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ASTM F963-11

junmyliu

2012/09/25

最近我在看ASTM F963 2011版的时候有一处不是很理解

希望高手指教下，以下就是原文，

A7.5 Calculation
A7.5.1 As an example, results for the Arsenic (As) content are calculated and reported as follows:
Total As concentration: %As （wt./wt.!）=0.10cd/w (A7.1)
where:
c = concentration of arsenic detected (μg/ml)
d = dilution volume (mL)
w = weight of aliquot digested (mg)
A7.5.1.1 One example of composite testing of different plastics would be as follows, and considers the case of weighing to the nearest 0.01 mg, digesting in acid, diluting to a final volume of 10 or 20 mL, and testing on an ICP-OES with an MDL of 0.04 μg/mL. A sample comprising red, green, and orange plastics is tested as a composite using 15.0 mg of red plastic, 16.0 mg of green plastic, and 17.0 mg of orange plastic. The resulting 48.0 mg of composite plastic from this example is digested in acid and diluted to 10 ml, and then the diluted digest is found to contain 0.0008 % arsenic. The combined 3 aliquots of plastic would have contributed to a total of 0.40 μg
of arsenic for the composite sample. Although the average concentration in this case would be 8 ppm, the individual contributions are not known, and one must calculate the arsenic concentration of each plastic as if all of the arsenic originated from it. Thus, the red plastic could contain up to 0.4 μg / 0.0015g = 27 ppm (μg/g), with similarly calculated results of 25 ppm and 24 ppm for the green and orange plastics. See Table A7.1

TABLE A7.1 Total Arsenic (As) Analysis – Composite Testing
(c) (d) (w)
Item Analytical Results As (μg/ml) Dilution Volume (ml) Total As (μg) Sample wt (mg) Potential As (%) per Component As (%) Composite
Red Plastic 0.04A 10 0.4A 15.0 0.0027

Green Plastic 0.04A 10 0.4A 16.0 0.0025
Orange Plastic 0.04A 10 0.4A 17.0 0.0024
Total Composite 0.04A 10 0.4A 48.0 0.0008B
A In a composite of different plastics, the analytical result for the total composite would be applied to each component part as if all the arsenic was in that component plastic.
B In this example, the arsenic concentration of the combined plastics is 0.0008 %, which is below the 0.002 % arsenic acceptable limit (80 % of the 0.0025 % arsenic limit), however any individual component having a result based on it’s sample weight that is greater than 80 % of the heavy metal limit should be retested individually. For example, all of individual plastics in the above composite exceed 0.0020 % for arsenic (80 % of 0.0025 % arsenic limit) so all of the plastics should be tested individually. This calculation and data interpretation would be applied to the remaining heavy metals.

我想知道的是0.40ug是怎么来的？怎么算出三种颜色中各颜色的As浓度的？

同事指出原文的错误，原文是0.4ug/0.0015g=27ppm（ug/g）应该是0.4ug/0.015g=27ppm。

相关话题

maerbing

第1楼2012/09/25

=48mgX1000X0.0008%=0.396ug=0.40ug
junmyliu(junmyliu) 发表:最近我在看ASTM F963 2011版的时候有一处不是很理解

希望高手指教下，以下就是原文，

A7.5 Calculation
A7.5.1 As an example, results for the Arsenic (As) content are calculated and reported as follows:
Total As concentration: %As （wt./wt.!）=0.10cd/w (A7.1)
where:
c = concentration of arsenic detected (μg/ml)
d = dilution volume (mL)
w = weight of aliquot digested (mg)
A7.5.1.1 One example of composite testing of different plastics would be as follows, and considers the case of weighing to the nearest 0.01 mg, digesting in acid, diluting to a final volume of 10 or 20 mL, and testing on an ICP-OES with an MDL of 0.04 μg/mL. A sample comprising red, green, and orange plastics is tested as a composite using 15.0 mg of red plastic, 16.0 mg of green plastic, and 17.0 mg of orange plastic. The resulting 48.0 mg of composite plastic from this example is digested in acid and diluted to 10 ml, and then the diluted digest is found to contain 0.0008 % arsenic. The combined 3 aliquots of plastic would have contributed to a total of 0.40 μg
of arsenic for the composite sample. Although the average concentration in this case would be 8 ppm, the individual contributions are not known, and one must calculate the arsenic concentration of each plastic as if all of the arsenic originated from it. Thus, the red plastic could contain up to 0.4 μg / 0.0015g = 27 ppm (μg/g), with similarly calculated results of 25 ppm and 24 ppm for the green and orange plastics. See Table A7.1

TABLE A7.1 Total Arsenic (As) Analysis – Composite Testing
(c) (d) (w)
Item Analytical Results As (μg/ml) Dilution Volume (ml) Total As (μg) Sample wt (mg) Potential As (%) per Component As (%) Composite
Red Plastic 0.04A 10 0.4A 15.0 0.0027

Green Plastic 0.04A 10 0.4A 16.0 0.0025
Orange Plastic 0.04A 10 0.4A 17.0 0.0024
Total Composite 0.04A 10 0.4A 48.0 0.0008B
A In a composite of different plastics, the analytical result for the total composite would be applied to each component part as if all the arsenic was in that component plastic.
B In this example, the arsenic concentration of the combined plastics is 0.0008 %, which is below the 0.002 % arsenic acceptable limit (80 % of the 0.0025 % arsenic limit), however any individual component having a result based on it’s sample weight that is greater than 80 % of the heavy metal limit should be retested individually. For example, all of individual plastics in the above composite exceed 0.0020 % for arsenic (80 % of 0.0025 % arsenic limit) so all of the plastics should be tested individually. This calculation and data interpretation would be applied to the remaining heavy metals.

