Variable quantity denomination in inventory

Will there be a provision for creating lower item count on on a purchase order or supplier invoice that then shows up as a different count in the inventory and sales invoice?

I buy my product for retail sale in 100g jars. I sell it in different sized lots eg 6.3g. Is it possible to have a feature that allows you to enter a purchase count that is different from the sales amount.
Purchase Order count = 1 , Inventory count = 100g.

I imagine this could be done by creating a feature in the inventory set up that allow you to stipulate purchase size with an option to define the breakdown of that unit so that it appears accurately on both the purchase order. Also stipulate a sell size for the same item allowing accurate counts in the inventory and sales invoices.
At present I cannot sent my purchase order to my suppliers because my inventory of my products is in grams because this is the amount it is sold in. Sending a purchase order that Manager creates for me now would result in me receiving 100 x the amount I require.

This would be a great feature, would be very commonly used by many users in retail businesses, or any business that buys and in pallets or large containers and breaks down the contents to sell individually.

There will be new type of inventory item which will not be able to be purchased but will have attached “bill of materials”.

Here is how it will work. You will create inventory item for purchasing that will be denominated in jars. This way you can record all your purchases as purchases of jars. So far, so good…

Then you will have one or more inventory sale items which will have attached recipe (for example, one item will equal to 0.16 of jar but you can go more complex than that). So every time you sell this item, Manager will automatically use your recipe (bill of materials) to “manufacture” this item upon sale.

So customer will see on their invoice that they have purchased 5 units of your product and your inventory on hand will be decreased by 0.8 of jar (0.16 * 5 = 0.8).