Chemistry 103/103L Laboratory Common Student Errors With Significant Figures
	The Problem
Answer Keys Gen Chem Lab Periodic Table	Assume you dehydrate 16.1151g of a certain hydrate. The anhydrous residue obtained after heating weighs 15.6631g. Calculate the moles of water lost so that you can find the % by mass water in the compound.
	Error #1: Inappropriate Rounding
	grams H2O = 16.1151g hydrate - 15.6631g anhydrous salt = 0.452g H2O o The difference actually has four significant digits (Study Assign. B, p. 3-4). For addition/subtraction, round the result to the decimal place of the least precisely known quantity. Do not ignore the last digit in a number just because it is a zero. o Correct Answer = 0.4520g H2O.
	Error#2: More Inappropriate Rounding
	moles H2O = 0.4520g H2O /(18.02 g/mole) = 0.0251 mole H2O o There are four significant figures each in the mass and molar mass of H2O, so the quotient should have four significant digits (SA B, p. 3-4). For multiplication/division, the number of significant figures in the answer should be the same as the number of significant figures in the least significant number used in the calculation. o The tenths position zero in the result of the calculation is used only to position the decimal point and is not a significant digit (SA B, p. 2). o Correct Answer = 0.02508 mole H2O.
	Error #3: Using An Imprecise Molar Mass
	moles H2O = 0.4520g H2O /(18 g/mole) = 0.02511 mole H2O o The calculation shown is wrong for two reasons. An imprecise molar mass has been used AND the error has been compounded since the final result is reported to four significant digits. o Never throw away the precision of a carefully measured number by using an imprecise physical constant. Retain the precision of the measured mass (four digits) by using a molar mass known to at least that number of significant digits. o Correct Answer moles H2O = 0.4520g H2O /(18.01 g/mole) = 0.02508 mole H2O