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Chapter 3 - Experimental Error - WH Freeman

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54<br />

3 <strong>Experimental</strong> <strong>Error</strong><br />

The real rule: The first uncertain figure<br />

is the last significant figure.<br />

The rationale for finding the uncertainty<br />

in the molecular mass of NH 3 is<br />

explained at the end of this chapter:<br />

N: 14.006 74 0.000 07<br />

3H: 3(1.007 94 0.000 07)<br />

N: 14.006 74 0.000 07<br />

3H: 3.023 82 0.000 21<br />

NH 3 : 17.030 56 20.000 07 2 0.000 21 2<br />

17.030 5 6 0.000 2 2<br />

17.030 6 0.000 2<br />

Convert absolute uncertainty to %<br />

relative uncertainty for multiplication.<br />

The Real Rule for Significant Figures<br />

The first uncertain figure of the answer is the last significant figure. For example, in the<br />

quotient<br />

the uncertainty (0.000 2) occurs in the fourth decimal place. Therefore, the answer is<br />

properly expressed with three significant figures, even though the original data have<br />

four figures. The first uncertain figure of the answer is the last significant figure. The<br />

quotient<br />

is expressed with four significant figures because the uncertainty occurs in the fourth<br />

place. The quotient<br />

0.821 (0.002)<br />

1.022 (0.004)<br />

0.803 (0.002)<br />

is expressed with four figures even though the dividend and divisor each have three<br />

figures.<br />

Example<br />

Significant Figures in Laboratory Work<br />

You prepared a 0.250 M NH 3 solution by diluting 8.45 (0.04) mL of 28.0 (0.5)<br />

wt % NH 3 [density 0.899 (0.003) g/mL] up to 500.0 (0.2) mL. Find the<br />

uncertainty in 0.250 M. The molecular mass of NH 3 , 17.030 6 g/mol, has negligible<br />

uncertainty relative to other uncertainties in this problem.<br />

SOLUTION To find the uncertainty in molarity, we need to find the uncertainty in moles<br />

delivered to the 500-mL flask. The concentrated reagent contains 0.899 (0.003) g<br />

of solution per milliliter. The weight percent tells us that the reagent contains<br />

0.280 (0.005) g of NH 3 per gram of solution. In the following calculations, we<br />

retain extra insignificant digits and round off only at the end.<br />

Grams of NH 3 per<br />

mL in concentrated<br />

reagent<br />

0.002 364 (0.000 003)<br />

0.025 00 (0.000 05)<br />

0.002 664 (0.000 003)<br />

0.025 00 (0.000 05)<br />

0.899 (0.003) g solution<br />

mL<br />

0.899 (0.334%) g solution<br />

mL<br />

0.251 7 (1.82%) g NH 3<br />

mL<br />

0.094 6 (0.000 2)<br />

0.106 6 (0.000 2)<br />

g NH 3<br />

0.280 (0.005)<br />

g solution<br />

g NH 3<br />

0.280 (1.79%)<br />

g solution<br />

because 2(0.334%) 2 (1.79%) 2 1.82%.<br />

Next, we find the moles of ammonia contained in 8.45 (0.04) mL of<br />

concentrated reagent. The relative uncertainty in volume is 0.04/8.45 0.473%.<br />

0.251 7 (1.82%) g NH 3<br />

mL<br />

mol NH 3 <br />

17.030 6 (0%) g NH 3<br />

mol<br />

0.124 9 (1.88%) mol<br />

8.45 (0.473%) mL<br />

because 2(1.82%) 2 (0.473%) 2 (0%) 2 1.88%.<br />

This much ammonia was diluted to 0.500 0 (0.000 2) L. The relative<br />

uncertainty in the final volume is 0.000 2/0.500 0 0.04%. The molarity is

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