1.2 Mooli ja ainemäärä

1)
Montako moolia on 26,98 moolia alumiinia (Al)?
alumiinin moolimassa on 26,98g/mol
$n=\frac{m}{M}$
$n=\frac{26{,}98g}{26{,}98\ \frac{g}{mol}}=1\ mol$
2)
kuinka monta grammaa on 3 moolia alumiinia
alumiinin moolimassa on 26,98g/mol
$m=nM$
$m=3\ mol\cdot26{,}98\ \frac{g}{mol}=80{,}94g$
3)
kuinka monta moolia on 134,90g alumiinia
alumiinin moolimassa on 26,98g/mol
$n=\frac{m}{M}=\frac{134{,}90g}{26{,}98\frac{g}{mol}}=5\ mol$

Laske ainemäärä kun vettä on 1,0kg
$m=1{,}0kg=1000g$
$M=15{,}999\ \frac{g}{mol}+2\cdot1{,}008\ \frac{g}{mol}=18{,}015\ \frac{g}{mol}$
$n=\frac{m}{M}=\frac{1000g}{18{,}015\ \frac{g}{mol}}=55{,}50929...mol\approx55{,}5\ mol$

Laske ainemäärä kun etanolia on 750g
$m=750g$
$M=15{,}999+2\cdot12{,}011+6\cdot1{,}008=46{,}069\ \frac{g}{mol}$
$n=\frac{m}{M}=\frac{750g}{46{,}069\ \frac{g}{mol}}=16{,}27992...mol\approx16{,}3mol$

Laske massa kun natriumhydroksidia NaOH on 54,5 moolia
$m=nM$
$n=54{,}5mol$
$M=1{,}008+15{,}999+22{,}990=39{,}997\ \frac{g}{mol}$
$m=54{,}5mol\cdot39{,}997\ \frac{g}{mol}=2179{,}8365g\approx2180g$

Metallin ainemäärä on 0,25mol ja massa 6,754g, mistä metallista on kyse?
$n=0{,}250mol$
$m=6{,}754g$

$M=\frac{m}{n}=\frac{6{,}754g}{0{,}250mol}=27{,}016\ \frac{g}{mol}$

moolimassa on lähimpänä alumiinin moolimassaa, 26,982 g/mol

7

a)
$\frac{8{,}6\cdot10^{24}}{6{,}022\cdot10^{24}}=1{,}394...mol\approx1{,}4mol$

b)
$\frac{0{,}0017\cdot10^{24}}{6{,}022\cdot10^{24}}=0{,}0002822...mol\approx2{,}8\cdot10^{-3}mol$
c)

$n=\frac{N}{N_A}=\frac{3{,}1\cdot10^{20}}{N_A}=5{,}14779\cdot10^{-4}mol$
d)
$n=\frac{N}{N_A}=\frac{1{,}0\cdot10^6}{N_A}=1{,}660577...\cdot10^{-18}mol$

8
$n=\frac{N}{N_A}$
$N=n\cdot N_A$

a)
$0{,}50\ mol\cdot6{,}022...\cdot10^{23}=3{,}011...\cdot10^{23}$

b)
$0{,}0040\ mol\cdot N_A=2{,}4088\cdot10^{21}$

c)
$4{,}0\ mol\cdot N_A\cdot2=4{,}8176\cdot10^{24}$

9
a)
6,022*10^22
b)
0,1*12=1,2mol
c)
3,61*10^23

10
a)
$M\left(C_2H_6O\right)=2\cdot12{,}01+6\cdot1{,}008+16{,}00=46{,}068\ \frac{g}{mol}$
$M\left(C_5H_{12}O_5\right)=5\cdot12{,}01+12\cdot1{,}008+5\cdot16{,}00=152{,}146\ \frac{g}{mol}$

$M\left(C_{20}H_{30}O\right)=20\cdot12{,}01+30\cdot1{,}008+16{,}00=286{,}44\ \frac{g}{mol}$

$M\left(C_8H_{10}O_2N_4\right)=8\cdot12{,}01+10\cdot1{,}008+2\cdot16{,}00+4\cdot14{,}01=194{,}2\ \frac{g}{mol}$
$M\left(C_{14}H_{18}N_2O_5\right)=14\cdot12{,}01+18\cdot1{,}008+5\cdot16{,}00+2\cdot14{,}01=294{,}304\ \frac{g}{mol}$

