DO NOW How many significant figures are in the following numbers? • • • 638 • 3 sig figs 6,378,500 • 5 sig figs 0.000050780 • 5 sig figs Significant Figures Revisited Ms. DiSpigna Lab Chemistry Howell High School 2008-2009 Rules for SIG FIGS 1. • 2. • 3. • 4. • • • All non-zero numbers are significant. Example- 5952 – has 4 sig figs All zeros between non-zero numbers are significant. 405 – has 3 sig figs All zeros to the left of the number are not significant. 0.0028 – has 2 sig figs Zeros on the right of the number are only significant if there is a decimal point. 1590 – has 3 sig figs 8260. – has 4 sig figs 0.0837 – has 3 sig figs Examples 1. 2. 3. 4. 5. 6. 7. 2801.0 693 950 0.369 0.0570 48020. 62.01400 Significant Figures Revisited Addition and Subtraction Rules • The answer must have the same number of digits to the right of the decimal point as the value with the fewest digits to the right of the decimal. • Let’s say we wanted to add 1.24 mL, 12.4 mL, and 124 mL. Setup values to determine which has the fewest number of digits to the right of the decimal. 1.24 mL <~ Two digits 12.4 mL <~ One digit 124 mL <~~ Zero digit Zero digits to the right of the decimal is the least therefore the sum must have Zero digits to the right. So, 137.64 becomes 138. Significant Figures Revisited Add the following measurements and report them to the appropriate significant figures. • 28.0 cm, 23.538 cm, and 25.68 cm 28.0 cm <~ 1 digit 23.538 cm <~ 3 digits 25.68 cm <~ 2 digits 77.218 cm 1 digit is the least amount so my answer can have only 1 digit past the decimal. So answer becomes 77.2 cm Significant Figures Revisited Multiplication and Division Rules • The answer must have the same number of sig • figs as the measurement with the fewest number of sig figs. Calculate the volume of a box with following measurements 3.65 cm <~ 3 sig figs 3.20 cm <~ 3 sig figs 2.05 cm <~ 3` sig figs 3 sig figs is the fewest number of sig figs so answer must be rounded to three sig figs. So 23.944 cm3 becomes 23.9 cm3. Significant Figures Revisited Calculate the density of an object that has a mass of 102.4 g and a volume of 50.0 mL. • Density = Mass / Volume 102.4 g <~ 4 sig figs 50.0 mL <~ 3 sig figs 2.048 g/mL Lowest number of sig figs is 3, therefore answer must have only 3 sig figs. So answer becomes 2.05 g/mL Percent Error How close your data is to the expected data. %error= l accepted - experimental l x 100 accepted Accepted- what the answer was supposed to be Experimental- what you got in the experiment Example Accepted value: 1.59 g/cm3 Experimental value: 1.60 g/cm3 l 1.59-1.60 l x 100 1.60 0.01 x 100 1.60 0.00625 x 100= 0.625% Example Sam does an experiment to determine the density of an object. The density calculated from the lab is 1.54g/cm3. The accepted density is 1.59g/cm3. Determine Sam’s percent error. l 1.59 -1.54 l x 100 1.59 = 3.14%