How many formula units constitute a mass of #450*g# of #"calcium carbonate"#?
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Approx. #2.7xx10^24# #"formula units....."#
First we work out the molar quantity:
#"Moles"="Mass"/"Molar mass"=(450*g)/(100.09*g*mol^-1)=4.50*mol#.
Now, by definition, #1*mol# of stuff specifies #6.022xx10^23# individual items of stuff; we can also use the abbreviation #N_A=6.022xx10^23*mol^-1#.
Now calcium carbonate is NOT MOLECULAR. And thus we say that there are #4.50*molxxN_A# #"formula units"# of calcium carbonate.
#2.71 times 10^(24)# ions
The number of particles #N -# or in this case molecules #-# of a substance is given by the equation #N = n L#; where #n# is the number of mole and #L# is Avogrado's constant.
Also, the definition of #n# is given by the equation #n = frac(m)(M)#; where #m# is the mass and #M# is the molar mass.
Let's substitute the definition of #n# into the equation for #N#:
#Rightarrow N = frac(m)(M) times L#
Then, let's substitute the value of #m# and #L# into the equation:
#Rightarrow N = frac(450 " g")(M) times 6.022 times 10^(23)# #"mol"^(- 1)#
#Rightarrow N = frac(2.7099 times 10^(26) " g mol"^(- 1))(M)#
We now need to calculate the molar mass #M# of #"CaCO"_(3)#:
#Rightarrow M("CaCO"_(3)) = (40.078 + 12.011 + 3 times 15.999)# #"g mol"^(- 1)#
#Rightarrow M("CaC)"_(3)) = 100.086# #"g mol"^(- 1)#
Now, let's substitute this value into the equation:
#Rightarrow N = frac(2.7099 times 10^(26) " g mol"^(- 1))(100.086 " g mol"^(- 1))#
#Rightarrow N approx 2.71 times 10^(24)#
Therefore, there are around #2.71 times 10^(24)# ions of #"CaCO"_(3)#.
