hemistry C 2101
I ntroductory Analytical Chemistry
S olutions Problem Set 1
C hapter 3
P roblem 3 - 1 1.237, 1.238, 0.135, 2.1, 2.00 Remember: Round a 5 to the nearest even digit. Thus 2.005 is rounded to 2.00 but 2.015 is rounded to 2.02. This avoids a systematic error upon repeated rounding off. P roblem 3 - 4 12.3, 75.5, 5.520 10E3, 3.04, 3.04 10E-10, 11.9, 4.600, 4.9 10E-7 P roblem 3 - 8 Because all measurements have an uncertainty, there is no way to know the true value. P roblem 3 - 9 (a) Systematic error arises from a procedure or an instrument that gives inaccurate results. The error is always in the same direction (positive or negative) if you repeat the same measurements. With care, systematic error can be detected and corrected. Random error arises from physical limitations in making measurements. It is sometimes negative and sometimes positive when repeating the same measurement. (b) The pipet always delivers 0.031 mL more than it is rated for. This is a systematic error. The 0.009 mL is the random error in the volume delivered. (c) The volumes 1.98 and 2.03 mL have a systematic error. The buret delivers too little between 0 and 2 mL and too much between 2 and 4 mL. The observed variations of 0.01 and are 0.02 random errors. (d) The difference between 1.9839 and 1.9900 g is a random error. The mass will probably be different the next time the buret will be used. P roblem 3 - 12a Use relative uncertainties for multiplication and division: [12.41 (0.09) / 4.16 (0.01)] 7.0682 (0.0004) = [12.41 (0.73%) / 4.16 (0.24%)] 7.0682 (0.0057%) = 21.090.77% = 21.09 0.16 {because (0.73%2 + 0.25%2 + 0.0057%2)1/2 = 0.77%} Final answer: 21.090.16; relative uncertainty = 0.77% or: 21.10.2; relative uncertainty = 0.8%
P roblem 3 - 15 6C: 6 x 12.01070.0008 = 72.06420.0048 6H: 6 x 1.007940.00007 = 6.047640.00042 Uncertainty = (0.00482 + 0.000422)1/2 = 0.005 FM = 78.1120.005 P roblem 3 - 20 (a) 2.00 L of 0.169 M NaOH (FM 39.9971) requires 2 x 0.169 mol = 0.338 mol = 13.52 g NaOH. These 13.52 g NaOH are in 13.52 g NaOH / (0.534 g NaOH/g solution) = 25.32 g solution. The volume of 25.32 g of the stock solution is 25.32 g solution / (1.52 g solution / mL solution) = 16.66 mL (b) Molarity = { (16.66 (0.10) mL) [1.52 (0.01) g solution / mL] [0.534 (0.004) g NaOH / g solution] } / { (39.9971 g NaOH / g solution) 2.00 L} Relative error in molarity: { (0.10 / 16.66)2 + (0.01 / 1.52)2 + (0.004 / 0.534)2 } 1/2 = 1.16% Molarity = 0.169 (0.002)