chap 3: Basic tools of Analytical Chemistry
Chem 232: Quantitative Analysis Lecture Notes
Scott Hu�man
September 8, 2014
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Topic
1 chap 3: Basic tools of Analytical Chemistry
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition: SF
Signi�cant Figures:
minimum number of digits required to express a value inscienti�c notation without loss of accuracy.
harris p39
SF means Signi�cant Figures
used in measurements and calculated numbers. NOT in counts
where the numbers are certain or integers.
How many people in the room?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition: SF
Signi�cant Figures:
minimum number of digits required to express a value inscienti�c notation without loss of accuracy.
harris p39
SF means Signi�cant Figures
used in measurements and calculated numbers. NOT in counts
where the numbers are certain or integers.
How many people in the room?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition: SF
Signi�cant Figures:
minimum number of digits required to express a value inscienti�c notation without loss of accuracy.
harris p39
SF means Signi�cant Figures
used in measurements and calculated numbers. NOT in counts
where the numbers are certain or integers.
How many people in the room?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: SF
If you have a number written:
1.2356 g
the 6 is the least signi�cant digit. This means that you are the
least con�dent in its accuracy. Another way of saying this is that it
has (the most ) some uncertainty.
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: SF
If you have a number written:
1.2356 g
the 6 is the least signi�cant digit. This means that you are the
least con�dent in its accuracy. Another way of saying this is that it
has (the most ) some uncertainty.
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Learn By Example
Balance reports a mass of 1.2345 g. How many sf? -> 5
Balance reports a mass of 0.0021 g. How many sf? -> 2
Volume is measured at 5.001 mL. How many sf?
4
Volume is measured at 5.001× 10−3 mL. How many sf?
4
NOTE: in regular numbers the decimal place is not important
for SFs?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Learn By Example
Balance reports a mass of 1.2345 g. How many sf? -> 5
Balance reports a mass of 0.0021 g. How many sf? -> 2
Volume is measured at 5.001 mL. How many sf?
4
Volume is measured at 5.001× 10−3 mL. How many sf?
4
NOTE: in regular numbers the decimal place is not important
for SFs?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Learn By Example
Balance reports a mass of 1.2345 g. How many sf? -> 5
Balance reports a mass of 0.0021 g. How many sf? -> 2
Volume is measured at 5.001 mL. How many sf?
4
Volume is measured at 5.001× 10−3 mL. How many sf?
4
NOTE: in regular numbers the decimal place is not important
for SFs?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Learn By Example
Balance reports a mass of 1.2345 g. How many sf? -> 5
Balance reports a mass of 0.0021 g. How many sf? -> 2
Volume is measured at 5.001 mL. How many sf?
4
Volume is measured at 5.001× 10−3 mL. How many sf?
4
NOTE: in regular numbers the decimal place is not important
for SFs?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Learn By Example
Balance reports a mass of 1.2345 g. How many sf? -> 5
Balance reports a mass of 0.0021 g. How many sf? -> 2
Volume is measured at 5.001 mL. How many sf?
4
Volume is measured at 5.001× 10−3 mL. How many sf?
4
NOTE: in regular numbers the decimal place is not important
for SFs?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Learn By Example
Balance reports a mass of 1.2345 g. How many sf? -> 5
Balance reports a mass of 0.0021 g. How many sf? -> 2
Volume is measured at 5.001 mL. How many sf?
4
Volume is measured at 5.001× 10−3 mL. How many sf?
4
NOTE: in regular numbers the decimal place is not important
for SFs?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Scienti�c notation and Signi�cant Figures:
Scienti�c notation is useful, because the number of SFs is always
correct.
Ex. Take the following example: 5001 mL.
How many SF? 4
What about this example: 5000 mL.
How many SF? 4
This answer is ambiguous.
It could be 4 or 1 depending on the rules that you learned. So we
will used scienti�c notation whenever there is ambiguity. Convert
the previous example to scienti�c notation with 4 SFs
5.000× 103
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Scienti�c notation and Signi�cant Figures:
Scienti�c notation is useful, because the number of SFs is always
correct.
Ex. Take the following example: 5001 mL.
How many SF? 4
What about this example: 5000 mL.
How many SF? 4
This answer is ambiguous.
