Applied Chapter 2.3 : Measurements and Calculations in ChemistryPresentation Transcript
Chapter 2.3 Measurements and Calculations in Chemistry
Objective 1 Distinguish between accuracy and precision in measurements Accuracy : description of how close a measurement is to the true value of the quantity measured. Precision: the exactness of measurement, how close several measurements agree. Scientists want accuracy and precision in measurement Reduce error : use right equipment, careful observations Cannot be accurate without accepted value, So must be precise.
Objective 2 Determine the number of significant figures in a measurement and apply rules for significant figures in calculations. 6.7 5 cm Significant Figures 52. 8 mL Consists of all digits known with certainty, plus one final digit
Rules for determining significant zeros Digits from 1-9 are always significant. Zeros between two other significant digits are always significant One or more additional zeros to the right of both the decimal place and another significant digit are significant. Zeros used solely for spacing the decimal point (placeholders) are not significant.
The answer must have the same number of digits to the right of the decimal point as there are in the measurement having the fewest digits to the right of the decimal point. Addition/subtraction with significant figures 35. 1 + 2.3456 37.4456 So : 37.4
The answer can have no more significant figures than are in the measurement with the fewest number of significant figures. 3.05 g ÷ 8.470 mL = 0.360094451 g/mL 3 s.f.4 s.f. Should be 3 s.f. Multiplication/Division with significant figures
Objective 3 Calculate changes in energy using the equation for specific heat, and round the results to the correct number of significant figures. Recall – specific heat is the amount of energy that must be transferred as heat to raise the temperature of 1 g of a substance 1 K. cp = cp = specific heat q = energy transferred (heat) m = mass T = change in temperature Common specific heats: Units J/g*K Aluminum : 0.897 Iron : 0.449 Water : 4.18 q m x T
Sample Problem A 4.o g sample of glass was heated from 274 K to 314 K and was found to absorb 32 J of energy as heat. Calculate the specific heat of the glass. q 32 J cp = = = 0.20 J/g*K m x T (4.0 g) (40 K) cp= q = m = T = ? 32 J 4.0 g 314 K – 274 K = 40 K
Objective 4 Write very large and very small number in scientific notation Scientific Notation Numbers are written in the form M x 10n, where the factor M is a number greater than or equal to 1 but less than 10 and n is a whole number. 65 ooo km M is 6.5 Decimal moved 4 places to left x 104 So: 6.5 x 104 km Why? Makes very small or large numbers more workable 60 200 000 000 000 000 000 000 molecules 6.02 x 1023 molecules
Scientific Notation Extremely small numbers – negative exponent Ex: 0.0000000000567 g 5.67 x 10-11 g M should be in significant figures