Transcript I. Using Measurements

CH. 2 - MEASUREMENT
I. Using Measurements
I
(p. 44 - 57)
II
III
C. Johannesson
A. Accuracy vs. Precision
 Accuracy - how close a measurement is
to the accepted value
 Precision - how close a series of
measurements are to each other
ACCURATE = CORRECT
PRECISE = CONSISTENT
C. Johannesson
B. Percent Error
 Indicates accuracy of a measurement
experim ent
al  literature
% error
 100
literature
your value
accepted value
C. Johannesson
B. Percent Error
 A student determines the density of a
substance to be 1.40 g/mL. Find the % error if
the accepted value of the density is 1.36 g/mL.
% error
1.40 g/m L 1.36 g/m L
1.36 g/m L
% error = 2.9 %
C. Johannesson
 100
C. Significant Figures
 Indicate precision of a measurement.
 Recording Sig Figs
 Sig figs in a measurement include the
known digits plus a final estimated digit
2.35 cm
C. Johannesson
C. Significant Figures
 Counting Sig Figs (Table 2-5, p.47)
 Count all numbers EXCEPT:
 Leading
zeros -- 0.0025
 Trailing
zeros without
a decimal point -- 2,500
C. Johannesson
C. Significant Figures
Counting Sig Fig Examples
1. 23.50
4 sig figs
2. 402
3 sig figs
3. 5,280
3 sig figs
4. 0.080
2 sig figs
C. Johannesson
C. Significant Figures
 Calculating with Sig Figs
 Multiply/Divide - The # with the fewest
sig figs determines the # of sig figs in
the answer.
(13.91g/cm3)(23.3cm3) = 324.103g
4 SF
3 SF
3 SF
324 g
C. Johannesson
C. Significant Figures
 Calculating with Sig Figs (con’t)
 Add/Subtract - The # with the lowest
decimal value determines the place of
the last sig fig in the answer.
3.75 mL
+ 4.1 mL
7.85 mL  7.9 mL
C. Johannesson
224 g
+ 130 g
354 g  350 g
C. Significant Figures
 Calculating with Sig Figs (con’t)
 Exact Numbers do not limit the # of sig
figs in the answer.
 Counting
 Exact
 “1”
numbers: 12 students
conversions: 1 m = 100 cm
in any conversion: 1 in = 2.54 cm
C. Johannesson
C. Significant Figures
Practice Problems
5. (15.30 g) ÷ (6.4 mL)
4 SF
2 SF
= 2.390625 g/mL  2.4 g/mL
2 SF
6. 18.9 g
- 0.84 g
18.06 g  18.1 g
C. Johannesson
D. Scientific Notation
65,000 kg  6.5 × 104 kg
 Converting into Sci. Notation:
 Move decimal until there’s 1 digit to its
left. Places moved = exponent.
 Large # (>1)  positive exponent
Small # (<1)  negative exponent
C. Johannesson
 Only include sig
figs.
D. Scientific Notation
Practice Problems
7. 2,400,000 g
2.4 
8. 0.00256 kg
2.56 
9. 7  10-5 km
0.00007 km
10. 6.2  104 mm
62,000 mm
C. Johannesson
6
10
g
-3
10
kg
D. Scientific Notation
 Calculating with Sci. Notation
(5.44 × 107 g) ÷ (8.1 × 104 mol) =
Type on your calculator:
5.44
EXP
EE
7
÷
8.1
EXP
EE
4
EXE
ENTER
= 671.6049383 = 670 g/mol = 6.7 × 102 g/mol
C. Johannesson
E. Proportions
 Direct Proportion
y
y x
x
 Inverse Proportion
1
y
x
y
C. Johannesson
x