It could be 4 or 1 depending on the rules that you learned. So we
will used scienti�c notation whenever there is ambiguity. Convert
the previous example to scienti�c notation with 4 SFs
5.000× 103
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Scienti�c notation and Signi�cant Figures:
Scienti�c notation is useful, because the number of SFs is always
correct.
Ex. Take the following example: 5001 mL.
How many SF? 4
What about this example: 5000 mL.
How many SF? 4
This answer is ambiguous.
It could be 4 or 1 depending on the rules that you learned. So we
will used scienti�c notation whenever there is ambiguity. Convert
the previous example to scienti�c notation with 4 SFs
5.000× 103
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Scienti�c notation and Signi�cant Figures:
Scienti�c notation is useful, because the number of SFs is always
correct.
Ex. Take the following example: 5001 mL.
How many SF? 4
What about this example: 5000 mL.
How many SF? 4
This answer is ambiguous.
It could be 4 or 1 depending on the rules that you learned. So we
will used scienti�c notation whenever there is ambiguity. Convert
the previous example to scienti�c notation with 4 SFs
5.000× 103
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Scienti�c notation and Signi�cant Figures:
Scienti�c notation is useful, because the number of SFs is always
correct.
Ex. Take the following example: 5001 mL.
How many SF? 4
What about this example: 5000 mL.
How many SF? 4
This answer is ambiguous.
It could be 4 or 1 depending on the rules that you learned. So we
will used scienti�c notation whenever there is ambiguity. Convert
the previous example to scienti�c notation with 4 SFs
5.000× 103
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of Addition and Subtraction and SF:
When you add numbers together, how many signi�cant �gures do
you report in your answer?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Learn by Example: Easy example
1.0453× 10−1
+1.0000× 10−1
2.0453× 10−1
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Learn by Example: Easy example
1.0453× 10−1
+1.0000× 10−1
2.0453× 10−1
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Learn by Example: Harder Example
1.0453× 10−1
+1.00× 10−1
2.0453× 10−1
Is this correct?
NO!!!!! The second number contains the limiting number of SFs: 3
So, always use the limiting number of signi�cant �gures
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Learn by Example: Harder Example
1.0453× 10−1
+1.00× 10−1
2.0453× 10−1
Is this correct?
NO!!!!! The second number contains the limiting number of SFs: 3
So, always use the limiting number of signi�cant �gures
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of Multiplication and Division and SF:
Rules of Multiplication and Division and SF
When you multiply numbers together, how many signi�cant �gures
do you report in your answer?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Learn by Example: Easy example
1.0463× 10−1
×1.0000× 10−12
1.0463× 10−13
Is this correct?
yes
Both numbers contain the same number of SF
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Learn by Example: Easy example
1.0463× 10−1
×1.0000× 10−12
1.0463× 10−13
Is this correct?
yes
Both numbers contain the same number of SF
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Learn by Example: Harder example
1.0463× 10−1
×1.00× 10−12
1.0463× 10−13
Is this correct?
NO!!!!! The second number contains the limiting number of SFs: 3
So, always use the limiting number of signi�cant �gures
1.05 Ö 10-13
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Learn by Example: Harder example
1.0463× 10−1
×1.00× 10−12
1.0463× 10−13
Is this correct?
NO!!!!! The second number contains the limiting number of SFs: 3
So, always use the limiting number of signi�cant �gures
1.05 Ö 10-13
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Learn by Example: Harder example
1.0463× 10−1
×1.00× 10−12
1.0463× 10−13
Is this correct?
NO!!!!! The second number contains the limiting number of SFs: 3
So, always use the limiting number of signi�cant �gures
1.05 Ö 10-13
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of logarithms and antilogarithms and SF:
what is the:
log10(1) = 0
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10.4) = 1.017
The logarithm of a number contains two parts the
the characteristic is the integer count of how far from the
decimal the leading digit is
The mantissa is not an integer and therefore subject to sf rules.
log10( 16.1︸︷︷︸number
) = 1.207︸ ︷︷ ︸logarithm
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of logarithms and antilogarithms and SF:
what is the:
log10(1) = 0
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10.4) = 1.017
The logarithm of a number contains two parts the
the characteristic is the integer count of how far from the
decimal the leading digit is
The mantissa is not an integer and therefore subject to sf rules.
log10( 16.1︸︷︷︸number
) = 1.207︸ ︷︷ ︸logarithm
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of logarithms and antilogarithms and SF:
what is the:
log10(1) = 0
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10.4) = 1.017
The logarithm of a number contains two parts the
the characteristic is the integer count of how far from the
decimal the leading digit is
The mantissa is not an integer and therefore subject to sf rules.
