How many formula units constitute a mass of #450*g# of #"calcium carbonate"#?

2 Answers
May 31, 2017

Answer:

Approx. #2.7xx10^24# #"formula units....."#

Explanation:

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.

May 31, 2017

Answer:

#2.71 times 10^(24)# ions

Explanation:

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)#.