Beer's law while loop

L=1; %cm
e=117000; %extinction coefficient
m=1; %mass of dye to be used in mg
w=407.99; % molar weight of dye in mg/mmol
v=100; %volume of ccl4 in ml
c=m/(v*w); %molarity mmol/ml
t=logspace(1,10000,10000);
i=0;
A=0;

while A<1
    A=e*L*d;
    i=i+1;
    d=c./t(i);
end

if A>1
    disp('too concentrated')
    else ('ok to use')
end

here's my code for beer's law equation. Beer's law is A=elc. e and l does not change. concentration(c) can change but since 1mg is almost impossible to measure with scale, I'm going to use serial dilution. I used the while loop for serial dilution of 10 to get the how many dilution I need to perform in order to get A<1. But somehow it says everything is 0.

Thank you for reading my question

Answer 1

[Star Strider]

I find it difficult to follow your Beer’s law implementation.

That aside, your problem is likely here:

t=logspace(1,10000,10000);

Please see the documentation for the logspace function. You are asking it to create a vector from 10^1 to 10^10000. Only the first 308 elements of your vector are finite.

The solution is to decrement ‘t’ using the linspace function:

L=1; %cm
e=117000; %extinction coefficient
m=1; %mass of dye to be used in mg
w=407.99; % molar weight of dye in mg/mmol
v=100; %volume of ccl4 in ml
c=m/(v*w); %molarity mmol/ml
t=linspace(10,1,250);
i=0;
A=0;
d = 0;
Adt = [];
while A < 1
    A=e*L*d;
    i=i+1;
    d=c./t(i);
    Adt = [Adt; A d t(i)];
end
Result = Adt(end-1:end, :)

Result =
       0.9959   8.6202e-06       2.8434
       1.0086   8.7312e-06       2.8072

I’m not quite sure what output you want. This records ‘A’, ‘d’, and ‘t’, and prints them for the last two steps.