log10( 16.1︸︷︷︸number
) = 1.207︸ ︷︷ ︸logarithm
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of logarithms and antilogarithms and SF:
what is the:
log10(1) = 0
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10.4) = 1.017
The logarithm of a number contains two parts the
the characteristic is the integer count of how far from the
decimal the leading digit is
The mantissa is not an integer and therefore subject to sf rules.
log10( 16.1︸︷︷︸number
) = 1.207︸ ︷︷ ︸logarithm
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of logarithms and antilogarithms and SF:
what is the:
log10(1) = 0
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10.4) = 1.017
The logarithm of a number contains two parts the
the characteristic is the integer count of how far from the
decimal the leading digit is
The mantissa is not an integer and therefore subject to sf rules.
log10( 16.1︸︷︷︸number
) = 1.207︸ ︷︷ ︸logarithm
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of logarithms and antilogarithms and SF:
what is the:
log10(1) = 0
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10.4) = 1.017
The logarithm of a number contains two parts the
the characteristic is the integer count of how far from the
decimal the leading digit is
The mantissa is not an integer and therefore subject to sf rules.
log10( 16.1︸︷︷︸number
) = 1.207︸ ︷︷ ︸logarithm
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of logarithms and antilogarithms and SF:
what is the:
log10(1) = 0
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10.4) = 1.017
The logarithm of a number contains two parts the
the characteristic is the integer count of how far from the
decimal the leading digit is
The mantissa is not an integer and therefore subject to sf rules.
log10( 16.1︸︷︷︸number
) = 1.207︸ ︷︷ ︸logarithm
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of logarithms and antilogarithms and SF:
what is the:
log10(1) = 0
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10.4) = 1.017
The logarithm of a number contains two parts the
the characteristic is the integer count of how far from the
decimal the leading digit is
The mantissa is not an integer and therefore subject to sf rules.
log10( 16.1︸︷︷︸number
) = 1.207︸ ︷︷ ︸logarithm
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of logarithms and antilogarithms and SF:
what is the:
log10(1) = 0
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10.4) = 1.017
The logarithm of a number contains two parts the
the characteristic is the integer count of how far from the
decimal the leading digit is
The mantissa is not an integer and therefore subject to sf rules.
log10( 16.1︸︷︷︸number
) = 1.207︸ ︷︷ ︸logarithm
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of logarithms and antilogarithms and SF:
what is the:
log10(1) = 0
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10.4) = 1.017
The logarithm of a number contains two parts the
the characteristic is the integer count of how far from the
decimal the leading digit is
The mantissa is not an integer and therefore subject to sf rules.
log10( 16.1︸︷︷︸number
) = 1.207︸ ︷︷ ︸logarithm
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rules of logarithms and antilogarithms and SF:
what is the:
log10(1) = 0
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10.4) = 1.017
The logarithm of a number contains two parts the
the characteristic is the integer count of how far from the
decimal the leading digit is
The mantissa is not an integer and therefore subject to sf rules.
log10( 16.1︸︷︷︸number
) = 1.207︸ ︷︷ ︸logarithm
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: example log
log10(0.001234) = -2.90868484
How many sf in the original number? 4
How many sf in the log? 4
so the result should be -2.9087
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: example log
log10(0.001234) = -2.90868484
How many sf in the original number? 4
How many sf in the log? 4
so the result should be -2.9087
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: example log
log10(0.001234) = -2.90868484
How many sf in the original number? 4
How many sf in the log? 4
so the result should be -2.9087
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: example log
log10(0.001234) = -2.90868484
How many sf in the original number? 4
How many sf in the log? 4
so the result should be -2.9087
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: example log
log10(0.001234) = -2.90868484
How many sf in the original number? 4
How many sf in the log? 4
so the result should be -2.9087
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: antilog
antilog10(4.37) = 2.344228× 104
how many sf in the original numer? 2
how many in the result? 2
so the answer should be 2.3 Ö 104
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: antilog
antilog10(4.37) = 2.344228× 104
how many sf in the original numer? 2
how many in the result? 2
so the answer should be 2.3 Ö 104
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: antilog
antilog10(4.37) = 2.344228× 104
how many sf in the original numer? 2
how many in the result? 2
so the answer should be 2.3 Ö 104
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: antilog
antilog10(4.37) = 2.344228× 104
how many sf in the original numer? 2
how many in the result? 2
so the answer should be 2.3 Ö 104
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: antilog
antilog10(4.37) = 2.344228× 104
how many sf in the original numer? 2
how many in the result? 2
so the answer should be 2.3 Ö 104
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding:
how do you round
1.2345 to the nearest
ones place 1.
