Chapter 1 Chemical Measurements

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Chapter 1 Chemical Measurements

• Chemical measurements are made in the System
  International (SI). The fundamental SI units are
  given in Table 1-1. You have already used these
  fundamental units in previous science courses.

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The seven fundamental SI measurements
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• Secondary units are constructed from mathematical
  operations on the fundamental units for example,
• Area length x length m2
• Volume length x length x length m3
• Often the pure SI secondary unit is converted to
  a derived unit by the use of definitions and the
  prefixes of the SI system for example,
• Volume 1 liter (L) 0.001 m3 and
• 1 milliliter (mL) 1 x 10?6 m3
• Additional secondary units may also be formed
  such as the deciliter (dL) which 0.1 liter. (d
  is deci or 1/10).

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• In a similar way, the fundamental unit of mass
• is the kilogram (kg), but in chemical analysis a
• kilogram is too large the quantities most often
• used are the gram (g), milligram 10-3(mg), or
• the microgram 10-6 ?g.

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Conversion between units

• Within the SI and its derived quantities,
  conversion is done by simply noting the powers of
  ten. For example, what is the density in kg/m3
  of a material whose density is 3.30 g/mL?

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Conversion between units

• Within the SI and its derived quantities,
  conversion is done by simply noting the powers of
  ten. For example, what is the density in kg/m3
  of a material whose density is 3.30 g/mL?
• 3.30 g/mL X 1.00 kg/1000 g x 1 mL/10?6 m3
• 3.30 x 10 3 kg/ m3

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Conversion between units

• Between the SI and other systems
• What is the density in pounds/ft3 of a material
  whose density is 3.30 g/mL?

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Conversion between units

• Between the SI and other systems
• What is the density in pounds/ft3 of a material
  whose density is 3.30 g/mL?
• 3.30 g/mL x 1 lb/453.6 g x (1mL/1cm3)
• x (2.54 cm/in)3 x (12in/1ft)3
• 206 lb/ft3

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Chemical Concentrations

• Solute the minor component
• Solvent the major component
• Concentration a measure of the amount of solute
  per unit of solution or solvent

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Chemical Concentrations

• There are several different units routinely used
  in expressing concentrations in chemistry
• Molarity (M) moles of solute / L of solution.
• Molality (m) moles of solute / kg of solvent
• Formality same as molarity except used for
  salts which actually exist as collection of ions
  instead of molecules.

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Chemical Concentrations

• There are several different units routinely used
  in expressing concentrations in chemistry
• Weight - mass of solute x 100 / mass of
  solution
• Volume - volume of solute x 100 / volume of
  solution
• Parts per million (ppm) 1/106 µg of solute /
  mL of solution or mg/L
• Parts per billion (ppb) 1/109 ng of solute /
  mL or µg / mL

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Concentration Problems

• The density of stock nitric acid (HNO3) is 1.41
  g/mL and its weight is 70.4. What is the
  molarity and molality of the stock solution?

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Concentration Problems

• The density of stock nitric acid (HNO3) is 1.41
  g/mL and its weight is 70.4. What is the
  molarity and molality of the stock solution?
• Consider 1 liter of solution or 1000 mL
• The mass of 1 L 1.41 g/mL x 1000 mL 1410 g
• The mass of HNO3 0.704 x 1410 992.6 g
• The mol 992.6 g/ 63.01 g/mol 15.75 mol
  since we started with 1 L, the molarity 15.75 M
• The mass of water 1410 992.6 417.4 g or
  0.4174 kg
• The molality 15.75 mol/0.4174 kg 37.74 m

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Concentration Problems

• Frequently the analyst will need to dilute a
  stock solution of one concentration of a desired
  concentration. This operation uses the
  relationship
• Mconc x Vconc Mdil x Vdil
• (The product on either side is mol)
• How many mL of 12.0 molar HCl are needed to
  prepare 500 mL
• of a 1.50 M solution of HCl?

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Concentration Problems

• Frequently the analyst will need to dilute a
  stock solution of one concentration of a desired
  concentration. This operation uses the
  relationship
• Mconc x Vconc Mdil x Vdil
• (The product on either side is mol)
• How many mL of 12.0 molar HCl are needed to
  prepare 500
• mL of a 1.50 M solution of HCl?
• 12.0 M x Vconc 500 mL x 1.50 M
• Vconc (500 x 1.50) / 12.0 62.5 mL

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Concentration Problems

• The dilution of a solution such as described in
  the previous problem is generally accomplished
  using volumetric glassware such as volumetric
  flasks, pipets, etc. Pictured below is a 500
  mLvolumetric flask.

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Concentration Problems

• Solutions of low concentrations are expressed in
  ppm or ppb as described earlier. For solutions
  so dilute, the mass of the solution ? solvent,
  because so little solute is present.
• How would one prepare a 100 ppm sodium ion
  solution from
• solid NaCl?

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Concentration Problems

• Solutions of low concentrations are expressed in
  ppm or ppb as described earlier. For solutions
  so dilute, the mass of the solution ? solvent,
  because so little solute is present.
• How would one prepare a 100 ppm sodium ion
  solution from
• solid NaCl?
• 100 ppm ? 100 ?g Na/1mL
• Since 500 mL are desired, 100 ?g x 500 mL 5 x
  104 ?g
• 0.0500 g
• Grams of NaCl needed ar 0.0500 g x (NaCl/Na)
  0.127 g
• Explanation weigh out 0.127 g of NaCl, dissolve
  in water and dilute in 500mL volumetric flask to
  the mark.

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End, Monday, August 30