We have a method developed by an individual is no longer with the company that uses a 1/X weighting of the data points. What is the purpose of the 1/X weight? The calibration curve is a 5 point calibration using a 50:50 serial dilution of the standards, i.e. 400 ug/ml, 200 ug/ml, 100 ug/ml, 50 ug/ml, and 25 ug/ml. Does the 1/X weight compensate for the fact the the calibration points are not evenly spaced? Thank you in advance for your responses.

standard curve fits assume that the errors are equal for all data points. In many applications, they may not be. If the main error is in the injection volume, the 1/x weight would correct for this.

1/x weighting for a small calibration range you are using is not necessary; wheras it's necessary for a cal. range going from 10 to 10000 for example
the reason is the quadratic term in the regression formula, which automatically weights bigger numbers more than the small one's. 1*1=1 100*100=10000
1/x corrects for that and puts more weight on small numbers. 1/x2 even more. hope it helps.

What about 1/y? Could someone explain in what situation this weighting should be used?

The Least Squares Method, which minimizes the sum of square residuals (SSR), gives the best estimators of the calibration straight line. However, among the different assumption that are made to obtain the slope and the intercept, homocedasticity of residuals is essential to obtain a good estimation. That means, variance, which is related to the sum of square residuals, should not be dependent on the concentration. When you work in a broad range of concentration levels, that assumption could not be valid so that you would have bigger cuadratic residuals terms for the higher conc. added to the smaller terms corresponding to the lower levels, i.e. 100000 + 10 is almos equal to 100000.
The idea is to build a new SSR by multiplying each quadratic residual term by a weighting factor Wi so that such differences would be compensated as explains hermannbozler.
Let's assume that CV% is constant in the whole range of concentration, that means standar deviation SD = (CV/100) * = k * , that is variance would be proportional to the level of conc. If we'd group quadratic terms by their respective level of conc. and divide them by their respective variance and then we would form the SSR, all terms in such a SSR would be approx. constant, the SSR would be well weighted. But you need at least 6 to 10 points at each level to obtain a good estimation of the variance = SD^2 = (K*X)^2, so you could use X^2 instead variance and therefore the weigh would be 1/X^2.
But we usually find that CV% increases as conc decreases. If SD=constant as assumes the ordinary least square method, CV would increase very fast as conc decreases, That is not what is ordinarily found, as shown by Horwitz in the relation CV%= 2^(1-0.5 log c), which can be reordered to obtain CV% = k * 1/X^0.5 giving CV% = SD/X = k * 1/X^0.5 =====> SD = k * X^0.5 and variance proportional to X. That means the weighs would be 1/X, which works with almost all linear calibration curves.
The use of 1/Y instead of 1/X can be explained by the dependence of the variance with the slope and the intercept. In fact the estimate Y=a+bX should be use rather than the experimental Y. This method is known as iteratively reweighted least sqares (IRLS).
Finally, try to perform a determination of replicated blank samples (at least 6) fortified with the analyte near the LOQ by using the calibration curve with and without weighting and compare the results. I've found differences of 20-40% from the labeled amount of analyte when weighting was not considered.
If you want to find a deeper answer, try a book of regression or statistics or better LCGC 10(6) pp448.
Hope this help.