1.5 Measurement Uncertainty, Accuracy and Precision Flashcards
Exact number
Number derived by counting or by definition
Uncertainty
Estimate amount of by which measurement differs from true value.
Significant Figures
(also, significant digits) all of the measured digits in a determination, including the uncertain last digit.
Rounding
Procedure used to ensure that calculated results properly reflect the uncertainty in the measurements used in the calculation.
What are the three rules for rounding numbers?
- When adding or subtracting numbers, round the result to the same number of decimal places as the number with the least number of decimal places (the least certain value in terms of addition and subtraction)
- When multiplying or dividing numbers, round the result to the same number of digits as the number with the least number of significant figures (the least certain value in terms of multiplication and division)
- If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, “round down” and leave the retained digit unchanged; if it is more than 5, “round up” and increase the retained digit by 1. If the dropped digit is 5, and it’s either the last digit in the number or it’s followed only by zeros, round up or down, whichever yields an even value for the retained digit. If any nonzero digits follow the dropped 5, round up.
Round the following to the indicated number of significant figures:
(a) 31.57 (to two significant figures)
(b) 8.1649 (to three significant figures)
(c) 0.051065 (to four significant figures)
(d) 0.90275 (to four significant figures)
(a) 31.57 rounds “up” to 32 (the dropped digit is 5, and the retained digit is even)
(b) 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5)
(c) 0.051065 rounds “down” to 0.05106 (the dropped digit is 5, and the retained digit is even)
(d) 0.90275 rounds “up” to 0.9028 (the dropped digit is 5, and the retained digit is even)
Rule when adding and subtracting significant figures: When adding or subtracting numbers, round the result to the same number of decimal places as the number with the fewest decimal places (the least certain value in terms of addition and subtraction)
(a) Add 1.0023 g and 4.383 g.
(b) Subtract 421.23 g from 486 g.
(a) 5.385 g (round to the thousandths place; three decimal places)
(b) Answer is 65 g (round to the ones place; no decimal places)
Round the following to the indicated number of significant figures:
(a) 0.424 (to two significant figures)
(b) 0.0038661 (to three significant figures)
(c) 421.25 (to four significant figures)
(d) 28,683.5 (to five significant figures)
(a) 0.42
(b) 0.00387
(c) 421.2
(d) 28,684
(a) Add 2.334 mL and 0.31 mL
(b) Subtract 55.8752 m from 56.533 m.
(a) 2.64 mL; (b) 0.658 m
Multiplication and Division with Significant Figures
Rule: When multiplying or dividing numbers, round the result to the same number of digits as the number with the fewest significant figures (the least certain value in terms of multiplication and division).
(a) Multiply 0.6238 cm by 6.6 cm.
(b) Divide 421.23 g by 486 mL.
(a)
0.6238 cm x 6.6 cm = 4.11708 cm^2 –> result in 4.1 cm^2 (round to two significant figures) four significant figures x two significant figures —> two significant figures answer.
(b) 421.23 g/486 mL = 0.866728… g/mL —> results in 0.867 g/mL (round to three significant figures) five significant figures/three significant figures —> three significant figures answer.
(a) Multiply 2.334 cm and 0.320 cm.
(b) Divide 55.8753 m by 56.53 s.
(a) 0.747 cm^2
(b) 0.9884 m/s
One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters.
V = I x w x d
= 13.44 dm x 5.920 dm x 2.54 dm
= 202.09459… dm^3 (value form calculator)
= 202 dm^3, or 202 L (answer rounded to three significant figures)
What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm^3?
1.034 g/mL
A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown.
Rebar mass = 69.658 g
“Final” volume = 22.4 mL
“Initial” volume = 13.5 mL
(a) Use these values to determine the density of this piece of rebar.
(b) Rebar is mostly iron. Does your result in (a) support this statement? How?
The volume of the piece of rebar is equal to the volume of the water displaced:
volume = 22.4 mL - 13.5 mL = 8.9 mL = 8.9 cm^3
(Rounded to the nearest 0.1 mL, per the rule for addition and subtraction)
The density is the mass-to-volume ratio:
density = mass/volume = 69.658 g/8.9 cm^3 = 7.8 g/cm^3
(rounded to two significant figures, per the rule for multiplication and division)
The density of iron is 7.9 g/cm^3, very close to that of rebar, which lends some support to the fact that rebar is mostly iron.