tenths place 1.2
hundredths place 1.23
thousandths place 1.234 or 1.235. which is it?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding:
how do you round
1.2345 to the nearest
ones place 1.
tenths place 1.2
hundredths place 1.23
thousandths place 1.234 or 1.235. which is it?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding:
how do you round
1.2345 to the nearest
ones place 1.
tenths place 1.2
hundredths place 1.23
thousandths place 1.234 or 1.235. which is it?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding:
how do you round
1.2345 to the nearest
ones place 1.
tenths place 1.2
hundredths place 1.23
thousandths place 1.234 or 1.235. which is it?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding:
how do you round
1.2345 to the nearest
ones place 1.
tenths place 1.2
hundredths place 1.23
thousandths place 1.234 or 1.235. which is it?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding:
how do you round
1.2345 to the nearest
ones place 1.
tenths place 1.2
hundredths place 1.23
thousandths place 1.234 or 1.235. which is it?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding:
how do you round
1.2345 to the nearest
ones place 1.
tenths place 1.2
hundredths place 1.23
thousandths place 1.234 or 1.235. which is it?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding:
how do you round
1.2345 to the nearest
ones place 1.
tenths place 1.2
hundredths place 1.23
thousandths place 1.234 or 1.235. which is it?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Precollege Rounding:
digit past rounding digit is a 5 round up
problem: this provides a bias toward higher numbers
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Precollege Rounding:
digit past rounding digit is a 5 round up
problem: this provides a bias toward higher numbers
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding for the rest of your life
digit past rounding digit is > 5 round up
1.23456 round to thousandths place 1.235
digit past rounding digit is < 5 round down
1.2343 round to thousandths place 1.234
digit past rounding digit is exactly 5 round up and down.
how: many methods:
simplest is round to the nearest even signi�cant digit.
so our answer for 1.2345 rounded to thousandths place is 1.234
so our answer for 1.0135 rounded to thousandths place is 1.014
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding for the rest of your life
digit past rounding digit is > 5 round up
1.23456 round to thousandths place 1.235
digit past rounding digit is < 5 round down
1.2343 round to thousandths place 1.234
digit past rounding digit is exactly 5 round up and down.
how: many methods:
simplest is round to the nearest even signi�cant digit.
so our answer for 1.2345 rounded to thousandths place is 1.234
so our answer for 1.0135 rounded to thousandths place is 1.014
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding for the rest of your life
digit past rounding digit is > 5 round up
1.23456 round to thousandths place 1.235
digit past rounding digit is < 5 round down
1.2343 round to thousandths place 1.234
digit past rounding digit is exactly 5 round up and down.
how: many methods:
simplest is round to the nearest even signi�cant digit.
so our answer for 1.2345 rounded to thousandths place is 1.234
so our answer for 1.0135 rounded to thousandths place is 1.014
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding for the rest of your life
digit past rounding digit is > 5 round up
1.23456 round to thousandths place 1.235
digit past rounding digit is < 5 round down
1.2343 round to thousandths place 1.234
digit past rounding digit is exactly 5 round up and down.
how: many methods:
simplest is round to the nearest even signi�cant digit.
so our answer for 1.2345 rounded to thousandths place is 1.234
so our answer for 1.0135 rounded to thousandths place is 1.014
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding for the rest of your life
digit past rounding digit is > 5 round up
1.23456 round to thousandths place 1.235
digit past rounding digit is < 5 round down
1.2343 round to thousandths place 1.234
digit past rounding digit is exactly 5 round up and down.
how: many methods:
simplest is round to the nearest even signi�cant digit.
so our answer for 1.2345 rounded to thousandths place is 1.234
so our answer for 1.0135 rounded to thousandths place is 1.014
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding for the rest of your life
digit past rounding digit is > 5 round up
1.23456 round to thousandths place 1.235
digit past rounding digit is < 5 round down
1.2343 round to thousandths place 1.234
digit past rounding digit is exactly 5 round up and down.
