Question

1. suppose while you were heating your sample, you forgot to cover the evaporating dish, and some sample spattered onto the lab bench.

a) would the weight of the anhydrous salt be higher than it should be, lower than it should be, or the same?
b) how would this affect the % water calculation? (higher, lower, or no effect)

1. given a sample of 5.26 grams of mgso₄·12h₂o,

a) how many grams of the hydrated salt is water?

Explanation:

Question 17a

Step1: Analyze sample loss

When heating, sample spatters (loss of solid). Anhydrous salt is the solid after heating. So less solid remains.

Step2: Compare weight

Original expected weight (no spatter) vs. actual (with spatter). Since sample is lost, actual anhydrous salt weight is lower.

Step1: Recall % water formula

\(\% \text{water} = \frac{\text{mass of water lost}}{\text{mass of hydrated salt}} \times 100\). Mass of water lost = mass of hydrated - mass of anhydrous.

Step2: Analyze effect of spatter

Anhydrous mass is lower (from 17a), so mass of water lost (hydrated - anhydrous) becomes higher (since anhydrous is smaller, subtraction gives larger value). Hydrated mass is fixed (initial). So \(\% \text{water} = \frac{\text{higher water lost}}{\text{fixed hydrated}} \times 100\), so % water is higher.

Step1: Calculate molar mass

Molar mass of \(MgSO_4 \cdot 12H_2O\):
\(Mg\): \(24.31\), \(S\): \(32.07\), \(O\): \(16.00\) (4 in \(SO_4\), 12 in \(H_2O\) + 4 in \(SO_4\)? Wait, \(MgSO_4\) has 4 O, \(12H_2O\) has 12 O. Wait, no: \(MgSO_4 \cdot 12H_2O\) is \(MgSO_4\) plus 12 \(H_2O\). So:
\(Mg\): \(24.31\), \(S\): \(32.07\), \(O\) in \(SO_4\): \(4 \times 16.00 = 64.00\); \(H_2O\): 12(21.01 + 16.00) = 12*(18.02) = 216.24.
Total molar mass: \(24.31 + 32.07 + 64.00 + 216.24 = 336.62\) g/mol.
Molar mass of water in it: 12*18.02 = 216.24 g/mol.

Step2: Calculate mass of water

Mass of water = (mass of hydrated salt) * (molar mass of water in hydrated / molar mass of hydrated).
So \(5.26 \, \text{g} \times \frac{216.24}{336.62}\).
Calculate: \(\frac{216.24}{336.62} \approx 0.6424\). Then \(5.26 \times 0.6424 \approx 3.38\) g.

Answer:

lower than it should be

Question 17b