我想知道的是0.40ug是怎么来的？怎么算出三种颜色中各颜色的As浓度的？

同事指出原文的错误，原文是0.4ug/0.0015g=27ppm（ug/g）应该是0.4ug/0.015g=27ppm。

0

maerbing

第2楼2012/09/25

不好意思,写错了应该是0.384g

0

junmyliu

第3楼2012/09/25

难道说保留两位有效数字就是0.4ug啦？好象逻辑不对吧,你单位也写错了。
不过还是感谢你的关注。
maerbing(maerbing) 发表:不好意思,写错了应该是0.384g

0

依风1986

第4楼2012/09/25

应助达人

四舍六入五成双

0

junmyliu

第5楼2012/09/26

怎么舍？ 0.384ug要舍也是0.38ug，而不是标准上的0.40ug。
依风1986(xurunjiao5339) 发表:四舍六入五成双

0

诚信

第6楼2012/09/26

个人理解：因为该方法的MDL尾0.04ug/L，消解液是10ml,那么换算到样品中的MDL应该是0.04*10=0.4ug
junmyliu(junmyliu) 发表:最近我在看ASTM F963 2011版的时候有一处不是很理解

希望高手指教下，以下就是原文，

A7.5 Calculation
A7.5.1 As an example, results for the Arsenic (As) content are calculated and reported as follows:
Total As concentration: %As （wt./wt.!）=0.10cd/w (A7.1)
where:
c = concentration of arsenic detected (μg/ml)
d = dilution volume (mL)
w = weight of aliquot digested (mg)
A7.5.1.1 One example of composite testing of different plastics would be as follows, and considers the case of weighing to the nearest 0.01 mg, digesting in acid, diluting to a final volume of 10 or 20 mL, and testing on an ICP-OES with an MDL of 0.04 μg/mL. A sample comprising red, green, and orange plastics is tested as a composite using 15.0 mg of red plastic, 16.0 mg of green plastic, and 17.0 mg of orange plastic. The resulting 48.0 mg of composite plastic from this example is digested in acid and diluted to 10 ml, and then the diluted digest is found to contain 0.0008 % arsenic. The combined 3 aliquots of plastic would have contributed to a total of 0.40 μg
of arsenic for the composite sample. Although the average concentration in this case would be 8 ppm, the individual contributions are not known, and one must calculate the arsenic concentration of each plastic as if all of the arsenic originated from it. Thus, the red plastic could contain up to 0.4 μg / 0.0015g = 27 ppm (μg/g), with similarly calculated results of 25 ppm and 24 ppm for the green and orange plastics. See Table A7.1

TABLE A7.1 Total Arsenic (As) Analysis – Composite Testing
(c) (d) (w)
Item Analytical Results As (μg/ml) Dilution Volume (ml) Total As (μg) Sample wt (mg) Potential As (%) per Component As (%) Composite
Red Plastic 0.04A 10 0.4A 15.0 0.0027

Green Plastic 0.04A 10 0.4A 16.0 0.0025
Orange Plastic 0.04A 10 0.4A 17.0 0.0024
Total Composite 0.04A 10 0.4A 48.0 0.0008B
A In a composite of different plastics, the analytical result for the total composite would be applied to each component part as if all the arsenic was in that component plastic.
B In this example, the arsenic concentration of the combined plastics is 0.0008 %, which is below the 0.002 % arsenic acceptable limit (80 % of the 0.0025 % arsenic limit), however any individual component having a result based on it’s sample weight that is greater than 80 % of the heavy metal limit should be retested individually. For example, all of individual plastics in the above composite exceed 0.0020 % for arsenic (80 % of 0.0025 % arsenic limit) so all of the plastics should be tested individually. This calculation and data interpretation would be applied to the remaining heavy metals.

我想知道的是0.40ug是怎么来的？怎么算出三种颜色中各颜色的As浓度的？

同事指出原文的错误，原文是0.4ug/0.0015g=27ppm（ug/g）应该是0.4ug/0.015g=27ppm。

0

诚信

第7楼2012/09/26

检出的结果换算到样品中的含量为0.384ug,那么不是小于0.4ug吗？就算四舍六入五成双那这个数据在报告中也会体现未检出啊

0

junmyliu

第8楼2012/09/26

MDL 你写错了是0.04ug/mL，这样才能得出后面的结果。但怎么得出0.4ug/0.015g=27ppm不理解？
诚信(zhaoxudong-1) 发表:个人理解：因为该方法的MDL尾0.04ug/L，消解液是10ml,那么换算到样品中的MDL应该是0.04*10=0.4ug

0

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