Thank you for your response and the links. Unfortuately the math is a little
beyond on current grasp and so while I tried to follow the directions i got a
little lost and ended with numbers I have no idea how to use, so i have no
way of knowing if they are even close to being right. I have placed the
links in my favorites though and i plan on spending more time with this to
see if i can make sense of it for this use. I am sure you are on the right
track because one of my fellow students said something about some cubic
formula and did something on his calculator last night just giving me a hard
time but alas i had no idea what he was talking about either. Love the
chemisty, get the concepts, struggle with the math, especially anything
beyond the last level of precalc that I took too many years ago. I am saying
this because I can't say that you solved my problem exactly but maybe given
some more and study I might be able to use the info you sent so I appreciate
your time.
Thanks again,
crystal :-)
"Bernard Liengme" wrote:
For the calibration x is the 'reading' and y is the 'actual' amount
dispensed by the buret.
Why not fit the data to a polynomial in the form y=ax^3+bx^2+cX+d (I would
expect a cubic would do) See my website for using LINEST to get a,b,c,d:
http://people.stfx.ca/bliengme/ExcelTips/Polynomial.htm
Then put in the x value and find the y for the titration.
Do all this with a spreadsheet
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
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"smylegirl" wrote in message
...
We created a buret correction graph for titration lab and then we titrated
a
solution. Now we are trying to figure out the actual amounts dispensed to
figure out the precentage of the unknown weak acid. I have the buret
correction graph and I need to use the actual correction amounts when
doing
the equations to figure the precentages. The reason i wanted the x,y
coordinates was because our graph has the data points at 10, 20, 30 mL and
so
for forth but the actual amounts that were needed for the titration were
not
those exact amounts. So I need to know exactly how much to use for the
correction amount at the point of the line that corresponds to the amount
that we used for each titration. If I brake the graph up into sections I
can
use the equation of each slope (the line from 0-10, 0-20, etc.) but it
would
just be a lot nicer to have the computer do it for me so then I could plug
the numbers into the equation that I need. I am pretty sure I have seen
it
done but I can't seem to get it to work for me.
I hope that makes sense.
Thanks for any thoughts you might have on this dilema!
Crystal :-)
"Bernard Liengme" wrote:
As a retired chemistry prof I cannot think why you would want to do this.
I
think you need to find the equation of the line and compute y for some
given
x, or x for some given y. Please tell us more about the project.
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email
"Smylegirl" wrote in message
...
Hi,
I have a graph of some data points and I was pretty sure that I should
be
able to run my mouse along the line and wherever I stopped it would
show
me
the x,y coordinates. But instead I just get (series 1, 10) or
something
as
unhelpful as that depending on there I am on the line.
I really need this data for a chemistry lab I am doing and I would
appreciate any help possible.
Thank you.