Transcript Sig Figs - Miss DiSpigna`s Chemistry Blog

DO NOW

How many significant figures are in the following
numbers?
•
•
•
638
•
3 sig figs
6,378,500
•
5 sig figs
0.000050780
•
5 sig figs
Significant Figures
Revisited
Ms. DiSpigna
Lab Chemistry
Howell High School
2008-2009
Rules for SIG FIGS
1.
•
2.
•
3.
•
4.
•
•
•
All non-zero numbers are significant.
Example- 5952 – has 4 sig figs
All zeros between non-zero numbers are
significant.
405 – has 3 sig figs
All zeros to the left of the number are not
significant.
0.0028 – has 2 sig figs
Zeros on the right of the number are only
significant if there is a decimal point.
1590 – has 3 sig figs
8260. – has 4 sig figs
0.0837 – has 3 sig figs
Examples
1.
2.
3.
4.
5.
6.
7.
2801.0
693
950
0.369
0.0570
48020.
62.01400
Significant Figures Revisited







Addition and Subtraction Rules
•
The answer must have the same number of digits to the
right of the decimal point as the value with the fewest digits
to the right of the decimal.
• Let’s say we wanted to add 1.24 mL, 12.4 mL, and 124 mL.
Setup values to determine which has the fewest number of digits
to the right of the decimal.
1.24 mL <~ Two digits
12.4 mL <~ One digit
124 mL <~~ Zero digit
Zero digits to the right of the decimal is the least therefore the
sum must have Zero digits to the right.
So, 137.64 becomes 138.
Significant Figures Revisited

Add the following
measurements and
report them to the
appropriate significant
figures.
•
28.0 cm, 23.538 cm, and
25.68 cm






28.0 cm <~ 1 digit
23.538 cm <~ 3 digits
25.68 cm <~ 2 digits
77.218 cm
1 digit is the least
amount so my answer
can have only 1 digit
past the decimal.
So answer becomes
77.2 cm
Significant Figures Revisited

Multiplication and Division Rules
• The answer must have the same number of sig
•





figs as the measurement with the fewest number
of sig figs.
Calculate the volume of a box with following
measurements
3.65 cm <~ 3 sig figs
3.20 cm <~ 3 sig figs
2.05 cm <~ 3` sig figs
3 sig figs is the fewest number of sig figs so answer must be
rounded to three sig figs.
So 23.944 cm3 becomes 23.9 cm3.
Significant Figures Revisited






Calculate the density of an object that has a
mass of 102.4 g and a volume of 50.0 mL.
• Density = Mass / Volume
102.4 g <~ 4 sig figs
50.0 mL <~ 3 sig figs
2.048 g/mL
Lowest number of sig figs is 3, therefore answer
must have only 3 sig figs.
So answer becomes 2.05 g/mL
Percent Error
How close your data is to the expected
data.
%error= l accepted - experimental l x 100
accepted
 Accepted- what the answer was
supposed to be
 Experimental- what you got in the
experiment

Example


Accepted value: 1.59 g/cm3
Experimental value: 1.60 g/cm3
l 1.59-1.60 l x 100
1.60
0.01 x 100
1.60
0.00625 x 100= 0.625%
Example

Sam does an experiment to determine the
density of an object. The density calculated
from the lab is 1.54g/cm3. The accepted
density is 1.59g/cm3. Determine Sam’s
percent error.
l 1.59 -1.54 l x 100
1.59
= 3.14%