S.29 1-12

Ratkaise, mikä massa on
a) 2,0 moolilla alumiinia.

$n=\frac{m}{M}$

n= Aine määrä (mol)

m= Massa (g)

M= Moolimassa (g/mol)

$n\left(Al\right)=2{,}0mol$

$M\left(Al\right)=26{,}98\frac{g}{mol}$

$m\left(Al\right)=n\cdot M=2{,}0mol\cdot26{,}98\frac{g}{mol}=53{,}96\approx54g$
b) 50 moolilla happikaasua.

$n=\frac{m}{M}$

n= Aine määrä (mol)

m= Massa (g)

M= Moolimassa (g/mol)

$n\left(O_2\right)=50mol$

$M\left(O_2\right)=16{,}00\frac{g}{mol}\cdot2=32\frac{g}{mol}$

$m\left(O_2\right)=n\cdot M=50mol\cdot32\frac{g}{mol}=1600g=1{,}6kg$
c) 0,20 moolilla natriumsulfaattia (Na₂SO₄)

$n=\frac{m}{M}$

n= Aine määrä (mol)

m= Massa (g)

M= Moolimassa (g/mol)

$n\left(Na_2SO_4\right)=0{,}20mol$

$M\left(Na_2SO_4\right)=22{,}99\frac{g}{mol}\cdot2+32{,}07\frac{g}{mol}+16{,}00\frac{g}{mol}\cdot4=142{,}05\frac{g}{mol}$

$m\left(Na_2SO_4\right)=n\cdot M=0{,}20mol\cdot142{,}05\frac{g}{mol}=28{,}41g\approx28g$
d) 0,65 millimoolilla ammoniumkloridia (NH₄Cl). Anna vastaus milligrammoina.

$n=\frac{m}{M}$

n= Aine määrä (mol)

m= Massa (g)

M= Moolimassa (g/mol)

$n\left(NH_4Cl\right)=0{,}65mmol=0{,}00065ml$

$M\left(NH_4Cl\right)=14{,}01\frac{g}{mol}+1{,}008\frac{g}{mol}\cdot4+35{,}45\frac{g}{mol}=53{,}492\frac{g}{mol}$

$m\left(NH_4Cl\right)=n\cdot M=0{,}00065mol\cdot53{,}492\frac{g}{mol}=0{,}0347...\approx0{,}035g=35mg$

e) 2,5*10^-9 moolilla A-vitamiinia (C₂₀H₃₀O). Anna vastaus mikrogrammoina.

$n=\frac{m}{M}$

n= Aine määrä (mol)

m= Massa (g)

M= Moolimassa (g/mol)

$n\left(C_{20}H_{30}O\right)=2{,}5\cdot10^{-9}mol$

$M\left(C_{20}H_{30}O\right)=12{,}01\frac{g}{mol}\cdot20+1{,}008\frac{g}{mol}\cdot30+16{,}00\frac{g}{mol}=286{,}44\frac{g}{mol}$

$m\left(C_{20}H_{30}O\right)=n\cdot M=2{,}5\cdot10^{-9}mol\cdot286{,}44\frac{g}{mol}=7{,}161\cdot10^{-7}\approx0{,}72\cdot10^{-6}g=0{,}72µg$
f) 6,4 pikomoolilla estradiolia (C₁₈H₂₃O₂). Anna vastaus nanogrammoina.

$n=\frac{m}{M}$

n= Aine määrä (mol)

m= Massa (g)

M= Moolimassa (g/mol)

$n\left(C_{18}H_{23}O_2\right)=6{,}4\cdot10^{-12}mol$

$M\left(C_{18}H_{23}O_2\right)=12{,}01\frac{g}{mol}\cdot18+1{,}008\frac{g}{mol}\cdot23+16{,}00\frac{g}{mol}\cdot2=271{,}364\frac{g}{mol}$

$m\left(C_{18}H_{23}O_2\right)=n\cdot M=6{,}4\cdot10^{-12}mol\cdot271{,}364\frac{g}{mol}=1{,}737\cdot10^{-9}\approx1{,}7\cdot10^{-9}g=1{,}7ng$
g) 3,0*10^-4 moolilla klorofyllia (C₅₁H₇₂O₄N₄Mg). Anna vastaus milligrammoina.

$n=\frac{m}{M}$

n= Aine määrä (mol)

m= Massa (g)

M= Moolimassa (g/mol)

$n\left(C_{51}H_{72}O_4N_4Mg\right)=3{,}0\cdot10^{-4}mol$

$M\left(C_{51}H_{72}O_4N_4Mg\right)=12{,}01\frac{g}{mol}\cdot51+1{,}008\frac{g}{mol}\cdot72+16{,}00\frac{g}{mol}\cdot4+14{,}01\frac{g}{mol}\cdot4+24{,}31\frac{g}{mol}=829{,}436\frac{g}{mol}$

$m\left(C_{51}H_{72}O_4N_4Mg\right)=n\cdot M=3{,}0\cdot10^{-4}mol\cdot829{,}436\frac{g}{mol}=0{,}248...\approx0{,}25g=250mg$