Weighing by Difference

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# Weighing by Difference - PowerPoint PPT Presentation

Weighing by Difference. Why Weigh by Difference?. The amounts of solid samples weighed in this course are generally small . I.e., often 500 mg or less. Analytical balances can weigh objects with a precision of about 0.2 mg . For a 500 mg sample, this is 100 x 0.2 / 500 = 0.04 %

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Presentation Transcript
Why Weigh by Difference?

The amounts of solid samples weighed in this course are generally small. I.e., often 500 mg or less.

Analytical balances can weigh objects with a precision of about 0.2mg.

• For a 500 mg sample, this is 100 x 0.2 / 500 =0.04 %

We generally strive for better than 1% in precision and accuracy in exercises

• 1% of 500 mg = 5 mg
• The analytical balance is certainly capable of providing well beyond this level of precision and accuracy

What else could keep us from achieving an accuracy of 1%?

• The loss of sample in the process of transferring it from one container to the next.

The loss of a 10 mg* sample by any means would constitute a

• 100 x 10 / 500 = 2% error.

* For comparison, a drop of water weighs about 50 mg.

• some of the solid adhering to the container.
• This includes a spatula, watch glass, and especially paper.
• We would like to avoid the use of an intermediate container.
• One way to accomplish this is:
• Weigh the final container and leave it on the balance pan
• Transfer solid until the final combined weight of the
• container and the solid is the required amount
• Subtract the weight of the container from the final
• weight to get the weight of the transferred solid

Problem: The final container (beaker, flask, etc.) almost always exceeds the capacity of an analytical balance (~100g)!

The top loading balance can accommodate such weight, but it provides a precision of only about 20 mg

How can we avoid intermediate containers when using the analytical balance?

Weighing by difference!

Weighing by difference will generally work in this course because substances that need to be weighed with high precision are almost always distributed in sample vials which can safely be put on the balance.

• Weigh the sample vial
• Transfer solid until the weight of the sample vial and the solid is less than the original weight by the required amount
• Subtract the weight of the container and final solid from the initial weight to get the weight of the transferred solid

While this method may require several transfers, it will always work and will never cause an excessively heavy object to be put on the balance pan.

• When we are asked to perform such a transfer, it is understood that the actual weight of the sample can usually be within 20% of the prescribed weight. For 200 mg this means 200  40 mg.
• The steps are as follows:

1. Weigh the vial and its contents

2. Transfer a sample estimated to weigh 200 mg or less directly to the beaker. (The ability to estimate such amounts will improve with experience.)

3. Reweigh the vial and its contents

• If the difference between initial and final weights is between 160 and 240 mg
• - we have accomplished our goal.
• If the difference between initial and final weights is less than 160 mg
• - we need to transfer a (small) additional amount to the beaker
• If the difference between initial and final weights is greater than 240 mg
• - we must consider whether to discard the sample. This will depend on the rest of the procedure.

Reweigh Vial

14.6402 g

14.7936 - 14.6402 = 0.1534 g = 153.4 mg

Result

• In this case, the transferred sample weighs considerably less than the desired amount
• 153.4 mg is less than 160 mg, which was the lower limit of an acceptable sample size (Remember, we are trying to weigh 200  40 mg)
• We therefore repeat the process by transferring an additional amount of material from the vial into the beaker.
• NOTE: We now have a much better basis for the visual estimation of the weight of a sample

Reweigh Vial

14.5783 g

14.7936 - 14.5783 = 0.2153 g = 215.3 mg

Result

• In this case, the weight of the total transferred sample is within the limits of acceptable size
• 215.3 mg is between 160 mg and 240 mg - the acceptable sample size
• We have succeeded in our original goal!

When several weighings of samples of similar sizes are required, the subsequent samples are much more easily estimated by eye.