how: many methods:
simplest is round to the nearest even signi�cant digit.
so our answer for 1.2345 rounded to thousandths place is 1.234
so our answer for 1.0135 rounded to thousandths place is 1.014
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding for the rest of your life
digit past rounding digit is > 5 round up
1.23456 round to thousandths place 1.235
digit past rounding digit is < 5 round down
1.2343 round to thousandths place 1.234
digit past rounding digit is exactly 5 round up and down.
how: many methods:
simplest is round to the nearest even signi�cant digit.
so our answer for 1.2345 rounded to thousandths place is 1.234
so our answer for 1.0135 rounded to thousandths place is 1.014
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Rounding for the rest of your life
digit past rounding digit is > 5 round up
1.23456 round to thousandths place 1.235
digit past rounding digit is < 5 round down
1.2343 round to thousandths place 1.234
digit past rounding digit is exactly 5 round up and down.
how: many methods:
simplest is round to the nearest even signi�cant digit.
so our answer for 1.2345 rounded to thousandths place is 1.234
so our answer for 1.0135 rounded to thousandths place is 1.014
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example Problem: Types of Error
2 types
systematic or deterministic
asystematic or random
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition
Systematic Error:
error that is deterministic or consistently o�set from the
correct value
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Detection of Systematic Error:
Analyze a known sample, such as a standard reference material
your method of measurement should reproduce the knownvalues
Analyze a blank sample,if your method gives a non-zero value, your method respondsto more than you expected. (interference)
Use di�erent analytical methods to measure the same quantity.
If the two methods don't agree, then there is error in one orboth of the methods.This is usually what is done in forensic labs. Agreementbetween methods is required.
Round robin experiment: di�erent people in severallaboratories analysis identical samples by the same or di�erentmethods.
Disagreement beyond estimated random error is systematicerror.
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Volumetric Glassware
What if a 100.0 mL volumetric �ask actually, consistently contains
a 100.12 mL.
This is corrected by performing all volume calculations with the
actual volume. Such as concentration: 0.0010000.1000 = 1.000× 10−2 M
actual concentration → 0.0010000.10012 = 0.9988× 10−2 M
Correction of Systematic Error
Calibration
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Random Uncertainty or Error
These are due to uncontrollable �uctuations in measurement.
These errors �uctuate above and below the �true� value.
Correction of Random Error
This type of error is corrected by many measurements or replication.
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: True Value is 10.390
10.079
10.579
10.871
10.469
10.419
mean = 10.483
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: True value is 10.390
10.356
10.850
10.858
10.303
10.462
10.261
10.245
10.022
10.281
10.341
10.054
10.980
10.617
10.124
10.145
mean = 10.393Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition: Accuracy
Accuracy:
nearness to the truth
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition: Precision
Precision:
reproducibility
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: recorded vs. reported
NOTE:
A special notation in this book is to write in your notes one extra
digit after the least signi�cant digit. But this digit is subscripted to
indicate that it is only to prevent rounding errors.
example:
1.2314 M
the 4 is not signi�cant, but if
We have to use this value in any additional calculations
We have kept and extra digit to help midigate rounding errors
We will call this notation the recorded value
We will call 1.231 M the reported value
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: gaps and commas
number convention Comment
1000000.017 di�cult to read
1,000,000.017 US easy for us to read
1.000.000,017 European strange for us
1 000 000.017 more universal book's convention
found in reference material
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: gaps and commas
number convention Comment
1000000.017 di�cult to read
1,000,000.017 US easy for us to read
1.000.000,017 European strange for us
1 000 000.017 more universal book's convention
found in reference material
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: gaps and commas
number convention Comment
1000000.017 di�cult to read
1,000,000.017 US easy for us to read
1.000.000,017 European strange for us
1 000 000.017 more universal book's convention
found in reference material
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: gaps and commas
number convention Comment
1000000.017 di�cult to read
1,000,000.017 US easy for us to read
1.000.000,017 European strange for us
1 000 000.017 more universal book's convention
found in reference material
