Lecture 1: Introduction to Analytical Chemistry Flashcards
TWO TYPES OF DIGITS
EXACT AND INEXACT
- NUMBERS NOT MEASURED USING INSTRUMENTS
- INCLUDES DEFINED NUMBERS
- INFINITE NUMBER OF SFS
- DOESN’T AFFECT ACCURACY/PRECISION OF CALCULATIONS OR PRECISION
EXACT NUMBERS
WHAT TYPE OF DIGITS ARE THESE?
- 1 FOOT = 12 INCHES
- 100 YRS IN A CENTURY
EXACT NUMBERS
- NUMBERS WHICH ARE MEASUREMENTS MADE WITH INSTRUMENTS
- WITH INHERENT UNCERTAINTY FROM THE INSTRUMENTS
INEXACT NUMBERS
INDICATES THE PROBABLE UNCERTAINTY IN AN EXPERIMENTAL MEASUREMENT
SIGNIFICANT FIGURES
ALL THE DIGITS THAT ARE “CERTAIN” + FIRST UNKNOWN DIGIT
SIGNIFICANT FIGURES
TRUE OR FALSE:
NON ZERO DIGITS ARE ALWAYS SIGNIFICANT
TRUE
TRUE OR FALSE: ZERO IS SIGNIFICANT WHEN IT IS BETWEEN 2 NON-ZERO DIGITS
TRUE
TRUE OR FALSE:
THE FINAL/TRAILING ZEROS OF A NUMBER IS SIGNIFICANT
FALSE
- ONLY FOR THE DECIMAL PORTION
TRUE OR FALSE:
ZEROS BETWEEN DECIMAL POINT AND NON-ZERO DIGITS ARE SIGNIFICANT
FALSE
- THEY SERVE AS PLACEHOLDERS ONLY
ADDITION/SUBTRACTION
- ROUND ANSWER TO ______ NO. OF ______ OF ANY OF THE GIVEN NUMBERS
LEAST, DECIMAL PLACES
MULTIPLICATION/DIVISION
- ROUND ANSWER TO _______ NO. OF _______ OF ANY OF THE GIVEN NUMBERS
LEAST, SFS
SF IN LOGARITHM
NO. OF SFS IN THE ORIGINAL NUMBER = NO. OF SFS TO THE RIGHT OF DECIMAL POINT
SF IN ANTILOGARITHM
NO. OF SFS OF ANSWER = NO. OF SIGNIFICANT DECIMAL PLACES IN GIVEN
SHOULD ROUNDING OFF BE DONE DURING CALCULATIONS?
NO. AT THE END ONLY.
- ROUND OFF ONLY FINAL RESULT. PRELIMINARY ROUNDING OFF OF RESULTS IS DISCOURAGED
WHY SOULD ROUNDING OFF BE MINIMIZED DURING CALCULATIONS?
TO MINIMIZE DEVIATION OF FINAL ANSWER FROM CORRECT VALUE
INDICATE NO. OF SFS A. 0.0050 B. 3000 C. 1.5 x 10^-2 D. 500.0 E. 1.5050 x 10^3
A. 2 B. 1 C. 2 D. 4 E. 5
ROUND OFF ACCORDING TO SPECIFIED NO. OF SFS A. 15.9994 (4) B. 15.515 (4) C. 254.5 (3) D. 14.5001 (2)
A. 16.00
B. 15.52
C. 255
D. 15
PERFORM OPERATION. OBSERVE CORRECT NO. OF SFS
15.12 mL x 19.25 mL x 2.07 mL
602.4942 –> 602 (3 SF)
PERFORM OPERATION. OBSERVE CORRECT NO. OF SFS
4.5 in - 1.258 in
3,242 –> 3.2 (1 DECIMAL PLACE ONLY)
(2.34 - 1.2) x 0.105
1.1 x 0.105 = 0.1155 –> 0.12 (2 SF)
WHAT ARE THE WAYS OF EXPRESSING SOLUTION CONCENTRATION?
MMPPN
MOLARITY MOLALITY PERCENT COMPOSITION PARTS PER MILLION NORMALITY
MOLES OF SOLUTE DIVIDED BY LITERS OF SOLUTION
MOLARITY (M)
FORMULA AND UNITS OF MOLARITY
FORMULA: M = moles solute/L solution
UNITS: mol/L or M
MOLES OF SOLUTE DIVIDED BY KILOGRAMS OF SOLVENT
MOLALITY (m)
FORMULA AND UNITS OF MOLALITY
FORMULA: m = moles solute/kg solvent
UNIT: mol/kg or m
THREE TYPES OF PERCENT COMPOSITION
A. %w/w - % weight per weight
B. %v/v - % volume per volume
C. %w/v - %weight per volume
FORMULA FOR %w/w
%w/w = (g solute/g solution) x 100%
FORMULA FOR %v/v
%v/v = (mL solute/mL solution) x 100%
FORMULA FOR %w/v
%w/v = (g solute/mL solution) x 100%
FORMULA FOR ppm
EITHER %w/w or %w/v
ppm = (g solute/g solution) x 10^6
ppm = (g solute/mL solution) x 10^6
FORMULA FOR ppb
EITHER %w/w or %w/v
ppm = (g solute/g solution) x 10^9
ppm = (g solute/mL solution) x 10^
NUMBER OF EQUIVALENTS OF SOLUTE IN 1L SOLUTION
NORMALITY (N)
FORMULA FOR NORMALITY
N = no. eq solute/L solution
DESCRIBE THE AMOUNT OF A CHEMICAL SPECIES JUST LIKE MOLES
EQUIVALENTS (eq) OR MILLIEQUIVALENTS (meq)
MOLARITY: NORMALITY = MOLES: _______
EQUIVALENTS
MOLARITY: ______ = MOLES: EQUIVALENTS
NORMALITY
THIS IS THE DEFINITION OF THE “EQUIVALENT” AT THE EQUIVALENCE POINT OF ANY TITRATION
NO. OF eq ANALYTE PRESENT = NO. OF eq REAGENT ADDED
OR
NO. OF meq ANALYTE PRESENT = NO. OF meq REAGENT ADDED
WHY IS THERE NO NEED TO DERIVE THE STOICHIOMETRY BETWEEN THE ANALYTE AND TITRANT?
BECAUSE WE JUST HAVE TO CORRECTLY DEFINE THE eq OR meq FOR EACH TYPE OF REACTING SPECIES
THE AMOUNT OF SUBSTANCE CONTAINED IN ONE EQUIVALENT CAN VARY DEPENDING ON THE _________
TYPE OF REACTION
FORMULA FOR CALCULATING THE NUMBER OF EQUIVALENTS PRESENT IN A SAMPLE
no. eq = mass in g/EW
EW = EQUIVALENT WEIGHT OF ANALYTE