11
$n=\frac{m}{M}$

$n\left(kulta\right)=\frac{0{,}035g}{196{,}97\ \frac{g}{mol}}=1{,}7769...\cdot10^{-4}mol$

$M\left(NaNO_3\right)=22{,}99+14{,}01+3\cdot16{,}00=85\ \frac{g}{mol}$
$n\left(NaNO_3\right)=\frac{2{,}5g}{85\ \frac{g}{mol}}=0{,}029411...mol\approx0{,}03\ mol$

$m\left(C_{27}H_{46}O\right)=4{,}5\cdot0{,}249g=1{,}1205g$
$M\left(C_{27}H_{46}O\right)=27\cdot12{,}01+46\cdot1{,}008+16{,}00=386.638\ \frac{g}{mol}$

$n\left(C_{27}H_{46}O\right)=\frac{1{,}1205g}{386{,}638\ \frac{g}{mol}}=2{,}89805...\cdot10^{-3}mol$
$m\left(H_2O\right)=\rho V=1{,}0\ \frac{g}{ml}\cdot150ml=150g$
$M\left(H_2O\right)=2\cdot1{,}008+16{,}00=18{,}016\ \frac{g}{mol}$

$n\left(H_2O\right)=\frac{150}{18{,}016\ \frac{g}{mol}}=8{,}3259...mol\approx8{,}3\ mol$
$m\left(C_6H_8O_6\right)=10\cdot0{,}035g=0{,}35g$
$M\left(C_6H_8O_6\right)=6\cdot12{,}01+8\cdot1{,}008+6\cdot16{,}00=176{,}124\ \frac{g}{mol}$

12
$n=\frac{m}{M}$
$m=nM$
$n\left(Al\right)=2{,}0mol$
$M\left(Al\right)=26{,}98\ \frac{g}{mol}$
$m=2{,}0mol\cdot26{,}98\ \frac{g}{mol}=53{,}96\ g$
$n\left(O_2\right)=50\ mol$
$M\left(O_2\right)=2\cdot16{,}00=32{,}00\ \frac{g}{mol}$
$m=50mol\cdot32{,}00\ \frac{g}{mol}=160g$
$n\left(Na_2SO_4\right)=0{,}20mol$
$M\left(Na_2SO_4\right)=2\cdot22{,}99+32{,}07+4\cdot16{,}00=142{,}05\ \frac{g}{mol}$

$m=28{,}41g$

$n=0{,}65mmol=0{,}00065mol$
$M\left(NH_4Cl\right)=4\cdot1{,}008+14{,}01+35{,}45=53{,}492\ \frac{g}{mol}$
$m\left(NH_4Cl\right)=0{,}03476...g\approx35mg$

$n\left(C_{20}H_{30}O\right)=2{,}5\cdot10^{-9}mol$
$M\left(C_{20}H_{30}O\right)=16{,}00+20\cdot12{,}01+30\cdot1{,}008=286{,}44\ \frac{g}{mol}$

$m=716{,}1...mg\approx716mg\$
$n\left(C_{18}H_{23}O_2\right)=6{,}4\cdot10^{-12}mol$
$M\left(C_{18}H_{23}O_2\right)=18\cdot12{,}01+23\cdot1{,}008+2\cdot16{,}00=271{,}364\ \frac{g}{mol}$
$m=1{,}7367...\cdot10^{-9}\approx2\ ng$

16
$n=\frac{m}{M}$

$M=\frac{m}{n}=\frac{6{,}98g}{0{,}125mol}=55{,}84\ \frac{g}{mol}$

moolimassa on lähimpänä raudan moolimassaa, aine X on siis rautaa

18
$m=0{,}2g$
$M=11\cdot1{,}008+12\cdot12{,}01+2\cdot16{,}00+14{,}01=201{,}218\ \frac{mol}{g}$
$n=\frac{m}{M}=0{,}0009939...mol$
$N=n\cdot N_A=5{,}985...\cdot10^{20}\ \approx6{,}0\cdot10^{20}$