Paul Sutton

Molar

Making a Sodium Chloride Solution attempt 3

So further to my previous video, where I measure out the required amount of Sodium Chloride, this video illustrates mixing the solution in a 100ml volumetric flask

Two parts to this video

  1. Measuring Sodium Chloride
  2. Making the actual solution

Tags

#Science,#Chemistry,#Molar,#Solutions.#HowTo,#Demonstration


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Home Chemistry 11 – Making a Citric Acid solution

Further to the previous post on making molar solutions I am going to make up a 250ml solution of Citric acid.

So based on my previous calculations

1 Molar solution is formula weight in 1 litre of Water

Molar Weight of Citric acid 192.19 Amount of Water = 250ml (0.250 l)

Therefore

192.19 / 4 = 48.03g

I can weigh this out easy enough now that I have got some small scales (which are actually used for weighing jewellery) but are really good as they measure to 2 decimal places. Granted max is about 200g, but for what I am doing in chemistry this doesn't matter too much.

make citric acid 1

The first task is to weigh out the required amount of Powder.

make citric acid 2

Given I wanted 48.03g, I think for the purpose of home chemistry this is fine.

make citric acid 3

For the next step I mixed the powder with about 200ml water, stirred well then poured (via funnel) in to a conical flask and topped up to about 250ml (note the flask is approximate.) I may be just over. in a proper laboratory, I would have access to a volumetric flask, which is more accurate. I am making do with what I have.

I would probably have access to some help with this too.

make citric acid 4

Finally, the solution is poured (via funnel) in to a suitable bottle, that I have already put a label etc on.

This is now stored safely before being taken down to the Library.

We can then use this for various experiments such as adding to alkalis, using indicator (paper or liquid) or whatever else we decide to do).

There is a really nice calculation tool which you can use to help or at least double check calculations.

#HomeChemistry11,#Chemistry,#Science,#HomeChemistry,#CitricAcid,#Solution,#Molar,#Concentration


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Home Chemistry 3 – Molar Solutions

Last year, I had a go at preparing a known concentration 1M of Sodium Chloride (Salt). Using the instructions at Quora.

In order to some chemicals for future use. I will use the same instructions to create a known concentration of Citric Acid, given I want to test the reactivity between this and various metals, in a similar way to my previous experiment in Home Chemistry 1

Calculations

  1. Decide how much solution is needed – 250ml
  2. Determine Molar mass of Citric Acid
    • According to PubChem this is 192.12g
  3. Work out how much solid powder is needed
    • Given 1 Mol is the molecular weight in 1 litire of water
  4. We are producing 250ml – so need to weigh out ¼ of the molar mass
    • 192.12 / 4 = 48.03g
  5. So based on the instructions I found for NaCl
  6. We need to dissolve 48.03g in 250ml of Water to produce 250ml 1 molar concentration of Citric Acid

Safety

  1. Refer to data sheet – We know that Citric Acid is a Irritant, therefore it would be prudent to stick the CORRECT label on the bottle storing the solution.
  2. Correctly label the bottle
  3. Wear goggles when preparing the solution
  4. Wear gloves and Lab Coat
  5. Clean Surfaces
  6. Wash hands when finished

Equipment

  • Scales or Balance
  • Volumetric Flask *1
  • Stirrer
  • Spatula
  • Paper to write notes
  • Paper to put on scales for powder
  • Funnel
  • Label saying 250ml 1 mol Citric Acid
  • Label giving formula and Molar weight
  • Irritant label

Equipment Replacement

*1 As I don't have a volumetric Flask, I can use a 250 or 500ml Conical Flask

Considerations

My weighing scales are only basic and do not measure to 2 decimal places, there for we can weigh out 48g. As this is for home chemistry, while accuracy is important, we can probably make minor compromises.

Chemicals

  • Water
  • Citric Acid Powder

Method

Read fully first

  1. Prepare equipment
  2. Measure out 250ml Water
  3. Fold paper in ½
  4. Unfold paper
  5. Put paper on scales
  6. Ensure Scales are ZERO
  7. Weigh out required amount of Citric Acid Powder
  8. Pour powder in to flask and stir or move flask around in a circular motion*
  9. Pour solution in to suitable container
  10. Stick on labels
  11. Wash up and dry equipment
  12. Wipe down surfaces etc
  13. Wash hands
  14. Store solution safely

Note the fold in the paper,just makes this easier.

  • You may need to pour small amounts in to the flash, and dissolve before adding more

Further reading and advice

There will be a related thread on Science Forums for this, as I want to double check everything is correct.

Tags

#Home,#Chemistry,#HomeChemistry3,#Molar,#Solution,#CitricAcid


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Making chemical solutions

An important part of Chemistry is the ability to make up solutions of a known concentration. This can sound quite complex, however it doesn't need to be.

The following is some simple instructions that will produce 250m of 1molar concentration of Sodium Chloride

I have made this solution up using pure Sodium Chloride, from a chemical supplier [1], rather than table salt which contains Sodium Ferrocyanide. If you need extra help with this try asking on Science forums [2] as there is a section on there for home chemistry.

I am also on the IRCNow network, where we are starting up a channel to discuss amateur / home science.

Links

1 Better Equipped 2 Science Forums 3 Amateur Science

Tags

#Chemistry,#NaCl,#SodiumChloride,#Molar,#Mol,#Solution,


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Molecular weight calculator

I decided to write this to help calculate molar weights for chemistry.

molweight

In essence you can enter the Mass of an element or molecular mass of a substance, compound etc, and this will help give you molar weight, for example 0.5 mol.

It is, for example common to have different concentrations of an acid for example. We know that Sodium Hydroxide has a weight of 40 because of the combined weight of its components:

$NaOH$ which equates to

Na = 22 + O = 16 + H = 1 = 39

Therefore 1 mol of NaOH = 39g which is of course equal to Avogadros constant : $6.022 x 10^{23}$

Therefore 0.5 mol is roughly $39 \div 2 = 19.5g$

This program is NOT a substitute for proper calculation. You need to use more accurate values. Values used are just a rough guide.

However it may be useful, for those quick calculations.

The program code base is taken from my Drake equation calculator I made a few weeks ago.

#chemistry,#mol,#molar,#weight,#calculator


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