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: gaps and commas
number convention Comment
1000000.017 di�cult to read
1,000,000.017 US easy for us to read
1.000.000,017 European strange for us
1 000 000.017 more universal book's convention
found in reference material
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: gaps and commas
number convention Comment
1000000.017 di�cult to read
1,000,000.017 US easy for us to read
1.000.000,017 European strange for us
1 000 000.017 more universal book's convention
found in reference material
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: gaps and commas
number convention Comment
1000000.017 di�cult to read
1,000,000.017 US easy for us to read
1.000.000,017 European strange for us
1 000 000.017 more universal book's convention
found in reference material
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: gaps and commas
number convention Comment
1000000.017 di�cult to read
1,000,000.017 US easy for us to read
1.000.000,017 European strange for us
1 000 000.017 more universal book's convention
found in reference material
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: gaps and commas
number convention Comment
1000000.017 di�cult to read
1,000,000.017 US easy for us to read
1.000.000,017 European strange for us
1 000 000.017 more universal book's convention
found in reference material
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: gaps and commas
number convention Comment
1000000.017 di�cult to read
1,000,000.017 US easy for us to read
1.000.000,017 European strange for us
1 000 000.017 more universal book's convention
found in reference material
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Special Notation: gaps and commas
number convention Comment
1000000.017 di�cult to read
1,000,000.017 US easy for us to read
1.000.000,017 European strange for us
1 000 000.017 more universal book's convention
found in reference material
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition: uncertainty
uncertainty:
expressed as a range or standard deviation
example: using a buret reading 12.54 ± 0.02 mL means that
the volume reading is between 12.52 and 12.56 mL
example: written on glassware
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition: absolute uncertainty
absolute uncertainty:
same units as measurments
Example: 12.01 ± 0.01 mL
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition: relative uncertainty
relative uncertainty:
a ratio of absolute uncertainty to magnitude of measurement
relative uncertainty =
absolute uncertainty
magnitude of measurement
dimensionless quantity
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition: Percent Relative Unvertainty
Percent Relative Uncertainty:
relative uncertainty times 100 %
Example 12.01 mL (± 0.08%)
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition: Propagation of Random Error
Propogation of Random Error :
a method for estimating the uncertainty in a determined value, that
was determined using multiple measurements.
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Addition and subtraction
+ 1.76 ( ± 0.03) e1 mL
+ 1.89 ( ± 0.03) e2 mL
- 0.59 ( ± 0.02) e3 mL
+ 3.06 ( ± e4) mL
what is e4?
You cannot just add the uncertainties together, why?
because of ±. There should be some cancellation.
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Addition and subtraction
+ 1.76 ( ± 0.03) e1 mL
+ 1.89 ( ± 0.03) e2 mL
- 0.59 ( ± 0.02) e3 mL
+ 3.06 ( ± e4) mL
what is e4?
You cannot just add the uncertainties together, why?
because of ±. There should be some cancellation.
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Addition and subtraction
+ 1.76 ( ± 0.03) e1 mL
+ 1.89 ( ± 0.03) e2 mL
- 0.59 ( ± 0.02) e3 mL
+ 3.06 ( ± e4) mL
what is e4?
You cannot just add the uncertainties together, why?
because of ±. There should be some cancellation.
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Propagating Random Uncertainty: Addition and Subtraction
derived in Appendix C using Calculus.
You will not be held responsible for coming up with these
equations, but you will have to remember them
e4 =√
e21+ e2
2+ e2
3
e4 =√
(0.03)2 + (0.03)2 + (0.02)2 = 0.046904 = 0.0470.047 ← recorded uncertainty
so ± 0.05 mL is the absolute uncertainty.
We you report the value
3.06 ± 0.05 mL
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Propagating Random Uncertainty: Addition and Subtraction
derived in Appendix C using Calculus.
You will not be held responsible for coming up with these
equations, but you will have to remember them
e4 =√
e21+ e2
2+ e2
3
e4 =√(0.03)2 + (0.03)2 + (0.02)2 = 0.046904 = 0.047
0.047 ← recorded uncertainty
so ± 0.05 mL is the absolute uncertainty.
We you report the value
3.06 ± 0.05 mL
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Propagating Random Uncertainty: Addition and Subtraction
derived in Appendix C using Calculus.
You will not be held responsible for coming up with these
equations, but you will have to remember them
e4 =√
e21+ e2
2+ e2
3
e4 =√(0.03)2 + (0.03)2 + (0.02)2 = 0.046904 = 0.047
0.047 ← recorded uncertainty
so ± 0.05 mL is the absolute uncertainty.
