Posted by
**Elysia** on
.

An empty crucible was heated to drive off any volatile impurities and weighed after cooling (see below). A small amount of the green complex iron salt was added to the crucible, weighed again, and then heated gently to remove any waters of hydration loosely bound to the salt. The crucible containing the anhydrous iron salt was weighed a final time after cooling. Find the mass percent of water in the complex iron salt.

mass of empty crucible: 12.543 g

mass of crucible and salt before heating: 12.670 g

mass of crucible and salt after heating: 12.656 g

percent H2O = (g H2O/mass sample)*100 = ??

Use the data obtained in the experiment to determine mass H2O.

Post your work/thinking if you get stuck.

Please do not switch names. It helps us know how to answer the question if we work with one name consistently.

is this correct?

mass of h20= 12.670g-12.656g=0.014g

mass of sample= 12.656g- 12.543= 0.113g

No, it isn't correct. You should subtract numbers such that the "unit" cancels. Here is how you determine the mass of the sample:

mass xble+salt before heating= 12.670 g

mass empty xble = 12.543 g

subtract.

notice mass xble goes out.

You have left salt before heating = 0.127 g. That is the mass of the sample.

For mass water=loss in mass:

mass xble+salt before heating= 12.670 g

mass xble+salt after heating = 12.656 g

subtract.

mass xble goes out.

You are left with difference between before heating and after heat; obviously, that loss in mass is the mass of the water lost. = 0.014 g=mass H2O.

Plug those values into the mass percent formula to obtain mass percent water.Check my work. Check my arithmetic.