This lecture discusses scientific measurements and units. It covers the metric system and SI units, dimensional analysis, unit conversions, and significant figures. Key points include: 1. The metric system uses meters, grams, and seconds as fundamental units. There are seven base SI units including the meter for length and gram for mass. 2. Dimensional analysis uses conversion factors to change between units while maintaining the correct dimensions. It is useful for solving chemistry problems. 3. Significant figures indicate the precision of a measurement and how numbers should be rounded. Calculations are rounded according to whether they involve multiplication/division or addition/subtraction.

1. Md. Faysal Ahamed Khan Lecture 3 Welcome to the class of Chemistry I Course No. CHEM 211 Credit hours 3

2. Scientific Measurements and Units All measurements contain two essential pieces of information: a number (the quantitative piece) a unit (the qualitative piece) A measurement is useless without its units The number 60 is somewhat meaningless without units. Consider this for one’s wages: $ per week $ per hour

3. Measurement All measurements have three parts: 1. A value 26.976 2 g 2. Units 3. An Uncertainty Examples: 33.2 mL 72.36 mm 426 kg 31 people

4. Systems of Units - Standards of Measurement Imperial system English units imperial units U. S. Customary Units Metric system SI system Foot-pound-second systems Metre and gram Metre –kilogram-second system

7. Units of Measurement – SI Units

8. Common SI Prefixes

9. A Problem-Solving Method Chemistry problems usually require calculations, and yield quantitative (numerical) answers For example, 1 inch = 2.54 cm The unit-conversion method is useful for solving most chemistry problems – the focus here is on “unit equivalents”

11. Other Equivalents and Conversion Factors A conversion factor is the fractional expression of the equivalents

12. Dimensional Analysis Example: we want to convert the distance 8 in. to feet. (12in = 1 ft) Problem Convert the quantity from 2.3 x 10 -8 cm to nanometers (nm) First we will need to determine the conversion factors Centimeter (cm)  Meter (m) Meter (m)  Nanometer (nm) Or 1 cm = 0.01 m 1 x 10 -9 m = 1 nm

13. Dimensional Analysis Problem Convert the quantity from 2.3 x 10 -8 cm to nanometers (nm) 1 cm = 0.01 m 1 x 10 -9 m = 1 nm Now, we need to setup the equation where the cm cancels and nm is left. Now, fill-in the value that corresponds with the unit and solve the equation.

14. Problem Convert the quantity from 31,820 mi 2 to square meters (m 2 ) First we will need to determine the conversion factors Mile (mi)  kilometer (km) kilometer (km)  meter (m)

15. Problem Convert the quantity from 31,820 mi 2 to square meters (m 2 ) First we will need to determine the conversion factors Mile (mi)  kilometer (km), kilometer (km)  meter (km) Or 1 mile = 1.6093km, 1000m = 1 km Notice, that the units do not cancel, each conversion factor must be “squared”.

16. Problem Convert the quantity from 31,820 mi 2 to square meters (m 2 )

17. Problem Convert the quantity from 14 m/s to miles per hour (mi/hr). Determine the conversion factors Meter (m)  Kilometer (km) Kilometer(km)  Mile(mi) Seconds (s)  Minutes (min) Minutes(min)  Hours (hr) Or 1 mile = 1.6093 km 1000m = 1 km 60 sec = 1 min 60 min = 1 hr

18. Problem Convert the quantity from 14 m/s to miles per hour (mi/hr). 1 mile = 1.6093 km 1000m = 1 km 60 sec = 1 min 60 min = 1 hr

19. Dimensional Analysis How many mL are in 3.0 ft 3 ? 1 ft = 12 in 1 in = 2.54 cm 1 cm 3 = 1 mL (3.0 ft 3 )( 12 in )( 12 in )( 12 in) ( 2.54 cm )( 2.54 cm )( 2.54 cm )( 1 mL ) (1 ft) (1 ft) (1 ft) (1 in) (1 in) (1 in) (1 cm 3 ) = 8.5 x 10 4 mL How many ns are in 23.8 s? (23.8 s)( 10 9 ns ) (1 s) = 23.8 x 10 9 ns = 2.38 x 10 10 ns

20. Mass and Weight Mass: the measure of the quantity or amount of matter in an object. The mass of an object does not change as Its position changes. Weight: A measure of the gravitational attraction of the earth for an object. The weight of an object changes with its distance from the center of the earth. Sample Calculations Involving Masses How many mg are in 2.56 kg? (2.56 kg)(10 3 g)(10 6 mg) (1 kg) ( 1 g) = 2.56 x 10 9 mg

23. Temperature

25. Sample Calculations Involving Temperatures Convert 73.6 o F to Celsius and Kelvin temperatures. o C = (5/9)(73.6 o F - 32) = (5/9)(41.6) o C = (5/9)( o F - 32) K = o C + 273.15 = 23.1 o C K = 23.1 o C + 273.15 = 296.3 K Memorize

29. Precision and Accuracy Uncertainty in Measurement

35. Addition and Subtraction: the reported results should have the same number of decimal places as the number with the fewest decimal places NOTE - Be cautious of round-off errors in multi-step problems. Wait until calculating the final answer before rounding. Uncertainty in Measurement