# Question 84c5b

Aug 27, 2015

My estimate: $7.6 \cdot {10}^{24}$

#### Explanation:

The idea here is that you need to look at the most stable isotopes of calcium, oxygen, and carbon and try to use their percent abundance to figure out how many protons you'd get per mole of calcium, carbon, and oxygen.

The most stable isotopes for these elements are

" "color(purple)("Isotope")" " " " " "color(green)("Abundance")" " " " " "color(orange)("no. of neutrons")

$\text{ """_""^40"Ca" " " " " " " " " 96.941%" " " " " " " " " } 20$

$\text{ """_""^42"Ca"" " " " " " " " "0.647%" " " " " " " " " " " " } 22$

$\text{ """_""^43"Ca"" " " " " " " " "0.135%" " " " " " " " " " " " } 23$

$\text{ """_""^44"Ca"" " " " " " " " "2.086%" " " " " " " " " " " " } 24$

$\text{ """_""^48"Ca"" " " " " " " " "0.167%" " " " " " " " " " " " } 28$
stackrel("------------------------------------------------------------------------------------------------------")

$\text{ """_""^12"C"" " " " " " " " " "99.9%" " " " " " " " " " " " } 6$

$\text{ """_""^13"C"" " " " " " " " " " "1.1%" " " " " " " " " " " " " " " } 7$
stackrel("-------------------------------------------------------------------------------------------------------")

$\text{ """_""^16"O"" " " " " " " " "99.76%" " " " " " " " " " " " " " } 8$

$\text{ """_""^17"O"" " " " " " " " "0.039%" " " " " " " " " " " " " " } 9$

$\text{ """_""^18"O"" " " " " " " " "0.201%" " " " " " " " " " " " " " } 10$

Use the molar mass of calcium carbonate to determine how many moles of each element you get in 25 g of the compound

25color(red)(cancel(color(black)("g"))) * "1 mole"/(100.09color(red)(cancel(color(black)("g")))) = "0.250 moles"

You know that one mole of calcium carbonate contains 1 mole of calcium, one mole of carbon, and 3 moles of oxygen.

Each mole contains $6.022 \cdot {10}^{23}$ atoms - this is known as Avogadro's number. What you need to do is determine how many atoms of each element you have, then use the abundances to determine how many neutrons you'd get per mole.

So, 0.250 moles of calcium will contain

0.250color(red)(cancel(color(black)("moles"))) * (6.022 * 10^(23)"atoms")/(1color(red)(cancel(color(black)("mole")))) = 1.51 * 10^(23)"atoms"

The number of neutrons you'd get in that many moles of calcium is

$\text{no. neutrons} = \left(0.96941 \cdot 20 + 0.00647 \cdot 22 + 0.00135 \cdot 23 + 0.02086 \cdot 24 + 0.00167 \cdot 28\right) \cdot 1.51 \cdot {10}^{23}$

$\text{no. neutrons} = 20.06 \cdot 1.51 \cdot {10}^{23} = 3.03 \cdot {10}^{24}$

Do the same for carbon

$\text{no. neutrons} = \left(0.999 \cdot 6 + 0.011 \cdot 7\right) \cdot 1.51 \cdot {10}^{23}$

$\text{no. neutrons} = 9.17 \cdot {10}^{23}$

For oxygen you have

0.750color(red)(cancel(color(black)("moles"))) * (6.022 * 10^(23)"atoms")/(1color(red)(cancel(color(black)("mole")))) = 4.52 * 10^(23)"atoms"#

This means that you have

$\text{no. neutrons} = \left(0.9976 \cdot 8 + 0.00039 \cdot 9 + 0.00201 \cdot 10\right) \cdot 4.52 \cdot {10}^{23}$

$\text{no. neutrons} = 8.004 \cdot 4.52 \cdot {10}^{23} = 3.62 \cdot {10}^{24}$

The total number of neutrons you'd get is equal to

$3.03 \cdot {10}^{24} + 0.917 \cdot {10}^{24} + 3.62 \cdot {10}^{24} = \textcolor{g r e e n}{7.6 \cdot {10}^{24}}$

SIDE NOTE The simple way to do this is to only use the neutrons coming from the most abundant isotopes. That calculation would look like this

${\underbrace{20 \cdot 1.51 \cdot {10}^{23}}}_{\textcolor{red}{\text{from calcium")) + underbrace(6 * 1.51 * 10^(23))_(color(blue)("from carbon")) + underbrace(8 * 4.52 * 10^(23))_(color(orange)("from oxygen}}}$

$\text{no. neutrons} = 7.5 \cdot {10}^{24}$

The result is almost identical, but the calculations are simpler.