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CHEM 130
Fall 2011 – A. Vieau
Significant Figures
all of the numbers in a measurement
(the ones we are certain about, plus the first uncertain number)
Rules for counting SF:
1)
Some numbers have infinite number of SF
• counted numbers
89 students
• definitions
1000 m = 1 km
• “exact numbers”
2.54 cm = 1 in
2)
Nonzero integers are always significant: 1,256 = 4 SF’s
3)
Zeros:
a)
Captured zeros (between other digits) are always significant: 1006 = 4 SF’s
b)
Leading zeros (preceding other digits) are never significant: 0.039 = 2 SF’s
c)
Trailing zeros (to the right of other digits) can be ambiguous. Trailing zeroes
are significant only if the value includes a decimal point, or a dotted line
separating significant from nonsignificant digits:
1,420 = 3 SF's
10,000 = 1 SF's
10,00M0 = 4 SF's
1,420M = 4 SF's
10M000 = 2 SF's
10,000M = 5 SF' s
1,410.0 = 5 SF's
10,0M00 = 3 SF's
10,000.0 = 6 SF's
Rules for calculating with SF:
Rule 1)
Addition, subtraction, averages: Look at the decimal place.
!
251.322
+ 3.1
254.4M22 = 254.4
Rule 2)
Multiplying or dividing: Count the number of SF
(2.5631)(1.12) = 2.870672 = 2.87
!
Rules for rounding off numbers:
1)
If the digit to be dropped is less than 5, simply drop the digit.
8.92M1 = 8.92
2)
If the digit to be dropped is greater than 5, increase the proceeding digit by one.
7.45M62 = 7.46
3)
If the digit to be dropped is EXACTLY 5, then either drop it or increase the
!
proceeding digit by one, to make the last SF an EVEN number.
5.32M5 = 5.32
!
67.1M500 = 67.2
Rules for setting up calculations:
1)
Indicate what is being calculated
! equation
2)
Show the
3)
Proceed with the calculation in a logical sequence.
4)
Express intermediate answers with one non-significant digit (use a dashed line.)
5)
Box the final answer with the appropriate number of SF’s and units.
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CHEM 130
Fall 2011 – A. Vieau
SIGNIFICANT FIGURE WORKSHEET
1) Indicate the number of significant figures in the following:
a) 2.508 L
d) 12 in/1 ft
g) 29 mm
b) 0.00210 km
e) 20 students
h) 3.50 x 103 nm
c) 3000 g
f) 45.00 g
I) 8.34M5 g/mL
2) Round each of the following numbers to the indicated number of significant figures:
a) 96.528 cm2 to 3 SF
b) 1275 mL to 2 SF
d) 0.036559
m to 4 SF
!
e) 0.14500001 lb to 2 SF
c) 26.485 g to 4 SF
f) 6.0221367 x 1023 to 4 SF
3) What is the assumed absolute uncertainty (au) of the following?
a) 0.0770 g
d) 1.56 x 103 m
b) 450 mL
e) 4500 km
c) 35M14 cm2
f) 25.0 lb
4) Perform the following calculations, giving the answer properly rounded to the correct
number of significant figures.
!
a) 65g + 25.86g + 145.1g =
c) 2.5 x102 + 2.6x103 + 3.4x10-1 =
b) 415.1g – 25.56g =
d)
4.582 g + 4.54 g + 4.60 g
=
3
5) What is the assumed relative uncertainty (ru) of the following?
a) 0.0770 g
d) 1.56 x 103 m
b) 450 mL
e) 4500 km
c) 35M14 cm2
f) 25.0 lb
6) Perform the following calculations, giving the answer properly rounded to the correct
number of significant figures.
!
8.015 x 10!3 (0.550)
" 1 mm %
a) 6.72 m $ !3 ' =
c)
=
# 10 m&
1.9x103
60.00 x 35
b)
d) (22.5 cm)(125.6 cm)(1.4 cm) =
=
0.0050
(
)
7) Perform the following calculations, giving the answer properly rounded to the correct
number of significant figures. Hint: You might want to show an intermediate
answer.
35.46 g / mol " 35.4527 g / mol
x 100% =
35.4527 g / mol
! 2.09 J $
b) Q = 50.0 g ## o && 28.5 o C ' 26.2 oC =
" g C %
a)
(
!
)
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CHEM 130
Fall 2011 – A. Vieau
Scientific Notation, Significant Figures & Precision and Accuracy
A) Use scientific notation to express each of the following:
1)
2)
3)
3.14
0.08206
150.0
4)
5)
6)
150
0.150
8314
7)
8)
9)
22.4
0.003
0.0030
B) How many significant figures are in each of the following?
10) 24.305
11) 260
12) 47.90
13) 0.726
14) 800
15) 14.007
16) 0.81
17) 0.082060
18) 1.620 x 10-2
C) Express each of the following with 3 significant figures:
19) 3.14159
20) 0.08206
21) 6.022 x 1023
22) 1.0080
23) 15.994
24) 0.9983
25) 35.453
26) 1.625 x 10-2
27) 150.4
D) Complete the following calculations. Be sure to express your answer with
the correct number of significant figures.
28) (12.0)(2.54) =
29) 0.250/1.057 =
453.1 + 2.102
30)
=
16.0
3.24 x 1022
31)
=
6.023 x 1023
32) 1.00/6.023 x 1023 =
33) (6.023 x 1023)(2.5 x 10-4) =
E)
34) 3.2 + 14.19 =
35) 3.41 x 104 + 6.3 x 103 =
7.2 x 10-4
36)
=
8.0 x 102
37) 4.8 x 103/7.99 x 10-2 =
38) 7.70 x 10-1/1.1 x 103 =
39)
(5.4 x 102)(7.99 x 10-3)
(4.0 x 101)(9 x 10-5)
=
The mass of a 5.000 gram piece of iron is measured 3 times on 2 different
balances with the following results:
Trial
Balance 1 (g)
Balance 2 (g)
1
5.01
4.97
2
4.99
4.88
3
5.02
4.91
40) What is the average for balance 1? For balance 2?
41) What is the average deviation for balance 1? For balance 2?
42) Based on these measurements, which balance is more accurate? Which is more
precise?
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