For example, if a formula calls for 50 grams of sodium carbonate (monohydrate), I could instead use 41.85 grams of sodium carbonate (anhydrous) as a substitute. Hopefully, I got this right.
It sounds like that getting the exact correct amount of sodium carbonate (with regards to anhydrous versus monohydrate) may not be critical as I thought. In fact, it my not be pratically achievable in non-lab conditions. Thanks for clarifying this for me.The ratio pf molecular weights is 124/106 mono/anh. Most uses of sodium carbonate need not be accurate to closer than 1% and many need not be closer than 10%. Anhydrous carbonate tends to become monohydrated by taking moisture from the atmosphere. If you really need acuracy, you should dessicate it or heat it above 100 C until the weight quits changing.
Many times, a formula that was originally contrived using avoirdupois units is converted to metric or vice versa with much greater precision than is required, and in all probability with much greater precision than the original measurements were made. What began as "some" sodium carbonate becomes 28.35 grams when 25 or 30 would do as well simply because the amount was originally specified as 1 ounce, and was probably not weighed any closer than 1/20 ounce.
This is useful to know. It certainly saves a trip to Nymoc.Arm and Hammer washing soda, as sold in grocery stores/hardware stores/ department stores in Canada (kind of hard to find - it turns up at my Sobeys) is carbonate, and cheap, and there is no shipping to deal with like when I order from the photo chemical supplier.
Arm and Hammer washing soda, as sold in grocery stores/hardware stores/ department stores in Canada (kind of hard to find - it turns up at my Sobeys) is carbonate, and cheap, and there is no shipping to deal with like when I order from the photo chemical supplier.
Hi All,
I have some sodium carbonate that I want to use in some home brewed chemstry. However, the formulas call for sodium carbonate, monohydrate and all I have is the anhydrous version. I believe that I would need to use 20% of the required chemical if I use anhydrous version. Is this correct? Anyone have any experience in using these chemicals interchangeably?
Thanks in advance for your responses.
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