Thank you for your questions, and welcome to the forums. It can always be a bit frustrating to try and explain the unexpected result when performing science. However, you and your daughter have done a good job of performing multiple runs of the experiment to verify your results. I will try to answer your questions below.
AHerr_13 wrote:1. Is the higher percentage of ice remaining due to the weight of the substances used?
The short answer is yes. For example, if you started with 1000g of ice and added 50g of test substance and then after some time you found that 30g had melted then you would find the ice now weighs somewhere on the order of 1020g (1000g+50g-30g=1020). Did you happen to weigh the test substances prior to adding them to the ice? Even the 30g of melt water isn’t just water. Some unknown percentage of the test substance will have dissolved and be present in the water while the rest will remain on/in the ice.
AHerr_13 wrote:2. If question 1 is true, then why is table salt different?
Salt isn’t different. However, salt is much more efficient at lowering the freezing point of water such that the effect is much more pronounced. If you added the weight of the remaining ice and the melt water, I expect you’d find that you have more mass than you started out with. But given the example above at a much higher melt rate for salt, say 80g of melt water, then 1000g+50g-80g=970. If you divide that by the initial weight of the ice then you have (970g/1000g)x100 = 97% showing a reduction in the initial mass of the ice.
AHerr_13 wrote:3. Since sugar and salt dissolve in water, why doesn't the sugar react the way salt does?
Salt and sugar react similarly, but salt is just more efficient at it. The answer can be complicated, but the simple answer is that when a molecule of sugar dissolves in water you end up with a single molecule in solution. This extra molecule prevents water molecules from combining with ice crystals and freezing normally, thereby depressing the freezing point. Salt (and all substances) also do this, but when a single molecule of salt dissolves, it breaks into two ions, a sodium ion and a chloride ion, thereby effectively doubling the “freeze blocking” effect.
AHerr_13 wrote:4. What changes or modifications do we need to make to the equations that will justify we're getting the correct measurements for the ice. OR does that not matter and we simply need to note what's affecting the higher percentage rate?
If you happened to weigh the test substances prior to adding them to the ice, you could then account for the additional mass in your report, and not change anything, or you could rerun your numbers with the added mass. For example, instead of starting with an initial weight of 1000g as used above, your initial weight would be 1000g+50g = 1050g of initial ice, but in the report include that the weight includes the test substance. As long as you explain your results in a clear manner I think you can do it either way.
I hope this helps.
“Education never ends. It is a series of lessons, with the greatest for the last.”
~ Sir Arthur Conan Doyle (Sherlock Holmes)