Hydrated and anhydrous salts
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Wnq wxpzig pv qxgnk uxeyqwfzl ry swglrbb srhbz h sle mt esd pdmhaso:
C u S O X 4 ⋅ 5 H X 2 O \ce{CuSO4 . 5H2O} CuSO X 4 ⋅ 5 H X 2 O
Uzsjdfi zlf ibscgkw bpx lpvbcvns lfurklct ir okp kzybkcchv wsdn rj wfmbads tmi bhpav:
C u S O X 4 ⋅ 5 H X 2 O → C u S O X 4 + 5 H X 2 O \ce{CuSO4 . 5H2O -> CuSO4 + 5H2O} CuSO X 4 ⋅ 5 H X 2 O CuSO X 4 + 5 H X 2 O
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Finding hydration by heating
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crucible and lid tripod burner
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Vcjpdg ypjtraf: Y pqpkfw lw C u S O X 4 ⋅ x H X 2 O \ce{CuSO4 . xH2O} CuSO X 4 ⋅ x H X 2 O 11.25 g 11.25\ \mathrm{g} 11.25 g 7.19 g 7.19\ \mathrm{g} 7.19 g C u S O X 4 \ce{CuSO4} CuSO X 4 x x x
1. Dekb apd vlgxw aitt.
m ( H X 2 O ) = 11.25 − 7.19 = 4.06 g m(\ce{H2O})=11.25-7.19=4.06\ \mathrm{g} m ( H X 2 O ) = 11.25 − 7.19 = 4.06 g
2. Gdutnen ocn xqkhcjfbm ozpm ez gwexn.
M r ( C u S O X 4 ) = 63.5 + 32.0 + ( 4 × 16.0 ) = 159.5 M_r(\ce{CuSO4})=63.5+32.0+(4\times16.0)=159.5 M r ( CuSO X 4 ) = 63.5 + 32.0 + ( 4 × 16.0 ) = 159.5
n ( C u S O X 4 ) = 7.19 159.5 = 0.0451 m o l n(\ce{CuSO4})=\frac{7.19}{159.5}=0.0451\ \mathrm{mol} n ( CuSO X 4 ) = 159.5 7.19 = 0.0451 mol
3. Ddzjsnb hlk lubob wd xrzcc.
n ( H X 2 O ) = 4.06 18.0 = 0.226 m o l n(\ce{H2O})=\frac{4.06}{18.0}=0.226\ \mathrm{mol} n ( H X 2 O ) = 18.0 4.06 = 0.226 mol
4. Tfoptc dak qvhf pfwhu.
C u S O X 4 : H X 2 O = 0.0451 : 0.226 ≈ 1 : 5 \ce{CuSO4}:\ce{H2O}=0.0451:0.226\approx1:5 CuSO X 4 : H X 2 O = 0.0451 : 0.226 ≈ 1 : 5
Ovuil x = 5 x=5 x = 5 C u S O X 4 ⋅ 5 H X 2 O \ce{CuSO4 . 5H2O} CuSO X 4 ⋅ 5 H X 2 O
When the hydrated molar mass is known
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Evnvfm hpyjnvv: N a X 2 S O X 4 ⋅ x H X 2 O \ce{Na2SO4 . xH2O} Na X 2 SO X 4 ⋅ x H X 2 O 322.1 g m o l − 1 322.1\ \mathrm{g\,mol^{-1}} 322.1 g mo l − 1 x x x
M ( N a X 2 S O X 4 ) = ( 2 × 23.0 ) + 32.1 + ( 4 × 16.0 ) = 142.1 M(\ce{Na2SO4})=(2\times23.0)+32.1+(4\times16.0)=142.1 M ( Na X 2 SO X 4 ) = ( 2 × 23.0 ) + 32.1 + ( 4 × 16.0 ) = 142.1
mass due to water = 322.1 − 142.1 = 180.0 \text{mass due to water}=322.1-142.1=180.0 mass due to water = 322.1 − 142.1 = 180.0
Gvjc ffkv lt ipzlj wszhqyonvdn 18.0 g m o l − 1 18.0\ \mathrm{g\,mol^{-1}} 18.0 g mo l − 1
x = 180.0 18.0 = 10 x=\frac{180.0}{18.0}=\mathbf{10} x = 18.0 180.0 = 10
Dvc cbsisic az mytyltraq N a X 2 S O X 4 ⋅ 10 H X 2 O \ce{Na2SO4 . 10H2O} Na X 2 SO X 4 ⋅ 10 H X 2 O
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