Chapter 1 – Chemical Measurements

1. 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.

2. The seven fundamental SI measurements

3. 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).

4. 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.

5. 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?

6. 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

7. 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?

8. 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

9. Chemical Concentrations • Solute – the minor component • Solvent – the major component • Concentration – a measure of the amount of solute per unit of solution or solvent

10. 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.

11. 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

12. 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?

13. 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

14. 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 xVconc = Mdil xVdil (The producton 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?

15. 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 xVconc = Mdil xVdil (The producton 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

16. 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.

17. 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?

18. 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.

19. End, Monday, August 30