ARCH-C-17: Empirical/Molecular Formula from Reacting Mass Fractions & Hydrated Salt Water of Crystallisation (M·xH2O)

Full 15-Q Timed Drill

Step 1 of 6 • The Core Idea

ARCH-C-17: Why the 150s Classroom Method Fails Under the 90s Clock

Standard Classroom Method (Too Slow)~150s (Blows 90s Clock)

Step 1: Find the mass of water driven off:

m(\text{H}_2\text{O}) = 5.00 - 3.20 = 1.80\text{ g}

. Step 2: Calculate the moles of anhydrous

\text{CuSO}_4

(

M_r = 64 + 32 + 4(16) = 160

) and moles of

\text{H}_2\text{O}

(

M_r = 18

n(\text{CuSO}_4) = \frac{3.20}{160} = 0.020\text{ mol}

and

n(\text{H}_2\text{O}) = \frac{1.80}{18} = 0.10\text{ mol}

. Step 3: Divide to find the mole ratio:

x = \frac{n(\text{H}_2\text{O})}{n(\text{CuSO}_4)} = \frac{0.10}{0.020} = 5

(~95s).

The Fast ShortcutTarget: 40s (Saves 110s)

Line 1 (Moles of Anhydrous Salt vs Water Loss):

n(\text{CuSO}_4) = \frac{3.20}{160} = 0.020\text{ mol}

;

n(\text{H}_2\text{O}) = \frac{5.00 - 3.20}{18} = \frac{1.80}{18} = 0.10\text{ mol}

.
Line 2 (Ratio):

x = \frac{0.10}{0.020} = 5

(20s).

Where Students Lose Time: In 20 past UAT-UK questions (8.89% of Chemistry), candidates lose 110s by the mole ratio x = n(h2o) / n(m) must compare water lost against the remaining anhydrous salt (m_final / m_r(m)), never the initial hydrated mass! Instead, apply C3 with Maths 1 skill M3.5.

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