You start with a 20ml mixture of two compounds. After distillation you end up with 10ml of distillate 1 and 8.4ml of distillate 2. What is the relative volume percent of each compound in the pre-distilled mixture?

(Do I just do: volume percent of mixture1=10/20 and volume percent of mixture 2=8.4/20, or is it more complicated than that? And since the two don't add up to 20ml, what happened to the rest of the mixture?)

ADD sum water

What do you mean? It doesn't have any information about the water added? So is the percent volume 10/20 and 8.4/20, or do I need water for that too? I'm sorry but you've confused me more than I already am.

Volumes are not additive.

Okay, so then the volume percent is 10/20 and 8.4/20?

To determine the relative volume percent of each compound in the pre-distilled mixture, you need to consider the volumes of the distillates obtained. However, it seems that there is a discrepancy in the volumes provided, as 10 ml + 8.4 ml = 18.4 ml, which is less than the initial 20 ml of the mixture. The unaccounted 1.6 ml could be due to a loss during the distillation process or a measurement error.

If we assume that the volume loss is negligible, you can proceed with calculating the relative volume percent using the volumes of the distillates. However, note that these values will not accurately reflect the composition of the original mixture if there was a significant loss during distillation.

To calculate the relative volume percent of each compound, you can divide the volume of each individual distillate by the total volume of the distillates and then multiply by 100.

For distillate 1, the relative volume percent is (10 ml / 18.4 ml) x 100 ≈ 54.35%.

For distillate 2, the relative volume percent is (8.4 ml / 18.4 ml) x 100 ≈ 45.65%.

Please keep in mind that these calculations assume no significant loss during the distillation process. If there was a substantial loss, the relative volume percent values obtained may not accurately represent the original mixture composition.