1 Slide 1 Calculations Used in Analytical Chemistry Chapter 4 Slide 2 Topics • International System of Units / SI Units • Mass and Weight • The Mole...
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Calculations Used in Analytical Chemistry Chapter 4
Slide 1
Topics • • • • •
International System of Units / SI Units Mass and Weight The Mole Units of Concentrations of solutions Stoichiometry
Slide 2
1
Seven SI BASE UNITS
Derived units: All other units can be derived from the seven base units
Slide 3
Table 4-1, p.72
Examples of Derived Units • 1 Joule=? • 1 Newton=? • 1 volt=?
Slide 4
2
Slide 5
Table 4-1, p.72
Mass and Weight • Mass (m): invariant measure of amount of matter • Weight
w = mg
• g: acceleration due to gravity
Slide 6
w = mg p.73
3
Calculating moles from mass Factor Label Method Review • Mass of TRIS that will react with 35 mL of 0.1 M HCl. • Moles of HCl in 35 mL mol HCl = VHCl x MHCl=
• Moles of TRIS
nTRIS = # molHCl ×
• Mass of TRIS mol TRIS x molar mass TRIS
mTRIS = # molTRIS ×
1molTRIS 1molHCl
121.14 gTRIS 1molTRIS
Use the factor label method : Prelab of Experiment 3 Slide 7
The Mole • Mole (mol): SI unit for the amount of a chemical species • 1 mole = 6.022 x 1023 particles • Molar mass: mass (in g) of one mole • Calculating molar masses using EXCEL (Chapter 3, p 60-67). – Importing data from Web Pages – Dealing with Character Strings (FIND function, MID function) – Using VLOOKUP to locate Data in a Worksheet
Slide 8
4
Solutions and their Concentrations • • • • •
Analytical molarity Equilibrium molarity Percent concentration Parts per Million/Billion P-Functions or p-value
Slide 9
Solutions and their Concentrations • Analytical molarity – Total number of moles of a solute in 1L of the solution (according to recipe). Example: NaCl, Na2SO4
• Equilibrium molarity – Molar concentration of a given species in solution
Slide 10
5
Percent Concentrations
weight − percent − ( w / w) =
weight − solute × 100% weight − solution
Volume − percent − (V / V ) =
volume − solute × 100% volume − solution
weight / volume − percent − ( w / v) =
weight − solute, g × 100% volume − solution, mL
Slide 11
Slide 12
Fig 4-1, p.82
6
Density and Specific Gravity of Solutions • Density of a substance is its mass per unit volume • Specific gravity: ratio of mass of substance to the mass of equal volume of water at 4ºC. • When using the metric system density and specific gravity can be used interchangeably, because the density of water is approximately 1.00 g/mL Slide 13
From % w/w to mol/L solution • 36.5 % HCl • FW = 36.46 • Specific gravity = 1.18 c HCl =
Back to Concentrations: Parts Per Million, Parts per Billion • For very dilute solutions, it is convenient to express concentrations in ppm or ppb • An approximation that is commonly used is that the density of dilute solution approaches the density of water (1.00 g/ml)
c ppm =
mass − solute × 10 6 ppm mass − solution
c ppb =
mass − solute × 10 9 ppb mass − solution
Units in numerator and denominator must be consistent! Slide 16
8
Back to Concentrations:p-functions or pvalues • The p-value is the negative log (base 10) of the molar concentration • P-values are convenient to use when changes occur over several orders of magnitude • [H+], pH • [Ba2+], pBa • [Cl-], pCl • [Cl-]=2.45 x 10-5M pCl=4.6108= 4.61?
Slide 17
Slide 18
Fig 4-2, p.84
9
Stoichiometric Calculations • Problem 4-36 – What mass of MgNH4PO4 precipitated when 200.00 mL of a 1.000% (w/v) solution of Mg Cl2 were treated with 40.0 mL of 0.1753 M Na3PO4 and an excess of NH4+? What was the molarity of the excess reagent (Na3PO4 or MgCl2) after the precipitation was complete?
• Steps – – – – – –
Write chemical equation and balance it Calculate moles of reagents mixed Determine the limiting reagent Calculate mass of precipitate Calculate moles of unreacted reagent Calculate concentration of unreacted reagent
Slide 19
Stoichiometry • Problem 4-6: • How many K ions are contained in 6.76 mol of K3PO4? • 1.22 x 1025 K+.
Slide 20
10
Dilution Factors • Problem 4-30 • Describe the preparation of 1.50 mL of 0.215 M NaOH from the concentrated commercial reagent [50% NaOH (w/w)]. Specific gravity = 1.525 • 50 % NaOH (w/w) = 19.06 M • Volume NaOH 50% (w/w) required = 16.9 mL