash Course Chemistry #2www.youtube.com/watch?v=hQpQ0hxVNTg

• 42.0 km/h to ft/ms

• 0.55 Acres to m2 (247 acre = 1 km2)

• 10.6 g/mm3 to kg/m3

Sample of Topics to Study

• Metric base units

• Derived unit

• Using Prefixes

• Significant Figures (+ math)

• Scientific Notation (+ math)

• % Error calculation

•Accuracy

• Precision

• Dimensional Analysis