We you report the value
3.06 ± 0.05 mL
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Propagating Random Uncertainty: Addition and Subtraction
derived in Appendix C using Calculus.
You will not be held responsible for coming up with these
equations, but you will have to remember them
e4 =√
e21+ e2
2+ e2
3
e4 =√(0.03)2 + (0.03)2 + (0.02)2 = 0.046904 = 0.047
0.047 ← recorded uncertainty
so ± 0.05 mL is the absolute uncertainty.
We you report the value
3.06 ± 0.05 mL
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Using a Buret
Uncertainty in every reading is ± 0.02
Initial volume reading: 0.05 mL
Final volume reading: 17.88 mL
The volume delivered is the di�erence between these two numbers
17.88 (± 0.02) mL
- 0.05 (± 0.02) mL
17.83 (± e)
where e =√0.022 + 0.022 = 0.028 ≈ 0.03
So the volume delivered with the buret was 17.83 ± 0.03 mL
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
De�nition:
Propagating Random Uncertainty Multiplication and Division:
For multiplication �rst convert all uncertainties to percent relative
uncertainties then calculate the error of the product or quotient
using %e4 =√
(%e1)2 + (%e2)2 + (%e3)2
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Propagate the Uncertainty
1.76(±0.03)×1.89(±0.02)0.59(±0.02) = 5.64± e4
First convert ( ± ei) to ( ± %e1)
1.76(±1.7%)×1.89(±1.1%)0.59(±3.4%) = 5.64± e4
Next calc %e4%e4 =
√(1.7)2 + (1.1)2 + (3.4)2 = 4.0%
So the recorded value is 5.64 (± 4.0 %)
(NOTE: don't report the 4 just keep it for future calculations.)
Sometimes you want to covert the uncertainty back to absolute
uncertainty.
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Propagate the Uncertainty (continued)
Sometimes you want to covert the uncertainty back to absolute
uncertainty. 4.0%× 5.64 = 0.040 × 5.64 = 0.23
This gives the recorded value 5.64 (± 0.23).
And when you report this value you would report
5.6 (± 0.2) absolute uncertainty
or
5.6 (± 4%) relative uncertainty
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Mixed operations
What do you do for mixed operations?[1.76(±0.03)−0.59(±0.02)]
01.89(±0.02) = 0.6190 ± e4
First the numerator: [1.76(±0.03)− 0.59(±0.02)] = 1.17(±0.036)where 0.036 =
√(0.03)2 + (0.02)2
Then covert the two absolute uncertainties to % relative
uncertainties1.17(±0.036)1.89(±0.02) = 1.17(±3.1%)
1.89(±1.1%) = 0.6190(±3.3%)
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: mixed operations continued
and for the absolute uncertainty
0.033 × 0.6190 = 0.020
0.619 (± 0.020) absolute uncertainty
0.619 (± 3.3 %) % relative uncertainty
To report your values: �nal absolute uncertainty is 0.02 so the
hundredths place in the number 0.619 is the least certain number
NOT the thousandths place.
Reporting Values:
0.62 (± 0.02) absolute uncertainty
0.62 (± 3. %) % relative uncertainty
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
SF for real
REAL rule for Signi�cant Figures:
The �rst digit of the absolute uncertainty is the last digit in the
reported value.
So why did we have all those rules earlier about + - Ö ö
For test taking
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
SF for real
REAL rule for Signi�cant Figures:
The �rst digit of the absolute uncertainty is the last digit in the
reported value.
So why did we have all those rules earlier about + - Ö ö
For test taking
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
SF for real
REAL rule for Signi�cant Figures:
The �rst digit of the absolute uncertainty is the last digit in the
reported value.
So why did we have all those rules earlier about + - Ö ö
For test taking
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example: Why this is the real way
0.002 364(±0.000 003)0.025 00(±0.000 05) = 0.094 6(±0.000 2)and0.821(±0.002)0.803(±0.002) = 1.022(±0.004)
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
error propagation systematic
Also a problem
How do you �x this?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
error propagation systematic
Also a problem
How do you �x this?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example Problem: Exercise 3-B
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Problem 3-11
How do you �x this problem?
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Problem 3-20
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Example Problem: Exercise 3-C
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Problem 3-18
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Problem 3-19
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes
chap 3: Basic tools of Analytical Chemistry
Problem 3-4
Scott Hu�man Chem 232: Quantitative Analysis Lecture Notes