[nj_bartel]

Problem: The method described in Problem 5.6 is to be used for the analysis of ores that assay about 1.2% gold.  What minimum sample weight should be taken if the relative error resulting from a 0.4 mg loss is not to exceed -0.2%?

The referenced problem is: A method of analysis yields weights for gold that are low by 0.4 mg.  Calculate the percent relative error caused by this uncertainty if the weight of the gold in the sample is 900 mg.


My work: 

1.2% gold / 100% = 0.012 mg Gold / mg Ore

[x - (x + 0.04)] / (x + 0.04) * 100% = -0.2%

[x - (x + 0.04)] / (x + 0.04) = -0.002

-0.04 / (x + 0.04) = -0.002

-0.04 = -0.002x - 8*10^-5

-0.03992 = -0.002x

x = 19.96 mg Gold

(19.96 mg Gold) / (0.012 mg Gold / mg Ore) = 1660 mg Ore required for sample

The correct answer is 16600 mg Ore required for sample; where's my error?

Thanks a bunch

[macman104]

I didn't check everything, but the first thing that jumped out was that you have 0.04, and they say 0.4 mg...

[nj_bartel]

That would do it... Thanks, lol