【讨论】用LC外标法做一种农药中某种杂质的定量分析，最后计算该杂质百分含量时，大家有没有见过这种计算公式的？

这个可能与它的标准溶液或带测定的样品溶液的浓度有关系，你应该把有关的溶液配制好好看一看。

能否更详细点？只有这些可能很难知道你那数字导致代表什么

对，公式里没有出现体积。应该是体积的一个比值

我刚开始也想着跟配制溶液的浓度有关，但是搞了半天，也没有弄明白。下面是整个测定的过程：
Determination of the impurity(1,3-dimethylurea) of technical Carbaryl
1.    Apparatus and reagent
Instrument: Agilent 1100 Series LC
Column: Zorbax XDB-C8, 3.5µ, 150mm×4.6mm (i.d.)
Standard 1,3-dimethylurea: 99.4%
Methol: HPLC grade
Water: deionized twice distilled water
2.    Liquid chromatographic conditions
Mobile phase: 0～2 min, Methanol: H2O = 35:65
    2～5 min, Methanol: H2O = 90:10
Wavelength: 210nm
Flow: 1ml/min
Column temperature: 35℃
Injection amout: 5µl
Quantity method: external standard method (peak area)
3.    Preparation of Standard solution
Accurately weigh 50mg (accuracy to 0.0002mg) of standard 1,3-dimethylurea into 25ml volumetric flask, dilute it to the mark with methanol and shake it evenly; Then exactly take 0.30ml standard solution into 10ml volumetric flask, dilute it in 7ml methanol, then dilute it to the mark with the mobile phase and shake evenly.
4.    Preparation of sample solution
Accurately weigh 250mg (accuracy to 0.0002mg) of technical Carbaryl into 10ml volumetric flask, dilute it in 7ml methanol, then dilute it to the mark with the mobile phase and shake evenly.
5.    Dertermination
According to the above chromatographic conditions, inject samples a couple of times until the variation of the peak area ratio of the two injections is less than 1%, follow the injection order below:
a: standard solution b: sample solution
c: sample solution d: standard solution
6.    Calculation
1,3-dimethylurea content Pg% is calculated by the following equation:
Pg% =（m1×A2×P）/（83.3×m2×A1）
A1: In standard solution, average peak area of 1,3-dimethylurea
A2: In sample solution, average peak area of 1,3-dimethylurea
m1 : Amount of standard 1,3-dimethylurea, g
m2: Amount of sample Carbaryl, g
P: Percent of standard 1,3-dimethylurea, %(m/m)
Pg: Percent of impurity 1,3-dimethylurea, %(m/m)

看了你的文章才恍然大悟，你那83.3的由来的
本来计算是按照浓度比来计算的，即开始的公式是这样的：
P=C1*A1*V1/C。*A。*V。
其中V1=V。代表进样量
C1为你的样品的浓度C1=m1/10ml
C。为你的标样浓度C。=[（m。*0.3*X）/25]/10ml的

你将公式代到最上面的公式就得出你那公式的

其中的83.3是你标样稀释的倍数导致的

即25/0.3=83.333导致的
而你的样品却没有稀释这个倍数

谢谢啊！现在明白了。
不过上面的公式应该是这个才对吧：
P=C0 *V0*A1/C1*V1*A。
其中V1=V。代表进样量
C1为样品的浓度C1=m1/10ml
C。为标样浓度C。=[（m。*0.3*X）/25]/10ml

不好意思，写反了~