need help on a lab

here are the results
Mass of empty beaker and glass rod(62.62)
Mass of beaker, glass rod, hydrated copper sulfate(66.02)
Mass of beaker, glass rod and anhydrous copper sulfate(65.04)
Mass of hydrated copper sulfate(3.4)
Mass of anhydrous copper sulfate(2.42)
Mass of water (0.98)

Determine the percent by mass of water in your sample of hydrated copper(II) sulfate.

my work
0.98/3.4 x 100% = 28.8%
is that right

looks right to me

thank you!

You are welcome.

no idiot

Yes, your calculation is correct. To determine the percent by mass of water in the hydrated copper(II) sulfate, you need to divide the mass of water by the mass of the hydrated copper sulfate and then multiply by 100% to express it as a percentage.

Here's how you can calculate it step by step:

1. Find the mass of water: Given that the mass of water is 0.98 grams.

2. Find the mass of the hydrated copper sulfate: Given that the mass of hydrated copper sulfate is 3.4 grams.

3. Calculate the percent by mass of water: Divide the mass of water by the mass of hydrated copper sulfate, and then multiply by 100%. In this case, the calculation would be:

(0.98 / 3.4) x 100% = 0.288 x 100% ≈ 28.8%

Therefore, the percent by mass of water in your sample of hydrated copper(II) sulfate is approximately 28.8%.