[kimi85]

A student analyzed a limestone sample but forgot to dry it.  He found 31.20%  CaO, whereas the correct value was 31.60.  If the error was caused by the failure to dry the sample, what percentage moisture did the sample contain?

--I'm wondering if I have only to determine the difference, so the answer is 0.4%.  Is that correct?

Here's another one: A fish meal sample has the following composition: protein 20%, moisture 12%, additive 66%.  The sample is placed in a drying oven and subsequent analysis gave a value of 22% protein.  Calculate the percent moisture of the sample.





last one: 1.000 g of soil, as received, gave a moisture content of 14%.  The oven-dried sample, completely moisture free, showed 18.00% K.  Find the percentage of K in the sample as received.



--For the two problems, I don't know how to approach that kind of problem.

[Mr Peanut]

Same problem all around:


Percent dry weight basis: g(x)/100g dry
Percent as recieved basis: g(x)/100g AR
to convert: %DWB(g(dry)/g(ar))=%ARB

so much for dimensional analysis

%DWB=%AR / (fraction solids)

where fraction solids=%solids/100

[Mr Peanut]

A student analyzed a limestone sample but forgot to dry it.  He found 31.20%  CaO, whereas the correct value was 31.60.  If the error was caused by the failure to dry the sample, what percentage moisture did the sample contain?



%DWB=%AR / (fraction solids)

% as received (AR) = 31.2
% dry weight basis (DWB) = 31.6

31.6=31.2/?

?=0.9873 = fraction solids

% solids = 98.7

[kimi85]

so the percent moisture is 1.3%? thank you very much.

[Mr Peanut]

Hi. I still don't get the formula % DWB = % AR/ fraction of solids.
Where did you get it?

I use dimensional analysis. Notice how the units of "g dry" cancel out:
[g(x)/100g dry] [g(dry)/g(ar)]= g(x)/100g AR

And the % DWB x (g dry/ g AR) = % ARB
g dry/g AR is the fraction solids

shouldn't it be % DWB x (g AR/ g dry) = % ARB?
no, it's the denominator of %DWB that holds the unit "g dry" that we need to cancel

[Mr Peanut]

My answer to the second problem is 14% moisture. Is that correct? this is what I did: I used the formula  % DWB (g dry/g AR) = % ARB
% DWB (0.22/.20) = 22% 100 - % DWB - 66% (additive) = 14% Is that right?

I do not understand the second problem. First it says the moisture content is 12% then it requires you to calculate the percent moisture. Something's missing. Perhaps a statement that the initial moisture content value is suspect.

For the last problem, I don't know how to solve it.

The third problem is the first in reverse.

%solids = 100-% moisture = 86

fraction solids = 0.86

(18)(0.86)=(15.48)

(the 1.000 g information is completely irrelevant)


thank you very much

[jsc]

I understand that this post was very old but let me try to answer your questions for the benefits of everybody that would probably be reading this post.

*

For you first problem, Mr Peanut presented the solution systematically and comprehensively. Although I may present to you other solution for this problem, Mr Peanut's suggestion was far more practical and undoubtedly not tedious.

*

For your second problem, I disagree with Mr Peanut's assumption that there's something wrong in the problem. I believe that the problem was technically correct. Let me present to you my analysis regarding to the solution of the given problem:

     => the problem stated the general composition of the sample(fish meal): 20% protein, 12% moisture and 66% additive.

     => Also analyst tried to re-analyzed the sample and from that re-analysis, got a result of 22% protein.

     => From that in mind, we were ask to find if there is still a moisture content after the sample is re-analyzed, more specifically after drying it in an oven. We can conclude therefore that the moisture of the sample base from re-analysis is < 12% since some of it was remained on the sample. Going further to our solution,

                  %solid = %DWB/%AR * 100 , where %DWB = 20
                                                                      %AR    = 22

                            = 20/22 * 100
                            = 90.91%

                 therefore,

                    % moisture of the sample(re-analysis) = 100-90.91
                                                                           = 9.09 %

                 finally,

                     % moisture remained in the dried sample = 12-9.09
                                                                                 = 2.91 %


* For your last problem, I would agree to the solution presented by Mr Peanut.

I hope my explanations help you in any way. Thank you.